Selina Class 10 Chemistry Solutions – ML Aggarwal Solutions

Selina Concise Chemistry Class 10 ICSE Solutions Organic Chemistry

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Selina Concise Chemistry Class 10 ICSE Solutions Organic Chemistry

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Selina ICSE Solutions for Class 10 Chemistry Chapter 12 Organic Chemistry

Exercise 12(A)

Solution 1.

(a) 2,2- dimethylpropane
(b) 2-methyl butane
(c) Prop-1-ene
(d) 2,2- dimethyl pentane
(e) Pent-2-yne
(f) 3-methyl but-1-yne
(g) 2,3-dichloropentane
(h) 3-methylheptane
(i) 2-methyl butane
(j) Hept-2-yne
(k) 2,2- dimethyl hexanal
(l) Pentan-2-ol
(m) 4-methylpentanoic acid
(n) 2-bromo2-methyl butane
(o) 1- bromo3-methyl butane

Solution 2.

The structure of the following compounds are:

(a) Prop-1-ene
CH₃-CH=CH₂

(b) 2,3-dimethylbutane
CH₃-CH(CH₃)-CH(CH₃)-CH₃

(d) 3-hexene
CH₃-CH₂-CH=CH-CH₂-CH₃

(e) Prop-1-yne
CH₃-C≡CH

(f) 2-methylprop-1-ene
CH₃-C(CH₃)=CH₂

(g) Alcohol with molecular formula C₄H₁₀O
CH₃-CH₂-CH₂-CH₂-OH

Solution 3.

(a) Correct answer: (iv)
C_nH_2n+1 is the formula for alkyl group. Hence it is C₅H_11.

(b) Correct answer: (i)
A hydrocarbon of general C_nH_2n is C₁₅H_30.

(c) Correct answer: (ii)
As the formula of Alkene is C_nH_2n.Thus n+2n = 72
3n = 72
n = 24
By filling value we get the molecular mass 72.

(d) (iv)
The total number of carbon chains that four carbon atoms form in alkane is 2. They are:
selina-icse-solutions-class-10-chemistry-organic-chemistry-12a-3-1

(e) Correct answer: (iv)
Alcohol and ether are functional isomers as they have same molecular formula but different functional groups.

(f) Correct answer: (ii)
selina-icse-solutions-class-10-chemistry-organic-chemistry-12a-3-2

Solution 4.

(a) Propane and ethane are homologues.
(b) A saturated hydrocarbon does not participate in a/an addition reaction.
(c) Succeeding members of a homologous series differ by CH₂.
(d) As the molecular masses of hydrocarbons increase, their boiling points Increase and melting point increase.
(e) C₂₅H₅₂ and C₅₀H₁₀₂ belong to the same homologous series.
(f) CO is an organic Compound.
(g) The physical and chemical properties of an organic compound are largely decided by the Functional group.
(h) CHO is the functional group of an aldehyde.
(i) The root in the IUPAC name of an organic compound depends upon the number of carbon atoms in Principal Chain.
(j) But-1-ene and but-2-ene are examples of position isomerism.

Exercise 12(B)

Solution 1.

Sources of alkane:
The principal sources of alkanes are Natural gas and petroleum.

Solution 2.

Methane is a primary constituent of natural gas. It absorbs outgoing heat radiation from the earth, and thus contributes to the green house effect and so it is considered as a green house gas.

Solution 3.

The general formula of alkane is :
C_nH _2n+2

Solution 4.

selina-icse-solutions-class-10-chemistry-organic-chemistry-12b-4

Solution 5.

selina-icse-solutions-class-10-chemistry-organic-chemistry-12b-5

Solution 6.

(a) Laboratory preparation of methane:
When the mixture of sodium ethanoate and soda lime is taken in a hard glass test tube and heated, the gas evolved is methane. It is collected by downward displacement of water.

(b) Laboratory preparation of ethane:
When the mixture of sodium propionate and soda lime is taken in the boiling tube and heated the ethane gas is evolved. It is also collected by downward displacement of water.

Solution 7.

When methyl iodide is reduced by nascent hydrogen at ordinary room temperature then methane is formed.
CH₃I + 2[H] → CH₄+ HI

When bromoethane is reduced by nascent hydrogen at ordinary room temperature then ethane is produced.
C₂H₅Br + 2[H] → C₂H₆+ HBr

Solution 8.

selina-icse-solutions-class-10-chemistry-organic-chemistry-12b-8

Solution 9.

(a) Sufficient air: When methane burns in sufficient air, then carbon dioxide and water vapors are formed.
CH₄ + 2O₂ → CO₂+2H₂O

(b) Insufficient air: When methane burns in insufficient air , then carbon monoxide and water is formed.
2CH₄ + 3O₂ → 2CO + 4H₂O

Solution 10.

(a) (i) When methane reacts with chlorine in the presence of sunlight or UV light, it undergoes substitution reaction to form Tetrachloromethane.

selina-icse-solutions-class-10-chemistry-organic-chemistry-12b-10

(ii) When it reacts with bromine it forms Tetrabromomethane
CH₄ + Br₂→ CH₃Br + HCl
CH₃Br + Br₂→ CH₂Br₂ + HCl
Dibromomethane
CH₂Br₂ + Br₂→ CHBr₃ + HCl
Tribromo methane
CHBr₃ + Br₂→ CBr₄ + HCl
Tetrabromomethane

(b) (i) When ethane reacts with chlorine it forms hexachoroethane.

C₂H₆ + Cl₂ → C₂H₅Cl + HCl
Chloroethane
C₂H₅Cl + Cl₂ → C₂H₄Cl₂+ HCl
Dichloroethane
C₂H₄Cl₂ + Cl₂ → C₂H₃Cl₃+ HCl
Trichloroethane
C₂H₃Cl₃ + Cl₂ → C₂H₂Cl₄ + HCl
Tetrachloroethane
C₂H₂Cl₄ +Cl₂ → C₂HCl₅ + HCl
Pentachloroethane
C₂HCl₅ + Cl₂ → C₂Cl₆ + HCl
Hexachloroethane

(ii) When ethane reacts with bromine it forms Hexabromoethane

C₂H₆ +Br₂ → C₂H₅Br + HBr
Bromoethane
C₂H₅B_r + Br₂ → C₂H₄Br₂+HBr
Dibromoethane
C₂H₄Br₂ +Br₂ → C₂H₃Br₃+HBr
Tribromoethane
C₂H₃Br₃ + Br₂ → C₂H₂Br₄ + HBr
Tetrabromoethane
C₂H₂Br₄ +Br₂ → C₂HBr₅ +HBr
Pentabromoethane
C₂HBr₅ +Br₂ → C₂Br₆ + HBr
HexaBromoethane

Solution 11.

selina-icse-solutions-class-10-chemistry-organic-chemistry-12b-11

Solution 12.

The decomposition of a compound by heat in the absence of air is called Pyrolysis. When pyrolysis occurs in alkanes, the process is termed cracking.

For example:
Alkanes on heating under high temperature or in the presence of a catalyst in absence of air broken down into lower alkanes, alkenes and hydrogen.

Solution 13.

selina-icse-solutions-class-10-chemistry-organic-chemistry-12b-13

Solution 14.

(a) Methane: Three uses of methane are:

Methane is a source of carbon monoxide and hydrogen
It is used in the preparation of ethyne, methanal, chloromethane, carbon tetrachloride.
It is employed as a domestic fuel.

(b) Ethane: Three uses of ethane are:

It is used in the preparation of ethene, ethanol, and ethanol.
It forms ethyl chloride, which is used to make tetraethyllead.
It is also a good fuel.

Solution 15.

(a) When a mixture of ethane and oxygen is compressed to about 120atm pressure and passed over copper tubes at 475K, ethyl alcohol is formed.

(b) When mixture of ethane and oxygen is passed through heated molybdenum oxide, the mixture is oxidized to Acetaldehyde.
selina-icse-solutions-class-10-chemistry-organic-chemistry-12b-15-2

Solution 16.

(a) Methane to methyl alcohol:
When a mixture of methane and oxygen is compressed to about 120atm pressure and passed over copper tubes at 475K, ethyl alcohol is formed.

(b) Methane to formaldehyde:
When mixture of methane and oxygen is passed through heated molybdenum oxide, the mixture is oxidized to Formaldehyde
selina-icse-solutions-class-10-chemistry-organic-chemistry-12b-16-2

Exercise 12(C)

Solution 1.

Selina ICSE Solutions for Class 10 Chemistry - Organic Chemistry-12c-1

Solution 2.

(a) n signifies the number of carbon atoms and 2n signifies the number of hydrogen atoms.
(b) The name of alkene when n=4 is Butene.
(c) The molecular formula of alkene when n=4 is C₄H_8.(d) The molecular formula of alkene when there are 10 H atom in it C₅H_10.(e) The structural formula of the third member of alkene is
Selina ICSE Solutions for Class 10 Chemistry - Organic Chemistry-12c-2
(f) Lower homologus of alkene which contain four carbons is C₃H_6.Higher homologus of alkene which contain four carbons is C₅H₁₀.

Solution 3.

The isomers of Butene are:
(i) CH₃-CH₂-CH=CH₂, But-1-ene
(ii) CH₃-CH=CH-CH₃ , But-2-ene
(iii) CH₂=C(CH₃)-CH₃ , 2-methyl propene

Solution 4.

Selina ICSE Solutions for Class 10 Chemistry - Organic Chemistry-12c-4

Solution 5.

Selina ICSE Solutions for Class 10 Chemistry - Organic Chemistry-12c-5

Solution 6.

When ethene and hydrogen are passed over finely divided catalyst such as platinum or palladium at ordinary temperature or nickel at 200^o C, the two atom of hydrogen molecule are added to the unsaturated molecule, which thus becomes a saturated one.

Solution 7.

Chlorine and bromine are added to the double bond of ethene to form saturated ethylene chloride and ethylene bromide respectively.
CH₂= CH₂ + Cl₂ → CH₂(Cl)-CH₂(Cl)
1,2-dichloro ethane
CH₂= CH₂+ Br₂ → CH₂(Br)-CH₂(Br)
1,2-dibromo ethane

Solution 8.

Selina ICSE Solutions for Class 10 Chemistry - Organic Chemistry-12c-8

Solution 9.

(a) Physical state: Ethene is a colourless and inflammable gas.
(b) Odour: It has faint sweetish odour.
(c) Density as compared to air: It has density less than one hence it is lighter than air.
(d) Solubility: It is sparingly soluble in water but highly soluble in organic solvents like alcohol, ether and chloroform.

Solution 10.

(a) Ethene into 1, 2 -dibromoethane: Ethene reacts with bromine at room temperature to form saturated ethylene chloride.
CH₂=CH₂+ Br₂ → CH₂(Br)-CH₂(Br)
1,2-dibromo ethane

(b) Ethene into ethyl bromide: When ethene is treated with HBr bromoethane is formed.
CH₂=CH₂+ HBr → CH₃-CH₂Br
Ethyl bromide

Solution 11.

Selina ICSE Solutions for Class 10 Chemistry - Organic Chemistry-12c-11

Solution 12.

Selina ICSE Solutions for Class 10 Chemistry - Organic Chemistry-12c-12

Solution 13.

Selina ICSE Solutions for Class 10 Chemistry - Organic Chemistry-12c-13

Solution 14.

When ethylene is passed through alkaline KMnO₄ solution 1, 2-Ethanediol is formed. The Purple color of KMnO₄ decolorizes.
CH₂=CH₂+H-O-H + [O] → CH₂(OH)-CH₂(OH)
Cold alkaline
KMnO₄ solution

Solution 15.

Three compounds formed by ethylene are:

Polythene
Ethanol
Epoxyethane

Uses of above compounds:

Polythene is used as carry bags.
Ethanol is used as a starting material for other products, mainly cosmetics and toiletry preparation.
Epoxyethane is used in the manufacture of detergents.

Exercise 12(D)

Solution 1.

Natural gas and Petroleum are sources for alkynes.
The general formula of alkynes are:
C_nH_2n-2

Solution 2.

selina-icse-solutions-class-10-chemistry-organic-chemistry-12d-2

Solution 3.

selina-icse-solutions-class-10-chemistry-organic-chemistry-12d-3

Solution 4.

Solution 5.

The following compounds can be classified as:

C₃H₄:- Alkynes
C₃H₈:- Alkanes
C₅H₈:- Alkynes
C₃H₆:- Alkenes

Solution 6.

selina-icse-solutions-class-10-chemistry-organic-chemistry-12d-6

Solution 7.

(a) Ethyne in an inert solvent of carbon tetrachloride adds chlorine to change into 1,2-dichloro ethene with carbon-carbon double bond, and then to an 1,1,2,2-tetrachloro ethane with carbon-carbon single bond.
1,2-dichloro ethene 1,1,2,2 -tetrachloro ethane
(b) Ethyne in an inert solvent of carbon tetrachloride adds bromine to change into 1,2-dibromo ethene and then to 1,1,2,2 -tetrabromo ethane.
selina-icse-solutions-class-10-chemistry-organic-chemistry-12d-7-2

Exercise 12(E)

Solution 1.

(a) Alcohols are the hydroxyl derivatives of alkanes and are formed by replacing one or more hydrogen atoms of the alkane with OH group.
Methanol is obtained from destructive distillation of wood while ethanol is obtained from fermentation of sugar.

(b) General formula of monohydric alcohol:
C_nH_{2n+ 1}OH

Solution 2.

selina-icse-solutions-class-10-chemistry-organic-chemistry-12e-2

Solution 3.

(a) By hydrolysis of ethene: When concentrated sulphuric acid is added to ethene at a temperature of 80^oC and pressure of 30 atm. ethyl hydrogen sulphate is produced. Ethyl hydrogen sulphate on hydrolysis with boiling water gives ethanol.
selina-icse-solutions-class-10-chemistry-organic-chemistry-12e-3-1

(b) By hydrolysis of alkyl halide: Alcohols can be prepared by the hydrolysis of alkyl halide with a hot dilute alkali.

Solution 4.

selina-icse-solutions-class-10-chemistry-organic-chemistry-12e-4

Solution 5.

selina-icse-solutions-class-10-chemistry-organic-chemistry-12e-5

Solution 6.

(a) The melting and boiling point of the successive members of the homologous series of alcohols increase with the increase in molecular mass.
(b) When ethanol reacts with acetic acid ethyl acetate is formed.

Solution 7.

Ethanol affects that part of the brain which controls our muscular movements and then gives temporary relief from tiredness. But it damages the liver and kidney too.

Solution 8.

(a) Absolute alcohol: Absolute alcohol may be obtained by distilling moist alcohol with benzene. The mixture of water and benzene distills off and anhydrous alcohol is left behind.
(b) Spurious alcohol: It is made by improper distillation. It contains large portions of methanol in a mixture of alcohols.
(c) Methylated spirit: Methylated spirit or denatured alcohol is ethyl alcohol with 5%methyl alcohol, a coloured dye and some pyridine.

Solution 9.

selina-icse-solutions-class-10-chemistry-organic-chemistry-12e-9

Solution 10.

S No	Formula	Common Name	IUPAC
1	C₃H₆	Propylene	Propene
2	C₂H₄	Ethylene	Ethene
3	C₂H₂	Acetylene	Ethyne
4	CH₃OH	Methyl alcohol	Methanol
5	C₂H₅OH	Ethyl alcohol	Ethanol

Solution 11.

selina-icse-solutions-class-10-chemistry-organic-chemistry-12e-11

Solution 12.

selina-icse-solutions-class-10-chemistry-organic-chemistry-12e-12

Solution 13.

(a) Used for illuminating country houses : Ethyne
(b) Used for making a household plastic material: ethyne
(c) Called ‘wood spirit’ : Methanol
(d) Poisonous: Methanol
(e) Consumed as a drink: Ethanol
(f) Made from water gas: Methanol

Exercise 12(F)

Solution 1.

An organic compound containing the carboxyl group(COOH) is known as carboxylic acid.
The general formula: C_nH_2n+1COOH

Solution 2.

Monocarboxylic acid:
Formula: HCOOH
Common name: Formic acid
IUPAC name: Methanoic acid
Dicarboxylic acid:
Formula: COOH-COOH
Common name : Oxalic acid
IUPAC name: Ethane-di-oic acid

Solution 3.

(a) First three members of carboxylic acids are:

Methanoic acid
Ethanoic acid
Propanoic acid

(b) Three compounds that can be oxidized directly or in stages to produce acetic acid are:

Ethanol
Acetylene
Ethanal

Solution 4.

Vinegar commonly called Sirka is a dilute solution of acetic acid. The presence of colouring matter gives it a greyish colour while the presence of some other organic acids and organic compounds impart it the usual taste and flavour.

Solution 5.

selina-icse-solutions-class-10-chemistry-organic-chemistry-12f-5

IUPAC name of acetic acid is: Ethanoic acid
Glacial acetic acid is the pure form of acetic acid. It does not contain water.

Solution 6.

(a) Ethanol
(b) Acetic acid
(c) Propanoic acid

Solution 7.

(a) It is prepared in the lab by the oxidation of ethanol with acidified potassium dichromate.

(b) Acetylene is first converted to acetaldehyde by passing through 40% H₂SO₄ at 60°C in the presence of 1% HgSO₄.
The acetaldehyde is then oxidised to acetic acid in the presence of catalyst manganous acetate at 70°C. selina-icse-solutions-class-10-chemistry-organic-chemistry-12f-7-2

Solution 8.

selina-icse-solutions-class-10-chemistry-organic-chemistry-12f-8

Solution 9.

selina-icse-solutions-class-10-chemistry-organic-chemistry-12f-9

Solution 10.

(a) When acetic acid and ethanol react it results in the formation of ethyl acetate.
(b) Lithum aluminium hydride(LiAlH₄) is used to convert acetic acid to ethanol.
(c) Phosphorous pentoxide(P₂O₅) is heated along with acetic acid to form acetic anhydride.

Solution 11.

Test to show that CH₃COOH is acidic are:
When litmus test is done, it turns blue litmus red.
It react with bases to form salt and water.

Solution 12.

selina-icse-solutions-class-10-chemistry-organic-chemistry-12f-12

Exercise Intext 1

Solution 1.

(a) Organic chemistry may be defined as the chemistry of hydrocarbons and its derivatives.
(b) Vital Force Theory is a theory made by the Scientist Berzelius in 1809 which assumed that organic compounds are only formed in living cells and it is impossible to prepare them in laboratories.
It was discarded because Friedrich Wohler showed that it was possible to obtain an organic compound(urea) in the laboratory.

Solution 2.

(a) Few sources of organic compounds are:
Plants, Animals, Coal, Petroleum and Wood.

(b) The various applications of organic chemistry is:

It is used in the production of soaps, shampoos, powders and perfumes.
Various fuels like natural gas, petroleum are also organic compounds.
The fabrics that we use to make various dresses are also made from organic compounds.

Solution 3.

Organic compounds are present everywhere. They are present in:

It is present in the production of soaps, shampoos, powders and perfumes.
It is present in the food we eat like carbohydrates, proteins, fats, vitamins etc.
Fuel like natural gas, petroleum are also organic compounds.
Medicines, explosives, dyes, insecticides are all organic compounds.

Thus we can say that organic compounds play a key role in all walks of life.

Solution 4.

The unique properties shown by carbon are:

Tetravalency of carbon
Catenation
Isomerism

Solution 5.

(a) Tetravalency: Carbon can neither lose nor gain electrons to attain octet. Thus it shares four electrons with other atoms. This characteristics of carbon by virtue of which it forms four covalent bonds, is called Tetravalency of carbon.
In structural form :

(b) Catenation: The property of self -linking of atoms of an element through covalent bonds in order to form straight chains, branched chains and cyclic chains of different sizes is known as catenation.
Carbon- carbon bond is strong so carbon can combine with other carbon atoms to form chains or rings and can involve single, double and triple bonds.

Solution 6.

Four properties of organic compound that distinguish them from inorganic compounds are:

Presence of carbon.
Solubility in the organic solvents.
Forming of covalent bonds.
Having low melting and boiling points.

Solution 7.

Due to the unique nature of carbon atom, it gives rise to formation of large number of compounds. Thus this demands a separate branch of chemistry.

Solution 8.

Hydrocarbons are compounds that are made up of only carbon and hydrogen.
Comparison of saturated and Unsaturated hydrocarbons:

Saturated Hydrocarbon	Unsaturated Hydrocarbon
1. Carbon atoms are joined only by single bonds.	Carbon atoms are joined by double or by triple bonds.
2. They are less reactive due to the non-availability of electrons in the single covalent bond.	They are more reactive due to presence of electrons in the double or the triple bond.
3. They undergo substitution reaction.	They undergo addition reaction.

Solution 9.

Due to presence of unique properties of carbon like Tetravalency, catenation and Isomerism large number of organic compounds are formed.

Solution 10.

selina-icse-solutions-class-10-chemistry-organic-chemistry-int-10

Solution 11.

selina-icse-solutions-class-10-chemistry-organic-chemistry-int-11

Solution 12.

The member of each of the following is:
(a) Saturated Hydrocarbon: Hexane (C₆H₁₄)
(b) Unsaturated Hydrocarbon: Hexene (C₆H₁₂)

Solution 13.

Substitution reaction: A reaction in which one atom of a molecule is replaced by another atom (or group of atoms) is called a substitution reaction.
Addition reaction: A reaction involving addition of atom(s) or molecules(s) to the double or the triple bond of an unsaturated compound so as to yield a saturated product is known as addition reaction.

Solution 14.

Chain isomerism
Chain isomerism arises due to the difference in arrangement of C atoms in the chain. For example, there are two isomers of butane, C₄H₁₀. In one of them, the carbon atoms lie in a “straight chain” whereas in the other the chain is branched.

selina-icse-solutions-class-10-chemistry-organic-chemistry-int-14-1

Position isomerism
It is due to the difference in position of functional groups.
For example, there are two structural isomers with the molecular formula C₃H₇Br. In one of them, the bromine atom is on the end of the chain, whereas in the other it is attached in the middle.
selina-icse-solutions-class-10-chemistry-organic-chemistry-int-14-2

Solution 15.

(a) Isomerism: Compounds having the same molecular formula but different structural formula are known as isomers and the phenomenon as isomerism.

Two main causes of isomerism are:
Difference in mode of linking of atoms.
Difference in the arrangement of atoms or groups in space.
selina-icse-solutions-class-10-chemistry-organic-chemistry-int-15

Solution 16.

A functional group is an atom or a group of atoms that defines the structure (or the properties of a particular family) of organic compounds.
The structural formula of
selina-icse-solutions-class-10-chemistry-organic-chemistry-int-16

Solution 17.

The functional group present in the following compounds are:

(a) CH₃OH :- Alcohol
(b) HCHO:- Aldehyde
(c) CH₃COOH:- Carboxyl

Solution 18.

selina-icse-solutions-class-10-chemistry-organic-chemistry-int-18

Solution 19.

(i) Physical properties: The alkyl group determines the physical properties.
(ii) Chemical properties: The functional group is responsible for the chemical properties.

Solution 20.

The alkyl radical and the functional group are:

Sr.No	Formula	Name of alkyl radical	Name of Functional group
a	CH₃OH	Methyl	Alcohol
b	C₂H₅OH	Ethyl	Alcohol
c	C₃H₇CHO	Propyl	Aldehyde
d	C₄H₉COOH	Butyl	Carboxyl

Solution 21.

(a) An alkyl group is obtained by removing one atom of hydrogen from an alkane molecule. Alkyl group is named by replacing the suffix ‘ane’ of the alkane with the suffix -yl.
selina-icse-solutions-class-10-chemistry-organic-chemistry-int-21

Solution 22.

selina-icse-solutions-class-10-chemistry-organic-chemistry-int-22

Solution 23.

(a) A homologous series is a group of organic compounds having a similar structure and similar chemical properties in which the successive compounds differ by a CH₂ group.

(b) The difference in molecular formula of any two adjacent homologues is
(i) It differs by 14 a.m.u in terms of molecular mass.
(ii) It differs by three atoms. The kind of atoms it differs is one carbon and two hydrogen.

Miscellaneous Exercise

Solution 1.

selina-icse-solutions-class-10-chemistry-organic-chemistry-mis-1

Solution 2.

selina-icse-solutions-class-10-chemistry-organic-chemistry-mis-2
They both are unsaturated compound. The structure (i) contains double bond where as structure (ii) contains triple bond.
(b) Both the compounds undergo addition reactions.

Solution 3.

selina-icse-solutions-class-10-chemistry-organic-chemistry-mis-3
The Saturated hydrocarbons undergo substitution reactions whereas unsaturated hydrocarbons undergo addition reactions.

Solution 4.

(a) CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂(b) When bromine in carbon tetrachloride is added to ethyne, the orange colour of the bromine disappears due to the formation of the colourless ethylene bromide.

Solution 5.

The alkanes form an (a) Homologous series with the general formula (b) C_nH_2n+2. The alkanes are (c) saturated (d) hydrocarbon which generally undergo (e) substitution reactions.

Solution 6.

(a) The conversion of ethanol into ethene is an example of Dehydration.
(b) Converting ethanol into ethene requires the use of Conc. H₂SO₄.
(c) The conversion of ethene into ethane is an example of hydrogenation.
(d) The catalyst used in the conversion of ethene into ethane is commonly nickel.

Solution 7.

(a) Ethyne is a highly reactive compound than ethene because of the presence of a triple bond between its two carbon atoms.
(b) Ethene is a highly reactive compound than ethane because of the presence of a double bond between its two carbon atoms.
(c) Hydrocarbons such as alkanes undergo combustion reactions with oxygen to produce carbon dioxide and water vapour. Alkanes are flammable which makes them excellent fuels.
Methane for example is the principal component of natural gas.
CH₄ + 2O₂ → CO₂ + 2H₂O

Solution 1 (2004).

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Solution 1 (2005).

selina-icse-solutions-class-10-chemistry-organic-chemistry-mis-1-2005

Solution 1 (2006).

(a) IUPAC name: Propanal
Functional group: -CHO
(b) IUPAC name: Propanol
Functional group: -OH

Solution 1 (2007).

selina-icse-solutions-class-10-chemistry-organic-chemistry-mis-1-2007

Solution 1a (2008).

selina-icse-solutions-class-10-chemistry-organic-chemistry-mis-1-2008a

Solution 1b (2008).

selina-icse-solutions-class-10-chemistry-organic-chemistry-mis-1-2008b

Solution 1c (2008).

selina-icse-solutions-class-10-chemistry-organic-chemistry-mis-1-2008c

Solution 1d (2008).

(i) Ethane undergoes substitution reaction.
(ii) Ethene undergoes addition reactions.

Solution 1e (2008).

(i) 2C₂H₆ + 7O₂ → 4CO₂+ 6H₂O
(ii) Ethane can be oxidized as follows:
When a mixture of ethane and oxygen in the ratio 9:1 by volume is compressed to about 120 atm pressure and passed over copper tubes at 475K, ethyl alcohol is formed. selina-icse-solutions-class-10-chemistry-organic-chemistry-mis-1-2008e

Solution 1f (2008).

(i) Pure acetic acid on cooling forms crystalline mass resembling ice and for this reason it is called glacial acetic acid.
(ii) When acetic acid reacts with alcohol, ester is formed.

Solution 2 (2004).

selina-icse-solutions-class-10-chemistry-organic-chemistry-mis-2-2004

Solution 2 (2005).

(a) Ethanol
(b) Ethanoic acid
(c) Ethene

Solution 2 (2006).

selina-icse-solutions-class-10-chemistry-organic-chemistry-mis-2-2006

Solution 2 (2007).

The homologous series of hydrocarbons are:

General Formula	C_nH_2n	C_nH_2n-2	C_nH_2n+2
IUPAC name of the homologous series	Alkenes	Alkynes	Alkanes
Characteristics bond type	Double bond	Triple Bond	Single Bond
IUPAC name of the first member of the series	Ethene	Ethyne	Methane
Type of reaction with chlorine	Addition	Addition	Substitution

Solution 3 (2005).

selina-icse-solutions-class-10-chemistry-organic-chemistry-mis-3-2005

Solution 3 (2006).

Alkenes are the (a) homologous series of (b) unsaturated hydrocarbons. They differ from alkanes due to presence of (c) single bonds. Alkenes mainly undergo (d) addition reactions.

Solution 4 (2006).

selina-icse-solutions-class-10-chemistry-organic-chemistry-mis-4-2006

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Selina Concise Chemistry Class 10 ICSE Solutions Study of Compounds – Nitric Acid

August 11, 2022July 20, 2017 by Veerendra

Selina Concise Chemistry Class 10 ICSE Solutions Study of Compounds – Nitric Acid

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 10 Nitric Acid. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Chemistry Chapter 10 Study of Compounds – Nitric Acid

Exercise 1

Solution 1.

(a) Aqua fortis: Nitric acid is called aqua fortis. Aqua fortis means strong water. It is so called because it reacts with nearly all metals.

(b) Aqua Regia: Conc. Nitric acid (1part by volume) when mixed with conc. Hydrochloric acid (3 parts by volume) gives a mixture called aqua regia. It means Royal water.
HNO₃ +3HCl → NOCl +2H₂O +2[Cl]

(c) Fixation of Nitrogen: The conversion of free atmospheric nitrogen into useful nitrogenous compounds in the soil is known as fixation of atmospheric nitrogen.

Solution 2.

During lightning discharge, the nitrogen present in the atmosphere reacts with oxygen to form nitric oxide.
selina-icse-solutions-class-10-chemistry-nitric-acid-2
The nitrogen dioxide dissolves in atmospheric moisture in the presence of oxygen of the air and forms nitric acid which is washed down by the rain and combines with the salt present on the surface of the earth.
4NO₂+ 2H₂O + O₂ → 4HNO₃

Solution 3.

selina-icse-solutions-class-10-chemistry-nitric-acid-3-1
(b) Concentrated hydrochloric acid cannot replace Conc. Sulphuric acid for the preparation of nitric acid because hydrochloric acid is volatile acid and hence nitric acid vapours will carry HCl vapours.

(c) Conc. Nitric acid prepared in the laboratory is yellow in colour due to the dissolution of reddish brown coloured nitrogen dioxide gas in acid. This gas is produced due to the thermal dissociation of a portion of nitric acid.
4HNO₃ → 2H₂O + 4NO₂ + O₂The yellow colour of the acid is removed:
If dry air or CO₂ is bubbled through the yellow acid, the acid turns colourless because it drives out NO₂ from warm acid which is further oxidized to nitric acid.
By addition of excess of water, nitrogen dioxide gas dissolves in water and thus the yellow colour of the acid is removed.

(d)The temperature of the mixture of concentrated sulphuric acid and sodium nitrate should not exceed 200^oC because sodium sulphate formed at higher temperature forms a hard crust which sticks to the walls of the retort and is difficult to remove. At higher temperature nitric acid may also decompose.

Solution 4.

Nitric acid forms a constant boiling mixture with water containing 68% acid. This mixture boils constantly at constant boiling point without any change in its composition. At this temperature, the gas and the water vapour escape together. Hence the composition of the solution remains unchanged. So nitric acid cannot be concentrated beyond 68% by distillation of dilute solution of HNO₃.

Solution 5.

Iron becomes inert when reacted with nitric acid due to the formation of extremely thin layer of insoluble metallic oxide which stops the reaction.
Passivity can be removed by rubbing the surface layer with the sand paper or by treating with strong reducing agent.

Solution 6.

(a) When carbon and conc. Nitric acid is heated the products formed are Carbon dioxide, Nitrogen dioxide and water.
C + 4HNO₃ → CO₂ + 2H₂O + 4NO₂

(b) Copper when reacts with dilute HNO₃ forms Copper nitrate, Nitric oxide and water.
3Cu + 8 HNO₃ → 3Cu(NO₃) ₂ + 4H₂O + 2NO

Solution 7.

(a) Reaction of nitric acid with non-metals:
C + 4HNO₃ → CO₂ + 2H₂O + 4 NO₂S + 6 HNO₃ → H₂SO₄ + 2H₂O + 6 NO₂

(b) Nitric acid showing acidic character:
K₂O + 2HNO₃ → 2KNO₃ + H₂O
ZnO + 2HNO₃ → Zn(NO₃)₂ + H₂O

Solution 8.

(a) When Sodium hydrogen carbonate is added to nitric acid sodium nitrate, carbon dioxide and water is formed.
NaHCO₃ + HNO₃ NaNO₃+ H₂O + CO₂

(b) When Cupric oxide reacts with dilute nitric acid, it forms Copper nitrate.
CuO + 2HNO₃ → Cu(NO₃)₂ + H₂O

(d) 4HNO₃ → 2H₂O + 4NO₂ + O₂

Solution 9.

selina-icse-solutions-class-10-chemistry-nitric-acid-9

Solution 10.

(a) Sodium nitrate:
NaOH + HNO₃ → NaNO₃ +H₂O
Sodium hydroxide reacts with nitric acid to form sodium nitrate.

(b) Copper nitrate:
CuO + 2HNO₃ → Cu(NO₃)₂ + H₂O
Copper oxide reacts with nitric acid to form copper nitrate.

(d) Magnesium nitrate:
Mg +2HNO₃ → Mg(NO₃)₂ + H₂Magnesium with dil. nitric acid to form magnesium nitrate.

(e) Ferric nitrate:
Fe + 6HNO₃ → Fe(NO₃)₃ + 3H₂O + 3NO₂Iron reacts with conc. nitric acid to form ferric nitrate.

(f) Aqua regia:
HNO₃ + 3HCl NOCl +2H₂O + 2[Cl]
Nitric acid reacts with hydrochloric acid to form a mixture called aqua regia.

Solution 11.

(a) HNO₃ is strong oxidizing agent.
(b) NaNO₃ gives NaNO₂ and oxygen on heating.
(c) Constant boiling nitric acid contains 68% nitric acid by weight.
(d) Nitric acid turns yellow solution when exposed to light.

Solution 12.

selina-icse-solutions-class-10-chemistry-nitric-acid-12

Solution 13.

The chemical name of the brown ring is Nitroso ferrous sulphate.
Formula: FeSO₄. NO

Solution 14.

Three important uses of Nitric acid and the property of nitric acid involved is:

S.NO.	Use	Property
1.	To etch designs on copper and brassware.	Nitric acid act as solvent for large number of metals.
2.	To purify gold.	Impurities like Cu, Ag, Zn, etc. dissolve in nitric acid.
3.	Preparation of aqua regia.	Dissolves noble metals.

Solution 15.

(a)KNO₃(b) FeSO₄(c) NO₂

Solution 16.

(a) Brown ring test
Procedure:

Add freshly prepared saturated solution of iron (II)sulphate to the aq. solution of nitric acid.
Now add conc. Sulphuric acid carefully from the sides of the test tube, so that it should not fall drop wise in the test tube.
Cool the test tube in water.
(iv) A brown ring appears at the junction of the two liquids.

selina-icse-solutions-class-10-chemistry-nitric-acid-16
(b) A freshly prepared ferrous sulphate solution is used because on exposure to the atmosphere, it is oxidized to ferric sulphate which will not give the brown ring.

Solution 17.

(a) Potassium nitrate
(b) Ammonium nitrate
(c) Lead nitrate

Solution 18.

(a) Aqua regia is a mixture of 3 parts Hydrochloric acid and one part Nitric acid.
(b) The catalytic oxidation of ammonia to nitric oxide is exothermic.
(c) Magnesium gives H₂ with very dilute nitric acid.
(d) Iron become passive in concentrated nitric acid

Solution 19.

selina-icse-solutions-class-10-chemistry-nitric-acid-19

Solution 1 (2004).

selina-icse-solutions-class-10-chemistry-nitric-acid-1-2004

Solution 1 (2005).

(a) Dilute acid is generally considered a typical acid except for its reaction with metals since it does not liberate hydrogen. It is a powerful oxidizing agent and the nascent oxygen formed oxidizes the hydrogen to water.
(b) 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 4H₂O +2NO

Solution 1 (2006).

(a) All glass apparatus are used because nitric acid vapours are highly corrosive in nature and corrodes cork and rubber etc.
(b) Nitric acid is kept in reagent bottle because nitric acid is a highly fuming liquid; it spreads in air and is highly corrosive.

Solution 1 (2007).

selina-icse-solutions-class-10-chemistry-nitric-acid-1-2007

Solution 1 (2008).

selina-icse-solutions-class-10-chemistry-nitric-acid-1-2008

Solution 2 (2006).

selina-icse-solutions-class-10-chemistry-nitric-acid-2-2006

Solution 3 (2006).

When ammonium nitrate is heated the products formed are nitrous oxide and steam.

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Selina Concise Chemistry Class 10 ICSE Solutions Study of Compounds – Study of Compounds – Ammonia

August 11, 2022July 19, 2017 by Veerendra

Selina Concise Chemistry Class 10 ICSE Solutions Study of Compounds – Study of Compounds – Ammonia

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 9 Study of Compounds Ammonia. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Chemistry Chapter 9 Study of Compounds – Ammonia

Exercise Intext 1

Solution 1.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-ina-1

Solution 2.

The different forms of ammonia:

Gaseous ammonia(dry ammonia gas)
Liquid ammonia
Liquor ammonia fortis
Laboratory bench reagent

Solution 3.

Formula of liquid ammonia is: NH₃.
Liquid ammonia is liquefied ammonia and is basic in nature. It dissolves in water to give ammonium hydroxide which ionizes to give hydroxyl ions.
selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-ina-3

Solution 4.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-ina-4

Solution 5.

(a) Lab preparation of ammonia:
2NH₄Cl + Ca(OH)₂ → CaCl₂ +2H₂O +2NH₃(b) The ammonia gas is dried by passing through a drying tower containing lumps of quicklime (CaO).
(c) Ammonia is highly soluble in water and therefore it cannot be collected over water.

Solution 6.

The drying agent used is CaO in case of ammonia.
Other drying agents like P₂O₅ and CaCl₂ are not used. As ammonia being basic reacts with them.
6NH₃ + P₂O₅ + 3H₂O → 2(NH₄)₃PO₄CaCl₂ +4NH₃ → CaCl₂.4NH₃

Solution 7.

The substance A is Ammonium chloride and ‘B’ is Ammonia.
Reaction:
2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2H₂O + 2NH₃

Solution 8.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-ina-8

Solution 9.

(a) Ammonium compounds being highly soluble in water do not occur as minerals.
(b) Ammonium nitrate is not used in the preparation of ammonia as it is explosive in nature and it decomposes forming nitrous oxide and water vapours.
(c) Conc. H₂SO₄ is not used to dry ammonia, as ammonia being basic reacts with them.
2NH₃ + H₂SO₄ → (NH₄)₂SO₄

Solution 10.

Preparation of Aqueous Ammonia: An aqueous solution of ammonia is prepared by dissolving ammonia in water. The rate of dissolution of ammonia in water is very high; therefore, back suction of water is possible. To avoid this, a funnel is attached to the outer end of the delivery tube with rubber tubing.

Procedure: Water is taken in a container and only a small portion of the mouth of funnel is dipped in water.

As ammonia dissolves in water at a higher rate than its production in the flask, the pressure in the funnel above water level decreases for a moment and water rushes into the funnel. As a result, the rim of the funnel loses its contact with water. Since, ammonia produced pushes the water down, the funnel comes in contact with water again. In this way, ammonia dissolves in water without back suction of water.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-ina-10

Exercise 1

Solution 1.

Physical properties of ammonia are:

Color: Colourless
Odour: Strong, Pungent chocking smell
Taste: Slightly bitter (alkaline ) taste
Physiological action: Non-Poisonous
Density: V.D = 8.5 Lighter than air
Nature: Alkaline
Liquefaction: easily liquefied at 10^oC by compressing it at 6 atm. Pressure
Boiling Point: Liquid ammonia boils at -33.5^oC
Freezing Point: Solid NH₃ melts at -77.7^oC
Solubility: Highly soluble in water, 1vol of water dissolves about 702 vols. of ammonia at 20^oC and 1 atm. pressure.
Reaction:
2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2H₂ + 2NH₃

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-1

Solution 2.

Ammonia is less dense than air. By Fountain Experiment we demonstrate the high solubility of ammonia gas in water.
Balanced equation for the reaction between ammonia and sulphuric acid is:
2NH₃ + H₂SO₄ → (NH₄)₂SO₄

Solution 3.

(a) Ammonia is basic in nature.
(b) Copper oxide because CuO is less reactive can be reduced by C, CO or by hydrogen whereas Al₂O₃, Na₂O, MgO are reduced by electrolysis.

Solution 4.

(a) The formula of the compound is Mg₃N_2.(b) Balanced equation :
Mg₃N₂ + 6 H₂O → 3Mg(OH)₂ + 2 NH₃(c) Ammonia is a reducing agent and reduces less active metal oxide to its respective metal.

Solution 5.

Reducing property.

Solution 6.

When a piece of moist red litmus paper is placed in a gas jar of ammonia it turns blue.

Solution 7.

(a) The gas is ammonia.
(b) The formula is NH_3.(c) Uses of ammonia:

It is used in the industrial preparation of nitric acid by Ostwald process.
It is used in the manufacture of fertilizers such as ammonium sulphate, ammonium nitrate, ammonium phosphate.
It is used in the manufacture sodium carbonate by Solvay process.
NaCl + NH₃ + CO₂ + H₂O → NaHCO₃ + NH₄Cl

Solution 8.

Equation:
CuSO₄ + 2NH₄OH → Cu(OH)₂↓ + NH₄]₂SO₄pale blue

Ammonia solution in water gives a blue precipitate when it combines with a solution of copper salt.

The pale blue precipitate of copper hydroxide dissolves in excess of ammonium hydroxide forming tetraamine copper[II] sulphate, an azure blue(deep blue)soluble complex salt.
Cu(OH)₂ +(NH₄)₂SO₄ +2NH₄OH → [Cu(NH₃)₄]SO₄ + 4H₂O

Solution 9.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-9

Solution 10.

(a) Liquid ammonia takes a lot of energy to vaporize .This heat is taken from the surrounding bodies which are consequently cooled down. Thus it is used as a refrigerant in ice plant.
(b) Ammonia emulsifies or dissolves fats, grease so it is used to remove grease from woolen clothes.
(c) Aqueous solution of ammonia gives pungent smell because of the presence of ammonia.
(d) Aqueous ammonia when dissolved in water breaks into ions which help in the conductance of electricity.

Solution 11.

(a) AlN + 3H₂O → Al(OH)₃ +NH₃(b) 2NH₃ + 3PbO → 3Pb + 3H₂O + N₂(c) 8NH₃ +3Cl₂→ N₂ + 6NH₄Cl
(d) 2NH₃ + CO₂ → NH₂CONH₂ + H₂O
(i) Ammonia act as reducing agent is explained by equation (c).
(ii) Urea the nitrogenous fertilizer is prepared from equation (d).

Solution 12.

(a) A Dirty green precipitate of Fe(OH)₂ is obtained when ammonium hydroxide is added to ferrous sulphate.
(b) Liquid ammonia is liquefied ammonia.
(c) Finely divided Iron is used in Haber process.
(d) Quicklime is a drying agent for NH₃.
(e) Ammonium salts when heated with caustic alkali.

Solution 13.

(a) Dirty green ppt. of Ferrous hydroxide is formed which is insoluble in excess of NH₄OH.
FeSO₄+ 2NH₄OH → [NH₄]₂SO₄ + Fe(OH)₂

(b) Reddish brown ppt. of ferric hydroxide is formed which is insoluble in ammonium hydroxide.
FeCl₃ + 3NH₄OH → 3NH₄Cl + Fe(OH)₃

(d) White gelatinous ppt. of Zinc hydroxide is formed which is soluble in NH₄OH.
Zn(NO₃)₂ + 2NH₄OH → 2NH₄NO₃ + Zn(OH)₂

Solution 14.

When correct amount of ammonium hydroxide is added drop wise to solutions of the metallic salts, ppts. (coloured generally) are formed. They help us to identify their metal ions.
Two equations:
FeSO₄ +2NH₄OH → (NH₄)₂SO₄ + Fe (OH)₂(Green) (Dirty green)
shows the presence of Fe⁺² ion.
FeCl₃ + 3NH₄OH → 3NH₄Cl + Fe (OH)₃(Brown) (Reddish brown)
shows the presence of Fe⁺³ ion.

Solution 15.

NH₄Cl on strong heating sublimes to form dense white fumes which condense to white powdery mass on cooler parts of the tube whereas no white fumes on heating NaCl.

(b) When ammonium hydroxide is added drop wise to solution to be tested.
Ferrous salt gives dirty green ppt.
Ferric salt gives reddish brown ppt of their hydroxides.

Solution 16.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-16

Solution 17.

(a) In the presence of Platinum at 800^oC, ammonia reacts with oxygen to give nitric oxide and water vapour.
Procedure:
Pass dry ammonia gas and oxygen gas through inlets over heated platinum placed in the combustion tube, which in the heated state emits reddish glow.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-17
Observation:
Reddish brown vapours of nitrogen dioxide are seen in the flask due to the oxidation of nitric oxide.
The platinum continues to glow even after the heating is discontinued since the catalytic oxidation of ammonia is exothermic.

(b) Two reactions to show reducing property of ammonia are:
8NH₃ +3Cl₂→ N₂ + 6NH₄Cl
2NH₃ +3CuO → 3Cu + 3H₂O +N₂

Solution 18.

(i) Neutralization
NH₃ + HCl → NH₄Cl

(ii) Thermal dissociation
NH₄Cl → NH₃ + HCl

(iii) Ammonia
NH₄Cl + NaOH → NH₃ + NaCl + H₂O

Solution 19.

(a) Ammonia
(b) Hydrogen chloride and chlorine gas.

(d) Acidic gas: HCl
Basic gas: Ammonia
Neutral gas: NH₄Cl

(e) Silver chloride
(f) Nitrogen
(g) Magnesium nitride
(h) Lead oxide
(i) Ammonium chloride

Solution 20.

CuSO₄ + 2NH₄OH → (NH₄)₂SO₄ + Cu(OH)₂ [Pale blue]
The cation present in solution B is Copper (Cu⁺²).
The colour of solution B is Blue.
The pale blue precipitate of copper hydroxide dissolves in excess of ammonium hydroxide forming tetraamine copper[II] sulphate, an azure blue(deep blue) soluble complex salt.
Cu(OH)₂ + (NH₄)₂SO₄ +2NH₄OH → [Cu(NH₃)₄]SO₄ + 4H₂O

Solution 21.

Three ways in which ammonia gas can be identified is:

It has a sharp characteristic odour.
When a glass rod dipped in HCl is brought in contact with the gas white colour fumes of ammonium chloride are formed.
It turns moist red litmus blue, moist turmeric paper brown and phenolphthalein solution pink.

Solution 22.

(a) Mg₃N₂ + 6H₂O 3Mg(OH)₂ + 2NH₃(b) 2NH₃ + 3CuO 3Cu + 3H₂O + N₂Ammonia acts as a reducing agent. It reduces metallic oxide to give metals, water vapour and nitrogen.
(c) 8NH₃ +3Cl₂ N₂ + 6NH₄Cl
(d) 4 NH₃ +5O₂ 6H₂O + 4NO +Heat
Ostwald process starts with the catalytic oxidation of ammonia to manufacture nitric acid in the presence of catalyst platinum.

Solution 23.

As the ‘A’ turns red litmus blue it is a base. Now the gas ‘A’ combines with ‘B’ in presence of Catalyst to give colourless gas Nitrogen monoxide. It reacts with oxygen to give brown gas which is Nitrogen dioxide.
A= NH₃B= O₂C=NO
D=NO₂

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-23

Solution 24.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-24

Solution 25.

(a) The main refrigerants used are Freon chlorofluorocarbons (CFC). They deplete ozone layer. The chlorofluorocarbons are decomposed by ultraviolet rays to highly reactive chlorine which is produced in the atomic form. selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-25
This causes depletion of ozone layer and chlorine monoxide so formed reacts with atomic oxygen and produces more chlorine free radicals.
ClO + O → Cl + O₂Again this free radical destroys ozone and the process continues thereby giving rise to ozone depletion.
(b) Liquid ammonia can be used as a refrigerant, as an alternative for chlorofluorocarbons.
(c) Advantages of ammonia as refrigerant:

Ammonia is environmentally compatible. It does not deplete ozone layer and does not contribute towards global warming.
It has superior thermodynamic qualities as a result ammonia refrigeration systems use less electricity.

Ammonia has a recognizable odour and so leaks are not likely to escape.

Solution 26.

Disadvantages of ammonia as a refrigerant are as follows:

It is not compatible with copper, so it cannot be used in any system with copper pipes.
It is poisonous in high concentration although it is easily detectable due to its peculiar smell and since it is less dense than air it goes up in the atmosphere not affecting the life too much on earth.

Solution 27.

(a) Explosive: ammonium nitrate
(b) Medicine: ammonium carbonate
(c) Fertilizers: ammonium sulphate
(d) Laboratory reagent: ammonia solution

Solution 28.

(a) Dry air free from carbon dioxide and dry ammonia from Habers process.
(b) The catalyst used in the process is Platinum.
(c) The oxidizing agent used in the process is oxygen.
(d) Ratio of ammonia and air is 1:10.
(e) Quartz is acid resistant and when packed in layers help in dissolving nitrogen dioxide uniformly in water.

Solution 29.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-29

Solution 30.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-30

Solution 1 (2003).

(a) Mg₃N₂ +6H₂O → 3Mg(OH)₂ + 2NH₃(b) Ammonia gas is collected in inverted gas jars by the downward displacement of air.
(c) Ammonia is not collected over water because it is highly soluble in water.
(d) Quicklime is used as a drying agent for ammonia.

Solution 1 (2004).

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-1-2004

Solution 1 (2005).

(a) It is the basic nature of ammonia molecule.
(b) Hydroxyl ion (NH₃+ H₂O → NH₄⁺ + OH^–)
(c) The red litmus paper turns blue in the solution.

Solution 1 (2006).

Pb(NO₃)₂+ NH₄OH →2NH₄NO₃+Pb(OH)₂The chalky white ppt. of lead hydroxide is formed.

Solution 1 (2007).

(a) HCl gas is more dense [V.D.=18.25,V.D. of ammonia =8.5] and it is collected by the upward displacement of air.
(b) NH₃ + HCl → NH₄Cl

Solution 2 (2005).

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-2-2005

Solution 2 (2007).

Balanced equation:

(a) 2NH₃ + 3CuO → 3Cu + 3H₂O + N₂(b) 2NH₃ + 3Cl₂ → N₂ + 6HCl

Solution 2 (2008).

Magnesium Nitride

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Selina Concise Chemistry Class 10 ICSE Solutions Study of Compounds – Hydrogen Chloride

August 11, 2022July 19, 2017 by Veerendra

Selina Concise Chemistry Class 10 ICSE Solutions Study of Compounds – Hydrogen Chloride

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 8 Study of Compounds Hydrogen Chloride. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Chemistry Chapter 8 Study of Compounds – Hydrogen Chloride

Exercise 1

Solution 1.

selina-icse-solutions-class-10-chemistry-study-compounds-hydrogen-chloride-1

Solution 2.

(a) Hydrogen chloride is dried by passing through conc. Sulphuric acid.
(b) Phosphorous pentoxide and CaO cannot be used to dry HCl because they react with HCl.
2P₂O₅+ 3HCl → POCl₃ + 3HPO₃CaO + 2HCl → CaCl₂ + H₂O

Solution 3.

(a) Anhydrous HCl is poor conductor due to the absence of ions in it whereas aqueous HCl is excellent conductor since it contains ions.

(b) When the stopper is opened HCl gas comes in contact with water vapors of air and gives white fumes due to the formation of hydrochloric acid.

(c) A solution of HCl in water gives hydronium ions and conducts electricity, but HCl is also soluble in dry toluene, but in that case it neither (i) turns blue litmus red (ii) nor does conducts electricity. This indicates the absence of H⁺ ions in toluene showing thereby that hydrogen chloride is a covalent compound.

(d) When ammonium hydroxide is brought near the mouth of HCl, dense white fumes are formed due to the formation of ammonium chloride.
HCl + NH₄OH → NH₄Cl + H₂O

(e) Dry hydrogen chloride is not acidic whereas moist Hydrogen chloride is acidic. In presence of a drop of water HCl gas dissolves in water and forms hydrochloric acid which turns blue litmus paper red.

(f) Hydrogen chloride is not collected over water as it is highly soluble in water.

Solution 4.

Difference between Hydrogen chloride gas and Hydrochloric acid is:

Hydrogen chloride gas

Hydrochloric acid

1. Dry hydrogen chloride gas does not turn blue litmus red due to non-acidic character.

2. Hydrogen chloride gas does not conduct electricity.

1. Being acidic it turns blue litmus red.

2. Hydrochloric acid is a good conductor of electricity.

Solution 5.

(a) Hydrochloric acid is prepared by this method.
(b) The reactants are sodium chloride and Sulphuric acid.
(c) The empty flask acts as Anti-Suction device. In case the back suction occurs the water will collect in it and will not reach the generating flask.
(d) The drying agent is Conc. Sulphuric acid. Sulphuric acid is chosen as drying agent because it does not react with HCl.
(e) The Inverted funnel :
Prevents or minimizes back suction of water.
Provides a large surface area for absorption of HCl gas.

Solution 6.

selina-icse-solutions-class-10-chemistry-study-compounds-hydrogen-chloride-6

Solution 7.

(a) Chlorine.
The compound formed which is strongly acidic in water, is HCl.
H₂ + Cl₂ → 2HCl

(b) A dilute aqueous solution of hydrochloric acid gets gradually concentrated on distillation, till the concentration of the acid reaches 22.2% HCl by weight which boils at 110⁰C.When this concentration is reached, no further increase in concentration of the acid becomes possible by boiling. This is because vapours evolved before 110⁰C are vapours of water but at temperature above 110⁰C vapours consist mostly of molecules of HCl.

Solution 8.

We can prove that hydrochloric acid contains both hydrogen and chlorine by the following experiment.
Take a voltameter used for electrolysis of water, fitted with platinum cathode and graphite anode.
Into the voltameter pour 4 molar HCl and pass direct current.
It is seen that a colourless gas is evolved at cathode and a greenish gas is evolved at anode.
When a burning splinter is brought near a colourless gas, it bursts into flame thereby proving that it is hydrogen gas.
When moist starch iodide paper is held in the greenish yellow gas, it turns blue black, thereby proving that the gas is chlorine.
2HCl → H₂ + Cl₂This experiment proves that hydrochloric acid contains both hydrogen and chlorine.

Solution 9.

(a) Manganese dioxide
(b) Hydrogen chloride and ammonia
(c) Hydrogen and oxygen
(d) AgCl(Silver chloride)
(e) Aqua regia
(f) Fountain experiment
(g) Hydrogen chloride gas

Solution 10.

(a) An aqueous solution of chlorine is acidic as it dissolves in water to form hydrochloric and hypochlorous acids.
(b) Silver nitrate reacts with hydrochloric acid to form thick curdy white ppt. of silver chloride whereas silver nitrate does not react with nitric acid.
AgNO₃ + HCl → AgCl + HNO₃(White ppt.)

Solution 11.

A is Silver nitrate
B is Hydrochloric acid
C is Silver chloride

Solution 12.

selina-icse-solutions-class-10-chemistry-study-compounds-hydrogen-chloride-12

Solution 13.

a. Sodium carbonate on treating withdil.HCl results in the formation of sodium chloride with the liberation of carbon dioxide gas.
Na₂CO₃+ 2HCl → 2NaCl + H₂O + CO₂ ↑
Sodium sulphite on treating with dil.HCl results in the formation of sodium chloride with the liberation of sulphur dioxide gas.
Na₂SO₃+ 2HCl → 2NaCl + H₂O + SO₂ ↑

b. Sodiumthiosulphate reacts with dil. HCl to produce sulphur dioxide gas and precipitates yellow sulphur.
Na₂S₂O₃ + 2HCl → 2NaCl + H₂O + SO₂ + S↓
Sulphur is not precipitated when sulphites are treated with dil.HCl.

Solution 14.

Three tests are:

HCl gas gives thick white fumes of ammonium chloride when glass rod dipped in ammonia solution is held near the vapours of the acid.
NH₃ + HCl NH₄Cl
With silver nitrate HCl gives white precipitate of silver chloride. The precipitate is insoluble in nitric acid but soluble in ammonium hydroxide.
AgNO₃ + HCl AgCl + HNO₃
A greenish yellow gas is liberated when concentrated hydrochloric acid is heated with oxidizing agent like manganese dioxide.
MnO₂ + 4HCl MnCl₂ +2H₂O + Cl₂

Solution 15.

MnO₂, PbO₂ and red lead react with conc. HCl acid to liberate Cl₂. This shows that hydrochloric acid is oxidized to chlorine by oxidizing agents.

Solution 16.

HCl dissolves both in water and toluene, when HCl dissolves in water it ionizes and forms hydronium and chloride ions. Whereas this ionization is not observed in toluene hence a solution of HCl in water can be used as an electrolyte.

Solution 17.

selina-icse-solutions-class-10-chemistry-study-compounds-hydrogen-chloride-17

Solution 18.

A mixture having three parts of conc. Hydrochloric acid and one part of conc. Nitric acid is called aqua-regia.
Nitric acid acts as oxidizing agent.

Solution 19.

selina-icse-solutions-class-10-chemistry-study-compounds-hydrogen-chloride-19

Solution 20.

selina-icse-solutions-class-10-chemistry-study-compounds-hydrogen-chloride-20

Solution 1 (2004).

Solution 1 (2005).

(a) (i) CuO +2HCl CuCl₂ + H₂O
(ii) MnO₂+ 4HCl MnCl₂ +2H₂O +Cl₂

(b) (i) The experiment is called Fountain Experiment.
(ii) This experiment shows that hydrogen chloride is highly soluble in water.
(iii) Red

Solution 1 (2007).

selina-icse-solutions-class-10-chemistry-study-compounds-hydrogen-chloride-1-2007

Solution 1 (2008).

When hydrogen chloride is collected by downward delivery or upward displacement, it shows that it is heavier than air.

Solution 2 (2008).

Hydrogen chloride is not collected over water as it is soluble in water.

Solution 3 (2008).

selina-icse-solutions-class-10-chemistry-study-compounds-hydrogen-chloride-3-2008

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Selina Concise Chemistry Class 10 ICSE Solutions Metallurgy

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Selina Concise Chemistry Class 10 ICSE Solutions Metallurgy

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Selina ICSE Solutions for Class 10 Chemistry Chapter 7 Metallurgy

Exercise 7(A)

Solution 1.

Three classes in which elements are classified are:
Metals, Non-metals and Metalloids
Copper was the first metal used by man.

Solution 2.

(a) The metal which is a constituent of blood pigment is Iron (Fe)
(b) The metal which is a constituent of plant pigment is Magnesium (Mg).

Solution 3.

(a) Nitrogen: It is used to preserve food.
(b) Hydrogen: It is used in the hydrogenation of vegetable oils to make ghee.
(c) Carbon: It is essential for the growth and development of living beings.

Solution 4.

The metal which is present in abundance in earths crust is aluminium.
The non-metal which is present in abundance in the earth crust is oxygen.

Solution 5.

Metals are defined as the elements which form positive ions by the loss of electrons.
Non-metals are the elements which form negative ions by the gain of electrons.

Solution 6.

(a) Alkali metals: They are placed in IA group, the first column on the left of the periodic table.
(b) Alkaline earth metal: They are placed in IIA group, the second column on the left of the periodic table.
(c) Iron and Zinc: Fe is placed in VIII group and Cu is placed in IB group.
(d) Aluminium: It is placed in IIIA group present on the right of periodic table.

Solution 7.

(a) Alkali metals:-
(i) Bonding: All alkali metal salts are ionic in nature.
(ii) Action of air: The react rapidly with oxygen and water vapour in the air.
(iii) Action of water: They react violently with water and produce hydrogen gas.
2M + 2H₂O → 2MOH + H₂(iv) Action of acid: They react violently with dil. HCl and dil. H₂SO₄ to produce hydrogen gas.
2M + 2HCl → 2MCl + H₂

(b) Alkaline earth metal:-
(i) Bonding: All alkaline earth metal salts except beryllium are ionic compounds.
(ii) Action of air: They are less reactive than alkali metals.
(iii) Action of water: They react with water to produce hydrogen gas.
M + 2H₂O → M(OH)₂ + H₂(iv) Action of acid: They react with dilute HCl and dil. H₂SO₄ to produce hydrogen gas.
M + 2HCl → MCl₂ + H₂

Solution 8.

Elements which show properties of both metals and non-metals are called metalloids.
For example: Silicon, Germanium.

Solution 9.

Hydrogen is placed with alkali metals as it has one electron similar to the alkali metals.

Solution 10.

(a) Bromine
(b) Lead
(c) Gallium
(d) Carbon
(e) Sodium
(f) Sodium
(g) Tungsten
(h) Carbon fibre
(i) Carbon
(j) Mercury

Solution 11.

(i) Ion formation: Metals form positive ions by loss of electrons whereas non-metals form negative ions by gain of electrons.

(ii) Discharge of ions: Metals are discharged at the cathode during electrolysis whereas non-metals are liberated at the anode during electrolysis.

(iii) Nature of oxide formed: Oxides of metals are usually basic. Soluble basic oxides dissolve in water forming an alkaline solution whereas oxides of non-metals are usually acidic. Soluble acidic oxides dissolve in water forming an acidic solution.

(iv) Oxidizing and reducing property: Metals ionize by loss of electrons and hence are reducing agents whereas non-metals ionize by gain of electrons and hence are oxidizing agents.

(v) Reaction with acids: Metals above hydrogen in activity series usually replace hydrogen from dilute non-oxidising acids whereas non-metals do not react with dilute hydrochloric acid or sulphuric acid.

Solution 12.

(a) Na – e^– → Na⁺(b) N + 3e^– → N^3-(c) Cl + e^– → Cl^–(d) Mg – 2e^–→ Mg²⁺(e) M + 2HCl → MCl₂ + H₂(f) Mg + H₂SO₄ → MgSO₄ + H₂

Solution 13.

(a) Fe₂O₃(b) PbO
(c) Mn₂O₇(d) NO

Solution 14.

Exercise 7(B)

Solution 1.

(a) Mercury and gallium
(b) Sodium and potassium
(c) Mercury
(d) Iodine
(e) Graphite
(f) Zinc
(g) Neon , Argon
(h) CrO₃ , Mn₂O₇(i) Al₂O₃,PbO
(j) Potassium , sodium
(k) Basic copper(II) sulphate
(l) Aluminium , Oxygen
(m) Hydrogen
(n) Carbon
(o) Iron

Solution 2.

(a) Occurrence of metals: The metals placed at the top of activity series are most reactive, so they always exist in the combined state whereas the metals placed below the activity series are least reactive, so they can be found in the isolated state also.

(b) Tendency to corrosion: The metals lying above the hydrogen in activity series can easily react with moisture and air and corrode easily whereas the metals such as gold and platinum do not corrode easily.

Potassium and sodium reacts with cold water whereas magnesium reacts with warm water and aluminium, zinc and iron reacts with steam.

(d) Reaction with acids: All the metals above hydrogen, in the activity series, reduce hydrogen ions from dil. hydrochloric or sulphuric acid and give out hydrogen gas. The rate of reaction decreases on moving down the series.

Solution 3.

selina-icse-solutions-class-10-chemistry-metallurgy-7b-3

Solution 4.

The metals placed higher in the activity series (i.e. Na and K) are stable to heat and soluble in water.
Whereas metals like Ca, Mg, Al, Zn, Fe, Pb, Cu decompose on heating with decreasing vigour to form metal oxide and carbon dioxide.
The metals which lie below in the activity series (i.e. Hg, Ag) decompose on heating to form metal, oxygen and carbon dioxide.

Solution 5.

(a) Alkali metals like sodium and potassium are kept in kerosene as they react with moisture and air.
(b) (i) Basic lead carbonate is a mixture of lead hydroxide and lead carbonate.
(ii) Brown powder is mainly hydrated iron(III) oxide (Fe₂O₃.xH₂O)

Solution 6.

Oxides of metals like Na, K, Ca, Mg, Al are stable to heat and so can be reduced only by electrolysis.
Zinc oxide can be reduced by coke only.
Oxides of iron, lead and copper are reduced by C, CO, H₂ and NH₃.
Oxides of mercury and silver decompose to give metal and oxygen.

Solution 7.

Metal A is more reactive than Metal B.
(a) Metal A is Na (Sodium). Metal B is Ca (Calcium).
Reaction with HCl:

selina-icse-solutions-class-10-chemistry-metallurgy-7b-7

(b) (i) Oxides: Sodium and calcium oxides are stable to heat.
(ii) Hydroxides: Sodium hydroxide is stable to heat whereas calcium hydroxide decomposes on heating to metal oxide and water vapour.
(iii) Carbonates: Sodium carbonate is stable to heat whereas calcium carbonates decompose on heating to form calcium oxide and carbon dioxide.
(iv) Nitrates: Sodium nitrate on heating form nitrite and oxygen whereas calcium nitrate decomposes on heating to form calcium oxide, nitrogen dioxide and oxygen.

Solution 8.

(a)

Metals	Non-metals
(i) Good conductors of heat (ii) Malleable (iii) Form positive ions (iv) Form basic oxides	Poor conductors of heat Non-Malleable Forms negative ions Form acidic oxides

(b) Valence electrons present in :
(i) Metals have 1, 2 or 3 valence electrons.
(ii) Non-metals have 5, 6 or 7 valence electrons.

Solution 9.

When the surface of metal is attacked by air, moisture or any other substance around it, the metal is said to corrode and the phenomenon is known as corrosion.
Necessary conditions for corrosion are:

Presence of oxygen and moisture.
Metals which are placed higher in the activity series corrode more easily.

Solution 10.

Conditions for increase of corrosion are:

Presence of oxygen and moisture.
Metals which are placed higher in activity series corrode more easily
Dissolved salts in water act as electrolyte and enhance the rate of corrosion.
The presence of pollutants like NO₂and CO₂increases rusting.

Solution 11.

Corrosion of metals is an advantage as it prevents the metal underneath from further damage. For example: On exposure to air, the surface of metal like aluminium and Zinc forms layers of their oxides which are very sticky and impervious in nature and hence act as protective layer. This layer protects the metal from further damage.

Solution 12.

Rusting is the slow oxidation of iron by atmospheric oxygen in the presence of water.
Equation:
4Fe + 3O₂ + 2x H₂O → 2Fe₂O₃.xH₂O

Solution 13.

Two conditions necessary for rusting of iron are:

Air
Water

Solution 14.

By painting an iron object, the iron do not come in contact with atmospheric reagents .This prevents rusting.

Solution 15.

Galvanisation is the process of applying a protective zinc coating to steel or iron, in order to prevent rusting.
The zinc coating does not allow iron to come in contact with air and moisture and thus protects it from rusting.

Solution 16.

Silver gets tarnished when exposed to the atmosphere which contains pollutant H₂S and forms a black coating of Ag₂S.
Copper forms a green deposit on its surface when exposed to moist air. This is usually basic copper (II) sulphate.

Solution 17.

Aluminium forms white colour oxide on exposure to the atmosphere. This white colour oxide prevents it from further corrosion whereas iron reacts with air to form hydrated oxide called rust. So, iron undergoes corrosion to greater extent.

Solution 18.

The noble metals such as gold and platinum do not corrode easily.

Solution 19.

Gold is the most unreactive metal so it does not react with air or water and other gases in atmosphere. So gold does not corrode. That is why gold look new after several years of use.

Exercise 7(C)

Solution 1.

The process used for the extraction of metals in their pure form from their ores is referred to as Metallurgy.
The processes involved in Metallurgy are:

Crushing and Grinding
Concentration
Roasting andcalcination
Reduction
Refining

Solution 2.

(a) A metal which occurs as sulphide is lead.
(b) A metal which occurs as halide is silver.
(c) A metal which occurs as carbonate is zinc.
(d) A metal which occurs as oxide is iron.

Solution 3.

(a) Minerals are naturally occurring compounds of metals which are generally present with other matter such as soil, sand, limestone and rocks. Ores are those minerals from which the metals are extracted commercially at low cost and comfortably. All ores are minerals, but all minerals are not necessarily ores.

(b) Ores are those minerals from which the metals are extracted commercially at low cost and with minimum effort. A metallic compound is a compound that contains one or more metal elements. Examples: AgNO₃ – Silver nitrate is a metallic compound.

Solution 4.

The metals that can be extracted from the following ores are:
(a) Bauxite- Aluminium
(b) Calamine- Zinc
(c) Haematite- Iron

Solution 5.

Three objectives achieved during the roasting of ores is:

It removes moisture from ores.
It makes the ore porous and more reactive.
It expels volatile impurities.
It convertssulphideores into oxides.

Solution 6.

(a) Hydraulic washing: The difference in the densities of the ore and the gangue is the main criterion.
(b) Forth floatation: This process depends on the preferential wettability of the ore with oil and the gangue particles by water.
(c) Electromagnetic separation: Magnetic properties of the ores.

Solution 7.

(a) The processes involved in

(i) Processes involved in concentration are:

Hydrolytic method
Magnetic Separation
Froth floatation
Leaching

(ii) Processes involved in Refining of ores are:

Distillation
Liquation
Oxidation
Electro- refining

(b) Potassium and sodium oxides cannot be reduced by carbon, carbon monoxide and hydrogen.

Solution 8.

(a) Flux: A flux is a substance that is added to the charge in a furnace to remove the gangue.
(b) Gangue: Earthly impurities including silica, mud etc., associated with the ore are called gangue.
(c) Slag: It is the fusible product formed when flux reacts with impurities during the extraction of metals.
(d) Smelting: Smelting is the process of reducing the roasted oxide ore and removing the gangue with the help of an appropriate flux added with the ore.

Solution 9.

Iron and zinc are quite reactive and hence they do not occur in the free state. The compounds of metals found in nature are their oxides, carbonate and sulphides.

Solution 10.

Solution 11.
selina-icse-solutions-class-10-chemistry-metallurgy-7c-11

Solution 12.

selina-icse-solutions-class-10-chemistry-metallurgy-7c-12

Solution 13.

Oxides of highly active metals like potassium, sodium, calcium, magnesium and aluminium have great affinity towards oxygen and so cannot be reduced by carbon or carbon monoxide or hydrogen.

Metals in the middle of activity series (iron, zinc, lead, copper) are moderately reactive and are not found in oxide form. These are found in nature as sulphides or carbonate. These are first converted into oxides and can be reduced by C, CO or H₂.
selina-icse-solutions-class-10-chemistry-metallurgy-7c-13
Metals low in the activity series is very less reactive and oxides of these metals are reduced to metals by heating alone.

Solution 14.

selina-icse-solutions-class-10-chemistry-metallurgy-7c-14

Solution 15.

Aluminium has a great affinity towards oxygen and so cannot be reduced by carbon or carbon monoxide. So it is extracted from its oxide by electrolysis.
Metals like copper, lead and iron are placed in the middle of the activity series and re moderately reactive and their oxides can be reduced by carbon, CO and hydrogen.
Mercury and silver are less reactive and are placed lower in the reactivity series. The oxides of these metals are reduced to metals by heating their oxides.

Solution 16.

The process used for the concentration of the ore is froth floatation process.

Solution 17.

(a) The purification depends upon:

Nature of metal.
Nature of impurities present in the metal.
Purpose for which metal is to be used

(b) Methods used for purification are:

Distillation
Liquation
Oxidation
Electro-refining

(c) The impure metal is made the anode, while a thin sheet of pure metal is made the cathode. Electrolyte used is a salt solution of a metal which is to be refined. Pure metal deposits at the cathode and impurities settle down forming anode mud.

selina-icse-solutions-class-10-chemistry-metallurgy-7c-17

Solution 18.

selina-icse-solutions-class-10-chemistry-metallurgy-7c-18

Exercise 7(D)

Solution 1.

Position in the Periodic Table: Period 3,Group IIIA(13)

Solution 2.

selina-icse-solutions-class-10-chemistry-metallurgy-7d-2

Solution 3.

Bauxite ore contains approximately 60% aluminium oxide. The rest being sand, ferric oxide and titanium oxide.

Solution 4.

Red mud consists of ferric oxide, sand etc. left after bauxite dissolves in NaOH forming sodium aluminate and is removed by filtration.

Solution 5.

As aluminium has great affinity for oxygen, so it is stable compound. It is not easily reduced by common reducing agents like carbon, carbon monoxide or hydrogen. Hence, electrolytic reduction is chosen as the method for reducing alumina.

Solution 6.

selina-icse-solutions-class-10-chemistry-metallurgy-7d-6

Solution 7.

(a) The process by which refining of aluminium is done is called Hoope’s electrolytic process.
(b) Molten impure aluminium forms the bottom layer. The bottom layer has carbon lining and serves as anode.
Pure molten aluminium with carbon electrodes serves as cathode in top layer.
(c) Reactions at the two electrodes are:
Anode: Al -3e^– → Al³⁺Cathode: Al³⁺ + 3e^– → Al

Solution 8.

Reaction of aluminium:

(a) Air: Aluminium forms oxide at room temperature.
Aluminium powder burns in air at about 800⁰C forming its oxide and nitride with a bright light.
4Al + 3O₂ → 2Al₂O₃2Al + N₂ → 2AlN

(b) Water: Water has no action on aluminium due to layer of oxide on it.
When steam is passed over pure heated aluminium, hydrogen is produced.
2Al + 3H₂O → Al₂O₃+ 3H₂

(c)Acid: It reacts with acids to produce salt and hydrogen.
2Al + 6HCl → 2AlCl₃ + 3H₂Dilute sulphuric acid reacts with metal to liberate hydrogen.
2Al + 3H₂SO₄ (dilute) → Al₂(SO₄)₃ + 3H₂

Concentrated sulphuric acid reacts with aluminium to produce sulphur dioxide.
2Al + 6H₂SO₄ → Al₂(SO₄)₃ + 6H₂O + 3SO₂Dilute and concentrated nitric acid does not attack the metal aluminium.

(d) Base: Aluminium reacts with boiling and dilute alkalies to produce meta aluminate while with fused alkali produce aluminate.
2Al+ 2NaOH +2H₂O → 2NaAlO₂ +3H₂(Sodium meta aluminate)
2Al + 6NaOH → 2NaAlO₃ +3H₂
(Sodium aluminate)

Solution 9.

The role of cryolite in the electrolytic reduction of alumina in Hall’s process is :

Lowers the fusion temperature from 2050⁰C to 950⁰C and enhances conductivity.
Increases its conductivity since pure alumina is almost a non-conductor of electricity.
Cryoliteacts as a solvent for the electrolytic mixture.

Solution 10.

(a) Aluminium is more active metal but it gets oxidized and forms a thin protective layer on its surface which prevents further corrosion.
(b) Aluminium vessels should not be cleaned with powders containing alkalis because it results in the formation of meta aluminates and hydrogen.
2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂

Solution 11.

selina-icse-solutions-class-10-chemistry-metallurgy-7d-11-1

(c) A layer of aluminium is formed on iron at high temperature during cooking and food becomes deficient in iron.

Solution 12.

(a) In the electrolytic reduction of alumina, the graphite (anode) is oxidized by oxygen to CO and further forms CO₂, so it is consumed and has to be replaced from time to time.
2C + O₂ → 2CO
2CO + O₂ → 2CO₂

(b) Roasting provides oxygen to convert metallic sulphides into metallic oxide and SO₂ which takes place when heated in excess of air.
Carbonate is converted into oxide by loss of CO₂ which takes place in the absence of air and when heated strongly.

(c) Aluminium has a great affinity towards oxygen and so cannot be reduced by carbon or carbon monoxide or hydrogen whereas lead oxide can be easily reduced to metal lead by carbon.
PbO + C → Pb + CO

Solution 13.

(a) Flux combines with the gangue to form a fusible mass called slag.
(b) It forms slag[CaSiO₃] with silica.
(c) It is removed from upper outlet, slag being lighter float on molten iron.

Solution 14.

(a) Froth flotation process: Zinc blende[ZnS]
(b) Magnetic Separation: Haematite[Fe₂O₃]

Solution 15.

Electrolytic Reduction

It is removal of oxide or halide from a metal.
Oxides of highly active metals like Na,K,Ca,Mg,Al are reduced by electrolytic reduction of their fused salts.
Oxides of these metals have great affinity for oxygen than carbon and cannot be reduced by carbon or CO or hydrogen.

Electrolytic refining of metals is the separation of residual impurities like Si and phosphorus.

Presence of other metals and non-metals like Si and phosphorus.
Unreduced oxides and sulphides of metals.

It depends upon:

Nature of metal
Purpose for which metal is to be obtained.
Nature of impurities present.
Impure metal is made anode while a thin sheet of pure metal is made cathode and electrolyte used is a salt of solution of a metal to be refined.

Solution 16.

The three ways in which metal zinc differs from the non-metal carbon is:

Zinc has avalency2 and carbon has valency 4.
Zinc does not form hydride but carbon does (CH₄).
Oxides of zinc areamphoteric(ZnO) whereas oxides of carbon are acidic (CO₂) and neutral (CO).

Exercise 7(E)

Solution 1.

(a) To prevent from rusting.
(b) Due to strong electropositive nature, it easily forms Zn⁺² ions.
(c) Antiseptic in face creams.

Solution 2.

(a) Aluminium:

Being a strong, light and corrosion resistant metal, it is used in alloys.
Aluminium is light, it has high tensile strength, is resistant to corrosion, good conductor of heat, unaffected by organic acids and has attractive appearance. So it is used for making cooking utensils, in building and construction work.
Aluminium has a strong affinity for oxygen so it is used as a deoxidizer in the manufacture of steel.

(b) Zinc:

Zinc has a strong electropositive character, so it is used for coating iron and steel sheets to prevent them from rusting and this process is known as galvanization.
Due to strong electropositive nature, it forms Zn⁺²ions, so it is used to make dry cell containers which act as negative electrode.
Zinc act as a reducing agent for many organic reductions and these reductions are employed in manufacturing drugs, dyes.

Solution 3.

(a) Zinc is electropositive metal than iron, gets oxidized and saves iron. Also zinc forms protective layer of ZnO on iron. This layer is sticky and impervious in nature and protects the iron metal underneath from rusting.

(b) A neutral gas other than oxygen which is formed at anode during electrolysis of fused alumina is carbon monoxide.

(c) Nitric acid can be stored in aluminium containers as the dilute and conc. nitric acid does not react with aluminum. It renders aluminium passive due to the formation of an oxide film on its surface.

Solution 4.

(a) Cast iron: It is used in drain pipes, gutter covers, weights and railings.
(b) Wrought iron: It is used in chains, horse shoes and electromagnets.
(c) Mild steel: It is to manufacture nuts, bolts etc.
(d) Hard steel: It is used to make tools.

Solution 5.

(a) Galvanized iron sheets
(b) Zinc
(c) Zinc

Solution 6.

(a) Aluminium being strong, light and corrosion resistant metal is used for making alloy.
(b) Aluminium is light, malleable and does not rust so it is used for wrapping chocolates.
(c) To prevent them from rusting.
(d) It is used in aluminothermy as it is a good reducing agent.
(e) As aluminium forms a film of aluminium oxide, it protects the ships from corrosion. So it is used for making ships.

Solution 7.

(a) A mixture of 3 parts of ferric oxide (Fe₂O₃) and one part of aluminium powder (Al).
(b) A mixture of Potassium chlorate and magnesium powder is the ignition mixture.
(c) Fe₂O₃ + 2Al → Al₂O₃ + 2Fe + heat

Solution 8.

Alloy is a homogeneous mixture of two or more metals or of one or more metals with certain non-metallic elements.
The properties of alloys are often greatly different from those of the components.
For example: Gold is too soft to be used without small percentage of copper.
A low percentage of molybdenum improves the toughness and wear resistance of steel.
Bell metal is more sonorous than copper or tin.
Alnico an alloy of aluminium, nickel and cobalt can lift 60 times its own mass.
These added elements improve hardness, wear resistance, toughness and other properties.

Solution 9.

Alloy’s name	Composition	Uses
1. Stainless steel	73% Fe, 18%Cr,8%Ni,1%C	Used for making utensils, cutlery, ornamental pieces and surgical instruments.
2. Manganese steel	85% Fe,1%C ,14%Mn	Used for making rock drills and armour plates.
3. Tungsten steel	84%Fe, 5%W, 1%C	Used for cutting tools for high speed lathes.

Solution 10.

The other element in Brass is Zinc.
The other elements in Bronze are Tin and Zinc.

Solution 11.

(a) Duralumin
(b) Solder
(c) Brass
(d) Zinc amalgam

Solution 12.

A mixture or an alloy of mercury with a number of metals or an alloy such as sodium, zinc, gold and silver as well as with some non-metals is known as amalgam.
Dental amalgam is a mixture of mercury and a silver tin alloy.

Solution 13.

(a) Two properties of brass that make it more useful than its components are:

It is malleable and ductile.
It resists corrosion.
Can be easily cast.

(b) A metal which forms a liquid alloy at ordinary temperature is sodium.

Solution 14.

Magnalium is an alloy of aluminium with composition 90-95% and magnesium with composition 10-5%. It is used for making aircrafts.

Solution 15.

The constituents of
(a) Duralumin are aluminium (95%), copper (4%), magnesium (0.5%) and manganese (0.5%).
(b) Solder are lead (50%) and tin (50%).
(c) Bronze are copper (80%), tin (18%) and zinc (2%).
(d) Invar are iron (63%), nickel (36%) and carbon (1%).

Miscellaneous Exercise

Solution 1.

(a) Bauxite: Aluminium is extracted from its main ore bauxite Al₂O₃.2H₂O. It contains 60% Al₂O_3.(b) Sodium hydroxide: Sodium hydroxide dissolves bauxite to form sodium meta aluminate, removes insoluble impurities from Al₂O₃ by forming red mud.

selina-icse-solutions-class-10-chemistry-metallurgy-mis-1

Solution 2.

(a) Copper
(b) Iron
(c) Zinc
(d) Magnesium

Solution 3.

Arrangement of metal in decreasing order of reactivity are:
Sodium > Magnesium > Zinc > Iron > copper

Solution 4.

selina-icse-solutions-class-10-chemistry-metallurgy-mis-4
(iii) In electrolytic process, the graphite acts as anode. The anode has to be replaced from time to time as it gets oxidized by the oxygen evolved at the anode.
(iv) The reaction that occurs at cathode is:
4Al³⁺ + 12e^–→ Al
(d) In construction the alloy of aluminium -duralumin is used because it is hard and resistant to corrosion.

Solution 5.

selina-icse-solutions-class-10-chemistry-metallurgy-mis-5
(d) In Aluminium thermite welding, the reduction with aluminium is highly exothermic and heat generated is sufficient to melt the metal.
Fe2O3 + 2Al → 2Fe + Al₂O₃ + Heat

Solution 1 (2005).

(i)

Zinc: Froth Flotation, ZincBlende,Coke
Aluminium: Bauxite,Cryolite, Sodium hydroxide solution

(ii)

Sodium hydroxide.
Cryolite

(iii) The formula of Cryolite is Na₃AlF_6.

Solution 1 (2006).

(a) Mercury
(b) Roasting
(c) CaSiO₃(d) Cryolite
(e) Graphite

Solution 1 (2007).

Acidic oxide(D)
Discharged at anode (F)
Covalent chlorides (I)
5,6,7 valence electrons (L)
Brittle(C)

Solution 1 (2008).

(i) A is made of carbon and B is thick graphite rod.
A → Cathode
B → Anode
(ii) Aluminium is formed at electrode A.
(iii) The two aluminium compound in the electrolyte C is Na₃AlF₆, Al₂O₃.
(iv) It is necessary to continuously replace electrode B from time to time as it gets oxidized by the oxygen evolved.

Solution 2 (2005).

(a) Stainless steel : Iron, Chromium
(b) Brass: Copper , Zinc

Solution 2 (2007).

selina-icse-solutions-class-10-chemistry-metallurgy-mis-2-2007

Solution 2 (2008).

Brass is an alloy of copper and Zinc.

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Selina Concise Chemistry Class 10 ICSE Solutions Electrolysis

August 11, 2022July 18, 2017 by Veerendra

Selina Concise Chemistry Class 10 ICSE Solutions Electrolysis

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 6 Electrolysis. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Chemistry Chapter 6 Electrolysis

Exercise Intext 1

Solution 1.

(a) Powdered sodium chloride (common salt) does not conduct an electric current, but it does so when dissolved in water or when melted.

(b) Molten lead bromide conducts electricity .It is called an electrolyte. It is composed of lead ions and bromide ions. The lead ions are positivelycharged and are called cations. The bromide ions are negatively charged and are called anions.

(d) The electron releasing tendency of zinc is more than that of copper.

(e) A solution of HCl gas in water conducts electricity because it ionizes, but a solution of HCl gas in toluene does not conduct an electric current because it does not ionize in toluene.

Solution 2.

(a) Electrolysis: It is the process of decomposition of a chemical compound in aqueous solutions or in molten state accompanied by a chemical change using direct electric current.

(b) Non-electrolyte: It is a compound which neither in solution nor in the molten state allows an electric current to pass through it.

(c) Cation and anion: Atoms which carry positive charge are called cations.
Atoms which carry negative charge are called anions.

(d) Weak electrolyte: Electrolytes which allow small amount of electricity to flow through them and are partially dissociated in fused or aqueous solution are called weak electrolyte.

Solution 3.

(a) Difference between Modern explanation and Arrhenius explanation for the theory of electrolysis:
Arrhenius considered that water ionizes electrolytes but Modern theory explained that electrolytes are ionic even in solid state and their ions are held by strong electrostatic forces which make them immobile. Water renders these ions mobility by breaking the electrostatic forces.

(b) Difference between electrolytic dissociation and ionization :

Ionisation	Dissociation
1. Formation of positively or negatively charged ions from molecules which are not initially in the ionic state.	1. Separation of ions which are already present in an ionic compound.
2. Polar covalent compounds show ionization. e.g. HCl, H₂CO₃, NH₄OH etc.	1. Electrovalent compounds show dissociation. e.g. Potassium chloride , lead bromide, etc.

Cation	Anion
1. Are positively charged ions.	Are negatively charged ions.
2. Migrate to cathode during electrolysis.	Migrate to anode during electrolysis.
3. Gain electron from the cathode and get reduced to become a neutral atom.	Lose electrons to the anode and get oxidized to become a neutral atom.

(d) Electrolytic dissociation and thermal dissociation:
Electrolytic dissociation is the dissociation of an electrovalent compound into ions in the fused state or in aqueous solution state.

Thermal dissociation: Reversible breakdown of a chemical compound into simpler substances by heating it. The splitting of ammonium chloride into ammonia and hydrogen chloride is an example. On cooling, they recombine to form the salt.

Solution 4.

(a) Sodium carbonate
(b) NH₄OH
(c) An inert electrode: graphite and Active electrode: silver
(d) H⁺(e) Electrode is cathode
(f) Graphite

Solution 5.

Electrolysis is a redox process. The reaction at the cathode involves reduction of cations as they gain of electrons while the reaction at anode involves oxidation of anions as they loss of electrons to become neutral.
Example: Dissociation of sodium chloride during electrolysis.
Cathode : Na⁺ + e^– → Na (reduction)
Cl^– – e^–→ Cl (oxidation)
Cl + Cl → Cl₂Overall reaction: 2NaCl → 2Na + Cl₂

Exercise Intext 2

Solution 1.

(a) Glucose, Kerosene
(b) NaCl and NaOH
(c) CH₃COOH and NH₄OH

Solution 2.

(a) Cane sugar is a compound which does not have ions even in solution and contains only molecules. Hence, it does not conduct electricity. On the other hand, sodium chloride solution contains free mobile ions and allows electric current to pass through it. This makes it a good conductor of electricity.

(b) Hydrochloric acid is a strong electrolyte and dissociates completely in aqueous solution. The solution contains free mobile ions which allow electric current to pass through it. Hence, hydrochloric acid is a good conductor of electricity.

(c) Hydrogen is placed lower in the electrochemical series and sodium is placed at a higher position. This is because H⁺ ions are discharged more easily at the cathode than Na⁺during electrolysis and gains electrons more easily.
Therefore, H⁺ ion is reduced at the cathode and not Na⁺ ion.

Solution 3.

(a) Zn occurs readily as ion whereas Cu occurs more readily as metal in nature.

(b) Copper is above silver in the electrochemical series and is thus more reactive than silver. So, copper displaces silver from silver nitrate. Hence, we cannot store AgNO₃ solution in copper vessel.
Cu +AgNO₃ → Cu(NO₃)₂ + 2Ag

Solution 4.

(a) By treating its salt with a more reactive metal.
(b) By supplying two electrons to Cu⁺²Cu⁺²+ 2e^–→ Cu

Solution 5.

In the aqueous state, the slightly negatively charged oxygen atoms of the polar water molecule exerts a pull on the positively charged sodium ions. A similar pull is exerted by the slightly charged hydrogen atoms of the water on the negatively charged chloride ions. Thus the ions become free in solution. These free ions conduct electricity.
In the molten state, the high temperatures required to melt the solid weakens the bond between the particles and the ions are set free.

Solution 6.

(a) Two anions are and OH^–.

(b) OH^– is discharged at anode and the main product of the discharge of OH^– is O₂Reaction is :
OH^–→ OH + e^–4OH → 2H₂O + O₂

(d) No change in colour is observed.

(e) Dilute sulphuric acid catalyse the dissociation of water molecules into ions, hence electrolysis of acidified water is considered as an example of catalysis.

Solution 7.

(a) Labelled diagram of electrolytic cell is:

(b) The ions present in the cell are Cu²⁺, H⁺, SO₄^2- , OH^–.
(c) SO₄^2- and OH^– ions both migrate towards anode.
(d) Both Cu²⁺ and H⁺ ions migrate towards cathode.
(e) SO₄^2- and H⁺ will not discharge at electrodes.
(f) Reaction at cathode:
Cu⁺² +2e^– → Cu
(g) Reaction at anode:
OH^– – e^– → OH
2OH + 2OH → 2H₂O + O₂(h) Sulphate ions are the spectator ions because they do not change in the reaction.

Solution 8.

(a) Reaction at anode during the electrolysis ofvery dilute sulphuric acid:
OH^– → OH + e^–4OH → 2H₂O + O₂

(b) Reaction at anode during the electrolysis of aqueous copper sulphate solution
4OH^– → 4OH + 4e^–4OH → 2H₂O + O₂

(d) Reaction at anode during the electrolysis of fused lead bromide
Br^– – e^–→ Br
Br + Br → Br₂

(e) Reaction at anode during the electrolysis of magnesium chloride (molten)
2Cl^– → Cl₂ +2e^–

(f) Concentrated HCl,
HCl in the pure liquid state is unionised and hence does not conduct electricity.

(g) Very dilute HCl,
Cl^– – e^– → Cl
Cl + Cl → Cl₂

Solution 9.

(a) Electrolyte
(b) Nickel
(c) Cathode
(d) Anode
(e) Cations

Exercise 1

Solution 1.

(a) During electrolysis of lead bromide, there is loss of electrons at anode by bromine and gain of electrons at cathode by lead. Thus oxidation and reduction go side by side. Therefore, it is a redox reaction.
PbBr₂ → Pb⁺² + 2Br^–

(b) The blue colour of copper ions fades due to decrease in Cu⁺² ions and finally the solution becomes colourless as soon as Cu⁺² ions are finished.

(c) Lead bromide dissociate into ions in the molten state whereas it does not dissociate in solid state. The ions become free when lead bromide is in molten state but in the solid state the ions are not free since they are packed tightly together due to electrostatic force between them. Therefore, lead bromide undergoes electrolytic dissociation in the molten state.

(d) Aluminium has great affinity towards oxygen, so it is not reduced by reducing agent. Therefore it is extracted from its oxide by electrolytic reduction.

(e) As per electrolytic reactions, 4H⁺¹ are needed at cathode and 4OH^– at the anode and two molecules of water are produced at the anode. Hence for every two molecules of water, two molecules of hydrogen and one molecule of oxygen are liberated at the cathode and anode respectively.

(f) This is because HNO₃ is volatile.

(g) Ammonia is a covalent compound. Therefore, it is unionized in the gaseous state but in the aqueous solution it gives NH₄OH which is a weak electrolyte and dissociates into ions.

(h) Graphite is unaffected by the bromine vapours.

(i) Silver nitrate is not used as electrolyte for electroplating with silver because the deposition of silver will be very fast and hence not very smooth and uniform.

(j) Carbon tetrachloride is a liquid and does not conduct electricity because it is a covalent compound and there are no free ions present and contain only molecules.

Solution 2.

(a) Strong electrolyte : Dilute hydrochloric acid, dilute sulphuric acid, ammonium chloride, sodium acetate
(b) Weak electrolyte: Acetic acid, ammonium hydroxide
(c) Non-electrolyte: Carbon tetrachloride

Solution 3.

(a) Molecules
(b) Will not

Solution 4.

Water is a non-conductor of electricity and consists entirely of molecules. It can be electrolytically decomposed by addition of traces of dilutesulphuric acid which dissociate as H⁺ and SO₄^2- ions and help in dissociating water into H⁺ and OH^–, water being a polar solvent.

Solution 5.

	Anode	Electrolyte	Cathode
Silver plating of spoon	Plate of pure clean silver	Solution of potassiumargentocyanide	Article to be electroplated
Purification of copper	Impure copper	Solution of coppersulphate and dilutesulphuric acid	Thin strip of pure copper

Solution 6.

Electricity, Chemical

Solution 7.

(b) CuSO₄ is preferred as an electrolyte.

(c) The copper anode continuously dissolves as ions in solution and is replaced periodically. The electrolyte dissociates into Cu⁺² ions which migrate towards the iron object taken as the cathode and are deposited as neutral copper atoms on the cathode.
Electrolyte: Aqueous solution of nickel sulphate
Dissociation: CuSO₄ → Cu²⁺ + SO₄^2-
H₂O → H⁺ + OH^–Electrodes:
Cathode: Article to be electroplated
Anode: Block of pure copper
Electrode reactions:
Reaction at cathode: Cu²⁺ + 2e^–→ Cu (deposited)
Reaction at anode: Cu – 2e^–→ Cu²⁺

Solution 1 (2004).

(a) X → X²⁺ + 2e^– , Y + 3e^–→ Y^3-(b) Y₂ + 3X → X₃Y₂(c) (i) It is used for the electroplating of metals.
(ii) It is also used in purification of metals.
(d) Cathode, Anode

Solution 1 (2005).

(a) Because Copper is an electronic conductor as it is a metal.
(b) In solid sodium chloride, Na⁺ and Cl^– ions are not free due to strong electrostatic forces of attraction among them. The ions, therefore are unable to move to any large extent when electric field is affected. Hence no current.

Solution 1 (2006).

(a) (i) The name of electrode A is Platinum anode and that of electrode B is platinum or copper cathode.
(ii) Anode act as oxidizing electrode.
(b) AgNO₃solution will turn blue.

Solution 1 (2007).

(i) Molten ionic compound: Strong electrolytes
(ii) Carbon tetrachloride: Non-electrolyte
(iii) An aluminium wire: Metallic conductor
(iv) A solution containing solvent molecules, solute molecules and ions formed by dissociation of solute molecules: Weak electrolyte
(v) A sugar solution with sugar molecules and water molecules: Non- electrolyte

Solution 1 (2008).

(a) The reaction takes place at anode. This is an example of oxidation.
(b) Cu⁺² will discharge easily at cathode.
Reaction at cathode:
Cu⁺² +2e^– → Cu
(c) Carbon tetrachloride is a non-electrolyte because it is a covalent compound. It does not ionize and hence do not conduct electricity.

Solution 2 (2004).

(a) Non-electrolyte contains molecules.
(b) Molecules of HX and H⁺ and X^– ions.
(c) Loss
(d) The electrolyte used for the purpose must contain the ions of metal which is to be electroplated on the article.
(e) The reaction at the cathode involves reduction of cations as they gain electrons to become neutral atoms while that at anode involves oxidation of anions as they lose electrons to become neutral.
Example: Dissociation of sodium chloride during electrolysis.
NaCl → Na⁺ + Cl^–At cathode: Na⁺ + e^– Na (Reduction)
At anode: Cl^– – e^– → Cl(oxidation)
Cl + Cl → Cl₂Overall reaction: 2NaCl → 2Na + Cl₂

Solution 2 (2005).

Hydrogen gas is released at cathode when acidulated water is electrolyzed.

Solution 2 (2008).

During the electrolysis of molten lead bromide. Lead is deposited at cathode.

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Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry

August 11, 2022July 17, 2017 by Veerendra

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 5 Mole Concept and Stoichiometry. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Chemistry Chapter 5 Mole Concept and Stoichiometry

Exercise 5(A)

Solution 1.

(a) Gay-Lussac’s law states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the gaseous product, provided that all the volumes are measured at the same temperature and pressure.

(b) Avogadro’s law states that equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.

Solution 2.

a) The number of atoms in a molecule of an element is called its atomicity. Atomicity of Hydrogen is 2, phosphorus is 4 and sulphur is 8.

b) N₂means 1 molecule of nitrogen and 2N means two atoms of nitrogen.
N₂ can exist independently but 2N cannot exist independently.

Solution 3.

(a) This is due to Avogadros Law which states Equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.
Now volume of hydrogen gas =volume of helium gas
n molecules of hydrogen =n molecules of helium gas
nH₂=nHe
1 mol. of hydrogen has 2 atoms of hydrogen and I molecule of helium has 1 atom of helium
Therefore 2H=He
Therefore atoms in hydrogen is double the atoms of helium.

(b) For a given volume of gas under given temperature and pressure, a change in any one of the variable i.e., pressure or temperature changes the volume.

(c) Inflating a balloon seems violating Boyles law as volume is increasing with increase in pressure. Since the mass of gas is also increasing.

Solution 4.

2H₂+ O₂ → 2H₂O
2 V 1V 2V

From the equation, 2V of hydrogen reacts with 1V of oxygen
so 200cm³ of Hydrogen reacts with = 200/2= 100 cm³Hence, the unreacted oxygen is 150 – 100 = 50cm³of oxygen.

Solution 5.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 1

Solution 6.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 2

Solution 7.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 3

Solution 8.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 4

Solution 9.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 5

Solution 10.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 6

Solution 11.

C₃H₈+ 5O₂ → 3CO₂ + 4H₂O
1 V 5 V 3 V

From equation, 5 V of O₂ required = 1V of propane
so, 100 cm³ of O₂will require = 20 cm³ of propane

Solution 12.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 7

Solution 13.

2CO + O₂ → 2CO₂2 V 1 V 2 V

2 V of CO requires = 1V of O₂so, 100 litres of CO requires = 50 litre of O₂

Solution 14.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 8

Solution 15.

H₂ + Cl₂→ 2HCl
1V 1V 2 V

Since 1 V hydrogen requires 1 V of oxygen and 4cm³ of H₂ remained behind so the mixture had com”>16 cm³ hydrogen and 16 cm³chlorine.
Therefore Resulting mixture is H₂ =4cm³,HCl=32cm³

Solution 16.

CH₄ + 2O₂→ CO₂ + 2H₂O
1 V 2 V 1 V

2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
2 V 5 V 4 V

From the equations, we can see that
1V CH₄ requires oxygen = 2 V O₂So, 10cm³ CH₄ will require =20 cm³ O₂Similarly 2 V C₂H₂ requires = 5 V O₂So, 10 cm³ C₂H₂will require= 25 cm³ O₂Now, 20 V O₂ will be present in 100 V air and 25 V O₂ will be present in 125 V air ,so the volume of air required is 225cm³

Solution 17.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 9

Solution 18.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 10

Solution 19.

This experiment supports Gay lussac’s law of combining volumes.
Since the unchanged or remaining O₂ is 58 cc so, used oxygen 106 – 58 = 48cc
According to Gay lussac’s law, the volumes of gases reacting should be in a simple ratio.

CH₄ + 2O₂ → CO₂ + 2H₂O
1 V 2 V
24 cc 48 cc
i.e. methane and oxygen react in a 1:2 ratio.

Solution 19.

According to Avogadro’s law, equal volumes of gases contain equal no. of molecules under similar conditions of temperature and pressure. This means more volume will contain more molecules and least volume will contain least molecules.
So,
(a) 5 litres of hydrogen has greatest no. of molecules with the maximum volume.
(b) 1 litre of SO₂ contains the least number of molecules since it has the smallest volume.

Solution 20.

Gas	Volume (in litres)	Number of molecules
Chlorine	10	x/2
Nitrogen	20	x
Ammonia	20	X
Sulphur dioxide	5	x/4

Solution 21.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 11

Exercise 5(B)

Solution 1.

a) This statement means one atom of chlorine is 35.5 times heavier than 1/12 time of the mass of an atom C-12.
b) The value of avogadro’s number is 6.023 × 10²³c) The molar volume of a gas at STP is 22.4 dm³at STP

Solution 2.

(a) The vapour density is the ratio between the masses of equal volumes of gas and hydrogen under the conditions of standard temperature and pressure.

(b) Molar volume is the volume occupied by one mole of the gas at STP. It is equal to 22.4 dm³.

(c) The relative atomic mass of an element is the number of times one atom of the element is heavier than 1/12 times of the mass of an atom of carbon-12.

(d) The relative molecular mass of an compound is the number that represents how many times one moleculae of the substance is heavier than 1/12 of the mass of an atom of carbon-12.

(e) The number of atoms present in 12g (gram atomic mass) of C-12 isotope, i.e. 6.023 x10²³ atoms.

(f) The quantity of the element which weighs equal to its gram atomic mass is called one gram atom of that element.

(g) Mole is the amount of a substance containing elementary particles like atoms, molecules or ions in 12 g of carbon-12.

Solution 3.

(a) Applications of Avogadro’s Law :

It explains Gay-Lussac’s law.
It determines atomicity of the gases.
It determines the molecular formula of a gas.
It determines the relation between molecular mass and vapour density.
It gives the relationship between gram molecular mass and gram molecular volume.

(b) According to Avogadro’s law under the same conditions of temperature and pressure, equal volumes of different gases have the same number of molecules.

Since substances react in simple ratio by number of molecules, volumes of the gaseous reactants and products will also bear a simple ratio to one another.This what Gay Lussac’s Law says.

H₂ + Cl₂ → 2HCl
1V 1V 2V (By Gay-Lussacs law)
n molecules n molecules 2n molecules (By Avogadros law)

Solution 4.

(a) (2N)28 + (8H)8 + (Pt)195 + (6Cl)35.5 x 6 = 444
(b) KClO₃ = (K)39 + (Cl)35.5 + (3O)48 = 122.5
(c) (Cu)63.5 + (S)32 + (4O)64 + (5H₂O)5 x 18 = 249.5
(d) (2N)28 + (8H)8 + (S)32 + (4O)64 = 132
(e) (C)12 + (3H)3 + (C)12 + (2O)32 + (Na)23 = 82
(f) (C)12 + (H)1+ (3Cl)3 x 35.5 = 119.5
(g) (2N)28 + (8H)8 + (2Cr)2 x 51.9+ (7O)7 x 16 = 252

Solution 5.

(a) No. of molecules in 73 g HCl = 6.023 x10²³x 73/36.5(mol. mass of HCl)
= 12.04 x 10²³

(b) Weight of 0.5 mole of O₂ is = 32(mol. Mass of O₂) x 0.5=16 g

(d) No. of moles in 10g of CaCO₃ = 10/100(mol. Mass CaCO₃)
= 0.1 mole

(e) Weight of 0.2 mole H₂ gas = 2(Mol. Mass) x 0.2 = 0.4 g

(f) No. of molecules in 3.2 g of SO₂ = 6.023 x 10²³ x 3.2/64
= 3.023 x 10²²

Solution 6.

Molecular mass of H₂O is 18, CO₂ is 44, NH₃ is 17 and CO is 28
So, the weight of 1 mole of CO₂ is more than the other three.

Solution 7.

4g of NH₃ having minimum molecular mass contain maximum molecules.

Solution 8.

a) No. of particles in s1 mole = 6.023 x 10²³So, particles in 0.1 mole = 6.023 x 10 ²³ x 0.1 = 6.023 x 10²²

b) 1 mole of H₂SO₄ contains =2 x 6.023 x 10²³So, 0.1 mole of H₂SO₄ contains =2 x 6.023 x 10²³ x0.1
= 1.2×10²³atoms of hydrogen

c) 111g CaCl₂ contains = 6.023 x 10²³molecules
So, 1000 g contains = 5.42 x 10²⁴molecules

Solution 9.

(a) 1 mole of aluminium has mass = 27 g
So, 0.2 mole of aluminium has mass = 0.2 x 27 = 5.4 g
(b) 0.1 mole of HCl has mass = 0.1 x 36.5 (mass of 1 mole)
= 3.65 g
(c) 0.2 mole of H₂O has mass = 0.2 x 18 = 3.6 g
(d) 0.1 mole of CO₂ has mass = 0.1 x 44 = 4.4 g

Solution 10.

(a) 5.6 litres of gas at STP has mass = 12 g
So, 22.4 litre (molar volume) has mass =12 x 22.4/5.6
= 48g(molar mass)
(b) 1 mole of SO₂ has volume = 22.4 litres
So, 2 moles will have = 22.4 x 2 = 44.8 litre

Solution 11.

(a) 1 mole of CO₂ contains O₂ = 32g
So, CO₂having 8 gm of O₂has no. of moles = 8/32 = 0.25 moles
(b) 16 g of methane has no. of moles = 1
So, 0.80 g of methane has no. of moles = 0.8/16 = 0.05 moles

Solution 12.

(a) 6.023 x 10 ²³atoms of oxygen has mass = 16 g
So, 1 atom has mass = 16/6.023 x 10²³= 2.656 x 10^-23g
(b) 1 atom of Hydrogen has mass = 1/6.023 x 10²³= 1.666 x 10^-24(c) 1 molecule of NH₃ has mass = 17/6.023 x10²³= 2.82 x 10^-23g
(d) 1 atom of silver has mass = 108/6.023 x 10²³=1.701 x 10^-22(e) 1 molecule of O₂ has mass = 32/6.023 x 10²³= 5.314 x 10^-23g
(f) 0.25 gram atom of calcium has mass = 0.25 x 40 = 10g

Solution 13.

(a) 0.1 mole of CaCO₃ has mass =100(molar mass) x 0.1=10 g
(b) 0.1 mole of Na₂SO₄.10H₂O has mass = 322 x 0.1 = 32.2 g
(c) 0.1 mole of CaCl₂ has mass = 111 x 0.1 = 11.1g
(d) 0.1 mole of Mg has mass = 24 x 0.1 = 2.4 g

Solution 14.

1molecule of Na₂CO₃.10H₂O contains oxygen atoms = 13
So, 6.023 x10²³molecules (1mole) has atoms=13 x 6.023 x 10²³So, 0.1 mole will have atoms = 0.1 x 13 x 6.023 x 10²³ =7.8 x 10²³

Solution 15.

3.2 g of S has number of atoms = 6.023 x10²³ x 3.2 /32
= 0.6023 x 10²³So, 0.6023 x 10²³ atoms of Ca has mass=40 x0.6023×10²³/6.023 x 10²³= 4g

Solution 16.

(a) No. of atoms = 52 x 6.023 x10²³ = 3.131 x 10²⁵(b) 4 amu = 1 atom of He
so, 52 amu = 13 atoms of He
(c) 4 g of He has atoms = 6.023 x10²³So, 52 g will have = 6.023 x 10²³ x 52/4 = 7.828 x10²⁴ atoms

Solution 17.

Molecular mass of Na₂CO₃ = 106 g
106 g has 2 x 6.023 x10²³ atoms of Na
So, 5.3g will have = 2 x 6.023 x10²³x 5.3/106=6.022 x10²²atoms
Number of atoms of C = 6.023 x10²³x 5.3/106 = 3.01 x 10²²atoms
And atoms of O = 3 x 6.023 x 10²³x 5.3/106= 9.03 x10²²atoms

Solution 18.

(a) 60 g urea has mass of nitrogen(N₂) = 28 g
So, 5000 g urea will have mass = 28 x 5000/60 = 2.33 kg
(b) 64 g has volume = 22.4 litre
So, 320 g will have volume = 22.4 x 320/64=112 litres

Solution 19.

(a) Vapour density of carbon dioxide is 22, it means that 1 molecule of carbon dioxide is 22 heavier than 1 molecule of hydrogen.
(b) Vapour density of Chlorine atom is 35.5.

Solution 20.

22400 cm³ of CO has mass = 28 g
So, 56 cm³ will have mass = 56 x 28/22400 = 0.07 g

Solution 21.

18 g of water has number of molecules = 6.023 x 10²³So, 0.09 g of water will have no. of molecules = 6.023 x 10²³ x 0.09/18 = 3.01 x 10²¹molecules

Solution 22.

(a) No. of moles in 256 g S₈ = 1 mole
So, no. of moles in 5.12 g = 5.12/256 = 0.02 moles

(b) No. of molecules = 0.02 x 6.023 x 10²³ = 1.2 x 10²²molecules
No. of atoms in 1 molecule of S= 8
So, no. of atoms in 1.2 x 10²²molecules = 1.2 x 10²²x 8
= 9.635x 10²²molecules

Solution 23.

Atomic mass of phosphorus P = 30.97 g
Hence, molar mass of P₄ = 123.88 g
If phosphorus is considered as P₄ molecules,
then 1 mole P₄≡ 123.88 g
Therefore, 100 g of P₄= 0.807 g

Solution 24.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 12

Solution 25.

No. of atoms in 12 g C = 6.023 x10²³So, no. of carbon atoms in 10^-12 g = 10^-12 x 6.023 x10²³/12
= 5.019 x 10¹⁰atoms

Solution 26.

Given:
P= 1140 mm Hg
Density = D = 2.4 g / L
T = 273 ⁰C = 273+273 = 546 K
M = ?

We know that, at STP, the volume of one mole of any gas is 22.4 L
Hence we have to find out the volume of the unknown gas at STP.

First apply Charle’s law.
We have to find out the volume of one liter of unknown gas at standard temperature 273 K.

V₁= 1 L T₁ = 546 K
V₂=? T₂ = 273 K
V₁/T₁ = V₂/ T₂V₂ = (V₁ x T₂)/T₁ = (1 L x 273 K)/546 K
= 0.5 L

We have found out the volume at standard temperature. Now we have to find out the volume at standard pressure.

Apply Boyle’s law.
P ₁ = 1140 mm Hg V₁ = 0.5 L
P₂ = 760 mm Hg V₂ = ?
P₁ x V₁= P₂ x V₂V₂ = (P₁ x V₁)/P₂ = (1140 mm Hg x 0.5 L)/760 mm Hg
= 0.75 L

Now, 22.4 L is the volume of 1 mole of any gas at STP, then 0.75 L is the volume of X moles at STP
X moles = 0.75 L / 22.4 L
= 0.0335 moles
The original mass is 2.4 g
n = m / M
0.0335 moles = 2.4 g / M
M = 2.4 g / 0.0335 moles
M= 71.6 g / mole
Hence, the gram molecular mass of the unknown gas is 71.6 g

Solution 27.

1000 g of sugar costs = Rs. 40
So, 342g(molar mass) of sugar will cost=342×40/1000=Rs. 13.68

Solution 28.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 13
Solution 29.

40 g of NaOH contains 6.023 x 10²³molecules
So, 4 g of NaOH contains = 6.02 x10²³ x 4/40
= 6.02 x10²² molecules

Solution 30.

The number of molecules in 18 g of ammonia= 6.02 x10²³So, no. of molecules in 4.25 g of ammonia = 6.02 x 10²³x 4.25/18
= 1.5 x 10²³

Solution 31.

(a) One mole of chlorine contains 6.023 x 10²³ atoms of chlorine.
(b) Under similar conditions of temperature and pressure, two volumes of hydrogen combined with one volume of oxygen will give two volumes of water vapour.
(c) Relative atomic mass of an element is the number of times one atom of an element is heavier than 1/12 the mass of an atom of carbon-12.
(d) Under similar conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules.

Exercise 5(C)

Solution 1.

Information conveyed by H₂O

That H₂O contains 2 volumes of hydrogen and 1 volume of oxygen.
That ratio by weight of hydrogen and oxygen is 1:8.
That molecular weight of H₂O is 18g.

Solution 2.

The empirical formula is the simplest formula, which gives the simplest ratio in whole numbers of atoms of different elements present in one molecule of the compound.
The molecular formula of a compound denotes the actual number of atoms of different elements present in one molecule of a compound.

Solution 3.

(a) CH (b) CH₂O (c) CH (d) CH₂O

Solution 4.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 14

Solution 5.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 15

Solution 6.

Molecular mass of KClO₃= 122.5 g
% of K = 39 /122.5 = 31.8%
% of Cl = 35.5/122.5 = 28.98%
% of O = 3 x 16/122.5 = 39.18%

Solution 7.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 98

Solution 8.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 17

Solution 9.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 18

Solution 10.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 19

Solution 11.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 20

Solution 12.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 20

Solution 13.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 35

Solution 14.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 21

Solution 15.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 37

Solution 16.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 38

Solution 17.

(a) G atoms of magnesium = 18/24 = 0.75 or g- atom of Mg
(b) G atoms of nitrogen = 7/14 = 0.5 or 1/2 g- atoms of N
(c) Ratio of gram-atoms of N and Mg = 1:1.5 or 2:3
So, the formula is Mg₃N₂

Solution 18.

Barium chloride = BaCl₂.x H₂O
Ba + 2Cl + x[H₂ + O]
= 137+ 235.5 + x [2+16]
= [208 + 18x] contains water = 14.8% water in BaCl₂.x H₂O
= [208 + 18 x] 14.8/100 = 18x
= [104 + 9x] 2148=18000x
= [104+9x] 37=250x
= 3848 + 333x =2250x
1917x =3848
x = 2molecules of water

Solution 19.

Molar mass of urea; CON₂H₄= 60 g
So, % of Nitrogen = 28 × 100/60 = 46.66%

Solution 20.

Element % At. mass Atomic ratio Simple ratio
C 42.1 12 3.5 1
H 6.48 1 6.48 2
O 51.42 16 3.2 1
The empirical formula is CH₂O
Since the compound has 12 atoms of carbon, so the formula is
C₁₂ H₂₄ O_12.

Solution 21.

(a) Now since the empirical formula is equal to vapour density and we know that vapour density is half of the molecular mass i.e. we have n=2 so, molecular formula is A₂B₄.

(b) Since molecular mass is 2 times the vapour density, so Mol. Mass = 2 V.D
Empirical formula weight = V.D/3
So, n = molecular mass/ Empirical formula weight = 6
Hence, the molecular formula is A₆B₆

Solution 22.

Atomic ratio of N = 87.5/14 = 6.25
Atomic ratio of H= 12.5/1 = 12.5
This gives us the simplest ratio as 1:2
So, the molecular formula is NH₂

Solution 23.

Element % at. mass atomic ratio simple ratio
Zn 22.65 65 0.348 1
H 4.88 1 4.88 14
S 11.15 32 0.348 1
O 61.32 16 3.83 11
Empirical formula of the given compound =ZnSH₁₄O₁₁Empiricala formula mass = 65.37+32+141+11+16=287.37
Molecular mass = 287
n = Molecular mass/Empirical formula mass = 287/287=1
Molecular formula = ZnSO₁₁H₁₄= ZnSO₄.7H₂O

Exercise 5(D)

Solution 1.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 39

Solution 2.
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 40

Solution 3.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 41

Solution 4.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 42

Solution 5.

Molecular mass of KNO₃= 101 g
63 g of HNO₃ is formed by = 101 g of KNO₃So, 126000 g of HNO₃ is formed by = 126000 x 101/63 = 202 kg
Similarly,126 g of HNO₃ is formed by 170 kg of NaNO₃So, smaller mass of NaNO₃is required.

Solution 6.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 43

Solution 7.
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 44

Solution 8.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 45

Solution 9.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 46

Solution 10.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 47

Solution 11.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 48

Solution 12.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 49

Miscellaneous Exercise

Solution 1.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 50

Solution 2.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 51

Solution 3.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 52

Solution 4.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 53

Solution 5.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 54

Solution 6.

Molecular mass of urea=12 + 16+2(14+2) =60g
60g of urea contains nitrogen =28g
So, in 50g of urea, nitrogen present =23.33 g
50 kg of urea contains nitrogen=23.33kg

Solution 7.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 55

Solution 8.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 56

Solution 9.

Mass of X in the given compound =24g
Mass of oxygen in the given compound =64g
So total mass of the compound =24+64=88g
% of X in the compound = 24/88 100 = 27.3%
% of oxygen in the compound=64/88 100 =72.7%
Element % At. Mass Atomic ratio Simplest ratio
X 27.3 12 27.3/12=2.27 1
O 72.7 16 72.2/16=4.54 2
So simplest formula = XO₂

Solution 10.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 57

Solution 11.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 58

Solution 12.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 59

Solution 13.

(a) Number of molecules in 100cm³ of oxygen=Y
According to Avogadros law, Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.Therefore ,number of molecules in 100 cm³ of nitrogen under the same conditions of temperature and pressure = Y
So, number of molecules in 50 cm³ of nitrogen under the same conditions of temperature and pressure =Y/100 50=Y/2

(b) (i) Empirical formula is the formula which tells about the simplest ratio of combining capacity of elements present in a compound.
(ii) The empirical formula is CH₃(iii) The empirical formula mass for CH₂O = 30
V.D = 30
Molecular formula mass = V.D 2 = 60
Hence, n =mol. Formula mass/empirical formula mass= 2
So, molecular formula = (CH₂O)₂ = C₂H₄O₂

Solution 14.

The relative atomic mass of Cl = (35 x 3 + 1 x 37)/4=35.5 amu

Solution 15.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 60

Solution 16.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 61

Solution 17.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 62

Solution 18.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 63
So, mass of CO₂ = 22 kg
(b) According to Avogadros law ,equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.
So, number of molecules of carbon dioxide in the cylinder =number of molecules of hydrogen in the cylinder=X

Solution 19.

(a) The volume occupied by 1 mole of chlorine = 22.4 litre
(b) Since PV=constant so, if pressure is doubled; the volume will become half i.e. 11.2 litres.
(c) V₁/V₂ = T₁/T₂22.4/V₂ =273/546
V₂ = 44.8 litres
(d) Mass of 1 mole Cl₂ gas =35.5 x 2 =71 g

Solution 20.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 64

Solution 21.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 65

Solution 22.

(a) The molecular mass of (Mg(NO₃)₂.6H₂O = 256.4 g
% of Oxygen = 12 x 16/256
= 75%

(b) The molecular mass of boron in Na₂B₄O₇.10H₂O = 382 g
% of B = 4 x 11/382 = 11.5%

Solution 23.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 66

Solution 24.

(a) 252 g of solid ammonium dichromate decomposes to give 152 g of solid chromium oxide, so the loss in mass in terms of solid formed = 100 g
Now, if 63 g ammonium dichromate is decomposed, the loss in mass would be = 100 x 63/252 = 25 g

(b) If 252 g of ammonium dichromate produces Cr₂O₃= 152 g
So, 63 g ammonium dichromate will produce = 63 x 152/252
= 38 g

Solution 25.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 67

Solution 26.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 68

Solution 27.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 69

Solution 28.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 70
Since the pressure (760mm) remains constant , but the temperature (273+273)=546 is double, the volume of the steam also gets doubled
So,Volume of steam produced at 760mm Hg and 273⁰C = 4.48 × 2 = 8.96litre

Solution 29.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 71

Solution 30.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 72

Solution 31.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 73

Solution 32.

V₁/V₂ = n₁/n₂So, no. of moles of Cl = x/2 (since V is directly proportional to n)
No. of moles of NH₃ = x
No. of moles of SO₂ = x/4

This is because of Avogadros law which states Equal volumes of all gases, under similar conditions of temperature and pressure, contain equal number of molecules.

So, 20 litre nitrogen contains x molecules
So, 10 litre of chlorine will contain = x × 10/20=x/2 mols.
And 20 litre of ammonia will also contain =x molecules
And 5 litre of sulphur dioxide will contain = x × 5/20 = x/4 mols.

Solution 33.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 74

Solution 34.

(a) Volume of O₂ = V
Since O₂and N₂have same no. of molecules = x
so, the volume of N₂= V
(b) 3x molecules means 3V volume of CO
(c) 32 g oxygen is contained in = 44 g of CO₂So, 8 g oxygen is contained in = 44 x 8/32 = 11 g
(d) Avogadro’s law is used in the above questions.

Solution 35.

(a) 444 g is the molecular formula of (NH₄)₂PtCl₆% of Pt = (195/444) x 100 = 43.91% or 44%

(b) simple ratio of Na = 42.1/23 = 1.83 = 3
simple ratio of P = 18.9/31 = 0.609 = 1
simple ratio of O = 39/16 = 2.43 = 4
So, the empirical formula is Na₃PO₄

Solution 36.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 75

Solution 37.

According to Avogadros law:
Equal volumes of all gases, under similar conditions of temperature and pressure ,contain equal number of molecules.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 76

Solution 38.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 77

Solution 39.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 78

Solution 40.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 79

Solution 41.

(i) D contains the maximum number of molecules because volume is directly proportional to the number of molecules.

(ii) The volume will become double because volume is directly proportional to the no. of molecules at constant temperature and pressure.
V₁/V₂ = n₁/n₂V₁/V₂ = n₁/2n₁So, V₂= 2V₁

(iii) Gay lussac’s law of combining volume is being observed.

(iv) The volume of D = 5.6 4 = 22.4 dm³, so the number of molecules = 6 x 10²³ because according to mole concept 22.4 litre volume at STP has = 6 x 10 ²³ molecules

(v) No. of moles of D = 1 because volume is 22.4 litre
so, mass of N₂O = 1 44 = 44 g

Solution 42.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 80

Solution 43.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 81

Solution 44.

(a) Element % Atomic mass Atomic ratio Simple ratio
K 47.9 39 1.22 2
Be 5.5 9 0.6 1
F 46.6 19 2.45 4
so, empirical formula is K₂BeF₄

(b) 3CuO + 2NH₃ → 3Cu + 3H₂O + N₂3 V 2 V 3 V 1V
3 x 80 g of CuO reacts with = 2 x 22.4 litre of NH₃so, 120 g of CuO will react with = 2x 22.4 x 120/80 x 3
= 22.4 litres

Solution 45.

(a) The molecular mass of ethylene(C₂H₄) is 28 g
No. of moles = 1.4/28 = 0.05 moles
No. of molecules = 6.023 x10²³ x 0.05 = 3 x 10²²molecules
Volume = 22.4 x 0.05 = 1.12 litres

(b) Molecular mass = 2 X V.D
S0, V.D = 28/2 = 14

Solution 46.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 82

Solution 47.
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 83

Solution 48.
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 85

Solution 49.
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 88

Solution 50.
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 90

Solution 51.
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 92

Solution 52.
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 94

Solution 42.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 95

Solution 48.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 97

Solution 49.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 97

Solution 50.

(a) (i) element % atomic mass at. ratio simple ratio
C 14.4 12 1.2 1
H 1.2 1 1.2 1
Cl 84.5 35.5 2.38 2
Empirical formula = CHCl₂(ii) Empirical formula mass = 12+1+71= 84 g
Since molecular mass = 168 so, n = 2
so, molecular formula = (CHCl₂)₂ = C₂H₂Cl₄

(b) (i) C + 2H₂SO₄ → CO₂ + 2H₂O + 2SO₂1 V 2 V 1 V 2 V
196 g of H₂SO₄ is required to oxidized = 12 g C
So, 49 g will be required to oxidise = 49 x 12/196 = 3 g
(ii) 196 g of H₂SO₄occupies volume = 2 x 22.4 litres
So, 49 g H₂SO₄will occupy = 2 x 22.4 x 49/196 = 11.2 litre
i.e. volume of SO₂ = 11.2 litre

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Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide

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Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide

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Selina ICSE Solutions for Class 10 Chemistry Chapter 4 Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide

Exercise 1

Solution 1.

(a) Ferrous salts : Light green
(b) Ammonium salts : Colourless
(c) Cupric salts : Blue
(d) Calcium salts : Colourless
(e) Aluminium salts : Colourless

Solution 2.

(a) Cu(OH)₂(b) ZnO
(c) NaOH
(d) NH₄OH
(e) Na⁺, Ca²⁺(f) Fe²⁺, Mn²⁺(g) Aluminium
(h) Zn(OH)₂ and Al(OH)₃(i) PbO
(j) Ammonium ion

Solution 3.

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry Uses of Ammonium Hydroxide And Sodium Hydroxide img 1

Solution 4.

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 2

Solution 5.

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 3

Solution 6.

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 4

Solution 7.

(a) ZnCl₂(b) Zn(OH)₂

Solution 8.

(a) PbO
(b) ZnO
(c) K₂ZnO₂

Solution 9.

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 5

Solution 10.

When freshly precipitated aluminum hydroxide reacts with caustic soda solution, whitesalt of sodium meta aluminate is obtained.

Solution 11.
Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 7

Solution 12.

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 8
With excess of NaOH, white gelatinous ppt. of Zn (OH)₂ is soluble. So, these two cannot be distinguished by NaOH alone. However white ppt. of Pb(OH)₂is readily soluble in acetic acid also.
Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 9
(i) On addition of NH₄OH to calcium salts no precipitation of Ca(OH)₂occurs even with addition of excess of NH₄OH because the concentration of OH^–ions from ionization of NH₄OH is so low that it cannot precipitate the hydroxide of calcium.
Pb(NO₃)₂+2 NH₄OHPb(OH)₂+2NH₄NO₃
Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 10
Solution 13.

Lead carbonate is dissolved in dilute nitric acid and then ammonium hydroxide is added to it. A white precipitate is formed which is insoluble in excess.
Zinc carbonate is dissolved in dilute nitric acid and then ammonium hydroxide is added to it. A white precipitate is formed which is soluble in excess.

Solution 14.

Reagent bottles A and B can identified by using calcium salts such as Ca(NO₃)₂.

On adding NaOH to Ca (NO₃)₂, Ca (OH)₂ is precipitated as white precipitate which is sparingly soluble in excess of NaOH.
Ca(NO₃)₂+2NaOH → Ca(OH)₂+ 2NaNO₃

Whereas, on addition of NH₄OH to calcium salts, no precipitation of Ca(OH)₂occurs even with addition of excess of NH₄OH because the concentration of OH^–ions from the ionization of NH₄OH is so low that it cannot precipitate the hydroxide of calcium.
So the reagent bottle which gives white precipitate is NaOH and the other is NH₄OH.

Intext Exercise

Solution 1.

(i) Analysis: The determination of chemical components in a given sample is called analysis.
(ii) Qualitative analysis: The analysis which involves the identification of the unknown substances in a given sample is called qualitative analysis.
(iii) Reagent: A reagent is a substance that reacts with another substance.
(iv) Precipitation: It is the process of formation of an insoluble solid when solutions are mixed. The solid thus formed is called precipitate.

Solution 2.

(i) Yellow
(ii) Colourless
(iii) PaleGreen
(iv) Colourless
(v) Colourless

Solution 3.

(i) Fe³⁺(ii) Cu²⁺(iii) Cu⁺²(iv) Mn²⁺

Solution 4.

(i) Ca(OH)₂(ii) Fe(OH)₂ and Cu(OH)₂(iii) Zn(OH)₂ and Pb(OH)₂

Solution 5.

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 11

Solution 6.

NH₄OH and NaOH can be distinguished by using calcium salts.

For example on adding NaOH to Ca(NO₃)₂, Ca(OH)₂ is obtained as white precipitate which is sparingly soluble in excess of NaOH.

Ca(NO₃)₂ + 2NaOH → Ca(OH)₂ + 2NaNO₃

On addition of NH₄OH to calcium salts, no precipitation of Ca(OH)₂ occurs even with the addition of excess of NH₄OH.This is because the concentration of OH^– ions from the ionization of NH₄OH is so low that it cannot precipitate the hydroxide of calcium.

Solution 7.

(i) Fe(OH)₂and Pb(OH)₂(ii) Cu(OH)₂andZn(OH)₂

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Selina Concise Chemistry Class 10 ICSE Solutions Sulphuric Acid

August 11, 2022July 15, 2017 by Veerendra

Selina Concise Chemistry Class 10 ICSE Solutions Sulphuric Acid

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 11 Sulphuric Acid. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Chemistry Chapter 11 Sulphuric Acid

Exercise 1

Solution 1.

(a) Sulphuric acid is called King of Chemicals because there is no other manufactured compound which is used by such a large number of key industries.
(b) Sulphuric acid is referred to as Oil of vitriol as it was obtained as an oily viscous liquid by heating crystals of green vitriol.

Solution 2.

(a) Two balanced equations to obtain SO₂ is:
(i) 4FeS₂ + 11O₂ → 2Fe₂O₃ +8SO₂(ii) S +O₂ → SO₂

(b) The conditions for the oxidation of SO₂ are:
(i) The temperature should be as low as possible. The yield has been found to be maximum at about 410⁰C-450^oC
(ii) High pressure (2 atm) is favoured because the product formed has less volume than reactant.
(iii) Excess of oxygen increases the production of sulphur trioxide.
(iv) Vanadium pentoxide or platinised asbestos is used as catalyst.

(c) Sulphuric acid is not obtained directly by reacting SO₃ with water because the reaction is highly exothermic which produce the fine misty droplets of sulphuric acid that is not directly absorbed by water.

(d) The chemical used to dissolve SO₃ is concentrated sulphuric acid. The product formed is oleum.

(e) Main reactions of this process are:

selina-icse-solutions-class-10-chemistry-sulphuric-acid-2

Solution 3.

Water is not added to concentrated acid since it is an exothermic reaction. If water is added to the acid, there is a sudden increase in temperature and the acid being in bulk tends to spurt out with serious consequences.

Solution 4.

Impurity of ARSENIC poisons the catalyst [i.e. deactivates the catalyst]. So, it must be removed before passing the mixture of SO₂ air through the catalytic chamber.

Solution 5.

Balanced reactions are:

(a) Acidic nature:

(i) Dilute H₂SO₄ reacts with basic oxides to form sulphate and water.
2 NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

(ii) CuO + H₂SO₄→ CuSO₄ + H₂O

(iii) It reacts with carbonate to produce CO₂.
Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂ ↑

(b) Oxidising agent:

H₂SO₄ → H₂O +SO₂ + [O]
Nascent oxygen oxidizes non-metals, metals and inorganic compounds.
For example,
Carbon to carbon dioxide
C+H₂SO₄→ CO₂ +H₂O +2SO₂

Sulphur to sulphur dioxide
S +H₂SO₄→ 3SO₂ +2H₂O

It has great affinity for water. It readily absorbs moisture from atmospheric air.

selina-icse-solutions-class-10-chemistry-sulphuric-acid-5

(d) Non-volatile nature:

It has a high boiling point (356^oC) so it is considered to be non-volatile. Therefore, it is used for preparing volatile acids like hydrochloric acid, nitric acid from their salts by double decomposition reaction.

NaCl + H₂SO₄→ NaHSO₄ + HCl

KCl + H₂SO₄→ KHSO₄ + HCl

Solution 6.

(a) Bring a glass rod dipped in Ammonia solution near the mouth of each test tubes containing dil. Hel and dil. H₂SO₄each. selina-icse-solutions-class-10-chemistry-sulphuric-acid-6

(b)

Dilute sulphuric acid treated with zinc gives Hydrogen gas which bums with pop sound.
Concentrated H₂SO₄ gives SO₂ gas with zinc and the gas turns Acidified potassium dichromate paper green.
Barium chloride solution gives white ppt. with dilute H₂SO₄, This white ppt. is insoluble in all acids.
Concentrated H₂SO₄ and NaCl mixture when heated gives dense white fumes if glass rod dipped in Ammonia solution is brought near it.

Solution 7.

selina-icse-solutions-class-10-chemistry-sulphuric-acid-7

Solution 8.

(a) Concentrated sulphuric acid is hygroscopic substance that absorbs moisture when exposed to air. Hence, it is stored in air tight bottles.

(b) Sulphuric acid is not a drying agent for H₂S because it reacts with H₂S to form sulphur.
H₂SO₄ + H₂S → 2H₂O + SO₂ + S

(c) Concentrated sulphuric acid has high boiling point (356^oC). So, it is considered to be non-volatile. Hence, it is used for preparing volatile acids like Hydrochloric acid and Nitric acids from their salts by double decomposition.
NaCl + H₂SO₄→ NaHSO₄ + HCl
NaNO₃ + H₂SO₄→ NaHSO₄ + HNO₃

Solution 9.

(a) Due to its reducing property. i.e, it is a non-volatile acid.

NaCl + H₂SO₄→ NaHSO₄ + HCl

selina-icse-solutions-class-10-chemistry-sulphuric-acid-9

(c) Magnesium is present above hydrogen in the reactivity series so sulphuric acid is able to liberate hydrogen gas by reacting with magnesium strip.

Mg + H₂SO₄→ MgSO₄ + H₂

(d) Due to its oxidizing character

Cu +H₂SO₄→ CuSO₄ +2H₂O +SO₂

(e) Due to its oxidizing property Hydrogen sulphide gas is passed through concentrated sulphuric acid to liberate sulphur dioxide and sulphur is formed.

H₂S + H₂SO₄→ S + 2H₂O + SO₂

Solution 10.

The name of the salt of
(a) Hydrogen sulphites and Sulphites.
(b) Sulphate and bisulphate.

Solution 11.

(a) Two types of salts are formed when sulphuric acid reacts with NaOH because sulphuric acid is dibasic.

NaOH + H₂SO₄→ NaHSO₄ + H₂O

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

(b) When hydrogen bromide reacts with sulphuric acid the bromine gas is obtained which produce red brown vapours.

2KBr + 3H₂SO₄ → 2KHSO₄ + SO₂ + Br₂ ↑ + 2H₂O

(d) When sulphuric acid is added to sodium carbonate it liberates carbon dioxide which produces brisk effervescence.

Na₂CO₃ + H₂SO₄→ Na₂SO₄+ H₂O + CO₂ ↑

Solution 12.

Column 1 Substance reacted with acid	Column 2 Dilute or concentrated acid	Column 3 Gas
Substance reacted with acid	Dilute or concentrated sulphuric acid	Gas
Zinc	Dilute sulphuric acid	Hydrogen
Calcium carbonate	Concentrated sulphuric acid	Carbon dioxide
Bleaching power CaOCl₂	Dilute sulphuric acid	only chlorine

Solution 1 (2004).

Hydrogen sulphide (H₂S) can be oxidized to sulphur.

Solution 1 (2006).

(a) The process used for the large-scale manufacture of sulphuric acid is Contact process.

(b) Sulphuric acid has great affinity for water. It readily removes element of water from other compound. Thus it acts as a dehydrating agent.

(c) Concentrated acid is non-volatile thus it is used for the preparation of volatile acids:
NaCl + H₂SO₄→ NaHSO₄ + HCl
Concentrated acid act as an oxidizing agent:
C + 2H₂SO₄→ CO₂ + 2H₂O + 2SO₂

Solution 1 (2007).

(i) B
(ii) D
(iii) C
(iv) A
(v) A

Solution 1 (2008).

(a) Concentrated sulphuric acid is non-volatile; hence it is used for the preparation of higher volatile acids.
(b) Due to its dehydrating nature sugar turns black in the presence of concentrated sulphuric acid.

Solution 2 (2004).

When sodium sulphide is added to solution of HCl, Hydrogen sulphide gas is produced. It has rotten egg like smell.

Solution 2 (2007).

(a) The acid formed when sulphur dioxide dissolves in water is sulphurous acid.
(b) Carbondioxide gas is released when sodium carbonate is added to solution of sulphur dioxide.

Solution 3 (2004).

(a) The catalyst which helps in the conversion of sulphur dioxide to sulphur trioxide in step C is Vanadium pentoxide.
(b) The two steps for the conversion of sulphur trioxide to sulphuric acid is:
(i) SO₃+ H₂SO₄→ H₂S₂O₇(ii) H₂S₂O₇ + H₂O → 2H₂SO₄(b) The substance that will liberate sulphur dioxide in step E is dilute H₂SO_4.(c) The equation for the reaction by which sulphur dioxide is converted to sodium sulphite in step F is:
SO₂ + 2NaOH → Na₂SO₃ + H₂O
Or Na₂O + SO₂ → Na₂SO₃

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Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts

August 11, 2022July 15, 2017 by Veerendra

Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 3 Acids, Bases and Salts. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Chemistry Chapter 3 Acids, Bases and Salts

Exercise Intext 1

Solution 1.
(a) Acids are defined as compounds which contain one or more hydrogen atoms, and when dissolved in water, they produce hydronium ions (H₃O⁺), the only positively charged ions.
(b) Hydronium ion
(c) H₃O⁺

Solution 2.
H2SO4 + H₂O ⇌ H₃O⁺ + HSO4-
HSO4- + H₂O ⇌ H₃O⁺ + SO4-2

Solution 3.
If water is added to a concentrated acid, the heat generated causes the mixture to splash out and cause severe burns. Thus, water is never added to acid in order to dilute it.

Solution 4.
Basicity: The basicity of an acid is defined as the number of hydronium ions (H₃O+) that can be produced by the ionization of one molecule of that acid in aqueous solution.
The basicity of following compounds are:
Nitric acid: Basicity= 1
Sulphuric acid: Basicity=2
Phosphoric acid: Basicity=3

Solution 5.
(a) Oxyacids: – HNO₃, H₂SO₄(b) Hydracid:- HCl, HBr
(c) Tribasic acid:- H₃PO₄, H₃PO₃(d) Dibasic acid: – H₂SO₄ , H₂CO₃

Solution 6.
(a) The anhydride of following acids are:

Sulphurous acid: SO₂
Nitric acid: N₂O₅
Phosphoric acid: P₂O₅
Carbonic acid : CO₂

(b) Acids present in following are:
Vinegar: Acetic acid
Grapes: Tartaric acid and Malic acid
Lemon: Citric acid
(c)

H⁺ ion turns blue litmus red.
OH^– ion turns red litmus blue.

Solution 7.
Acetic acid is a monobasic acid which on ionization in water produce one hydronium ion per molecule of the acid.

Solution 8.
2NO₂_(g) + H₂O_(l)→ HNO_2(aq) + HNO₃

Solution 9.
The strength of an acid is the extent to which the acid ionizes or dissociates in water. The strength of an acid depends on the degree of ionization and concentration of hydronium ions [H₃O⁺] produced by that acid in aqueous solution.

Solution 10.
(a) Carbonic acid is a dibasic acid with two replaceable hydrogen ions; therefore it forms one acid salt or one normal salt. Hydrochloric acid is a monobasic acid with one replaceable hydrogen ion and so forms only one normal salt.
(b) Strength of an acid is the measure of concentration of hydronium ions it produces in its aqueous solution. Dil. HCl produces high concentration of hydronium ion compared to that of concentrated acetic acid. Thus, dil. HCl is stronger acid than highly concentrated acetic acid.
(c) H₃PO₃ is not a tribasic acid because in oxyacids of phosphorus, hydrogen atoms which are attached to oxygen atoms are replaceable. Hydrogen atoms directly bonded to phosphorus atoms are not replaceable.
Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts img 1
(d) The salt produced is insoluble in the solution so the reaction does not proceed. Hence, we do not expect lead carbonate to react with hydrochloric acid.
(e) NO₂ is called double acid anhydride because two acids – nitrous acid and nitric acid – are formed when it reacts with water.
2NO₂ + H₂O → HNO₂ + HNO₃

Solution 11.
Acid rain is a by-product of a variety of human activities which release oxides of sulphur and nitrogen in the atmosphere. Burning of fossil fuels, coal, oil, petrol and diesel produces sulphur dioxide and nitrogen oxide which pollute the air. Polluted air also contains many oxidising agents which produce oxygen because of excessive heat. This oxygen combines with the oxides of sulphur and nitrogen and rain water to form acids.
2SO₂ + O₂ + 2H₂O → 2H₂SO₄4NO₂ + O₂ + 2H₂O → 4HNO₃

Solution 12.
(a) Acids are prepared from non-metals by their oxidation. For example :
Sulphur or phosphorus is oxidized by conc. Nitric acid to form sulphuric acid or phosphoric acid.
Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts img 2
(b) Acids are prepared from salt by the displacement reaction. For example :
Nitric acid is prepared by using H₂SO₄and sodium chloride.

Solution 13.
(a) SO₂ +H₂O H₂SO₃(b) P₂O₅ +3H₂O 2H₃PO₄(c) CO₂ + H₂O H₂CO₃

Solution 14.
4(a) Citric acid
(b) Carbonic acid
(c) Oxalic acid
(d) Boric acid

Exercise Intext 2

Solution 1.
An alkali is a basic hydroxide which when dissolved in water produces hydroxyl ions (OH^–) as the only negatively charged ions.
(a) Strong alkalis: Sodium hydroxide , Potassium hydroxide
(b) Weak alkalis: Calcium hydroxide , Ammonium hydroxide

Solution 2.
(a) An alkali and a base:

Alkalis are soluble in water whereas bases may be or may not be soluble in water.
All alkalis are bases but all bases are not alkalis.

(b) The chemical nature of an aqueous solution of HCl and an aqueous solution of NH₃

The aqueoussolution of HCl is acidic in nature. It can turn blue litmus to red.
The aqueoussolution of NH₃is basic in nature. It can turn red litmus to blue.

Solution 3.
(a) Hydroxyl ion (OH^–)
(b) H⁺

Solution 4.
(a) Barium oxide
(b) Sodium hydroxide
(c) Manganese oxide
(d) Cupper hydroxide
(e) Carbonic acid
(f) Ferric hydroxide
(g) Copper oxide
(h) Ammonia
(i) Ammonium hydroxide

Solution 5.
The test tube containing distilled water does not affect the red litmus paper.
The test tube containing acidic solution does not change the red litmus paper.
But the test tube containing basic solution turns red litmus paper blue.

Solution 6.
It is because HCl and HNO₃ ionize in aqueous solution whereas ethanol and glucose do not ionize in aqueous solution.

Solution 7.
(a) DryHCl gas does not contain any hydrogen ions in it, so it does not show acidic behaviour. Hence, dry HCl gas does not change the colour of dry litmus paper.
(b) Lead oxide is a metallic oxide which reacts with hydrochloric acid to produce lead chloride and water, but it is excluded from the class of bases, because chlorine is also produced.
PbO₂ + 4HCl → PbCl₂ + Cl₂ + 2H₂O
Thus, lead oxide is not a base.
(c)Yes, basic solutions have H⁺ions, but the concentration of OH^– ions is more than the H⁺ ions which makes the solution basic in nature.

Solution 8.
(a) We can obtain a base from another base by double decomposition. The aqueous solution of salts with base precipitates the respective metallic hydroxide.
FeCl₃ +3NaOH Fe(OH)₃ +3NaCl
(b) An alkali from a base

(c) Salt from another salt

Solution 9.
(a) Mg +2HCl MgCl₂ + H₂(b) HCl + NaOH NaCl + H₂O
(c) CaCO₃ +2HCl CaCl₂+H₂O + CO₂(d) CaSO₃ + 2HCl CaCl₂ + H₂O+ SO₂(e) ZnS + 2HCl ZnCl₂ + H₂S

Solution 10.
As we know that alkalis react with oil to form soap. As our skin contains oil so when we touch strong alkalis, a reaction takes place and soapy solution is formed. Hence we should wear gloves.

Solution 11.
Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts img 6

Solution 12.
pH represents the strength of acids and alkalis expressed in terms of hydrogen ion concentration. The solution with pH value 10 will give pink colour with phenolphthalein indicator.

Solution 13.
A = Strongly acidic
B= neutral
C=Weakly alkaline
D= Strongly alkaline
E= Weakly acidic
(a) Solution A (acidic solution) + MgH₂ + Mg salt
(b) SolutionA (acidic solution) + ZnH₂ + Zn salt

Solution 14.
(a) A common acid-base indicator and a universal indicator:
An acid-base indicator like litmus tells us only whether a given substance is an acid or a base. The universal indicator gives an idea as to how acidic or basic a substance is universal indicator gives different colours with solutions of different pH values.
(b) The acidity of bases and basicity of acids
The acidity of bases: The number of hydroxyl ions which can be produced per molecule of the base in aqueous solution.
Basicity of acid: The basicity of an acid is defined as the number of hydronium ions that can be produced by the ionization of one molecule of that acid in aqueous solution.
(c) Acid and alkali:
An acid is that substance which gives H⁺ ions when dissolved in water.
An alkali is that substance which gives OH^– ions when dissolved in water.

Solution 15.
Substances like chocolates and sweets are degraded by bacteria present in our mouth. When the pH falls to 5.5 tooth decay starts. Tooth enamel is the hardest substance in our body and it gets corroded. The saliva produced by salivary glands is slightly alkaline, it helps to increase the pH, to some extent, but toothpaste which contains basic substance is used to neutralize excess acid in the mouth.

Solution 16.
A universal indicator is a mixture of dyes which identify a gradual change of various colours over a wide range of pH, depending on the strength of the acid. When we use a universal indicator, we see solutions of different acids produce different colours. Indeed, solutions of the same acid with different concentration give different colours.
The more acidic solutions turn universal indicator bright red. A less acidic solution will only turn it orange-yellow. Colour differences can also be observed in case of vinegar which is less acidic and battery acid which is more acidic.

Solution 17.
(a)

The pH can be increased by adding a basic solution.
The pH can be increased by adding an acidic solution.

(a) The solution is basic in nature and the pH value will be more than 7.
(b)Less than 7

Solution 18.
(a) Solution P
(b) Solution R
(c)Solution Q

Exercise Intext 3

Solution 1.
(a) A normal salt: Normal salts are the salts formed by the complete replacement of the ionizable hydrogen atoms of an acid by a metallic or an ammonium ion.
(b) An acidic salt: Acid salts are formed by the partial replacement of the ionizable hydrogen atoms of a polybasic acid by a metal or an ammonium ion.
(c) A mixed salt: Mixed salts are those salts that contain more than one basic or acid radical.
Examples:
(a) A Normal salt: Na₂SO₄ , NaCl
(b) An acid salt: NaHSO₄ , Na₂HPO₄(c) A mixed salt: NaKCO₃ , CaOCl₂

Solution 2.
(a) Salt is a compound formed by the partial or total replacement of the ionizable hydrogen atoms of an acid by a metallic ion or an ammonium ion.
(b) An insoluble salt can be prepared by precipitation.
(c) A salt prepared by direct combination is Iron(III) chloride.
Reaction:
2Fe +3Cl₂ 2FeCl₃(d) By neutralizing sodium carbonate or sodium hydroxide with dilute sulphuric acid:
Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂

2 NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Solution 3.
(a) Copper sulphate crystals from a mixture of charcoal and black copper oxide:
The carbon in the charcoal reduces the black copper oxide to reddish-brown copper. The lid must not be removed until the crucible is cool or the hot copper will be re-oxidized by air.
Take dilute sulphuric acid in a beaker and heat it on wire gauze. Add cupric oxide in small quantities at a time, with stirring till no more of it dissolves and the excess compound settles to the bottom.
Filter it hot and collect the filtrate in a china dish. Evaporate the filtrate by heating to the point of crystallization and then allow it to cool and collect the crystals of copper sulphate pentahydrate.
Reaction:
CuO + H₂SO₄ CuSO₄ + H₂O
CuSO₄ + 5H₂O CuSO_4. 5H₂O
(b) Zinc sulphate crystals from Zinc dust:
Take dilute sulphuric acid in a beaker and heat it on wire gauze. Add some granulated zinc pieces with constant stirring. Add till the Zinc settles at the base of the beaker. Effervescences take place because of the liberation of hydrogen gas. When effervescence stops, it indicates that all the acid has been used up. The excess of zinc is filtered off. Collect the solution in a china dish and evaporate the solution to get crystals. Filter, wash them with water and dry them between the folds of paper. The white needle crystals are of hydrated Zinc sulphate.
Reaction:
Zn + H₂SO₄ ZnSO₄ + H₂ZnSO₄ +7 H₂O ZnSO_4. 7 H₂O
(c) Lead sulphate from metallic lead:
Metallic lead is converted to lead oxide by oxidation. Then lead sulphate is prepared from insoluble lead oxide, by first converting it into soluble lead nitrate. Then the lead nitrate solution is treated with sulphuric acid to obtain white ppt. of Lead sulphate.
Reaction:
PbO +2HNO₃ Pb(NO₃)₂ + H₂O
Pb(NO₃)₂ + H₂SO₄ PbSO₄ + 2HNO₃(d)Sodium hydrogen carbonate crystals:
Dissolve 5 grams of anhydrous sodium carbonate in about 25 ml of distilled water in a flask. Cool the solution by keeping the flask in a freezing mixture. Pass carbon dioxide gas in the solution. Crystals of sodium bicarbonate will precipitate out after some time. Filter the crystals and dry it in folds of filter paper.
Reaction:
Na₂CO₃ + CO₂ + H₂O 2NaHCO₃

Solution 4.

Anhydrous ferric chloride: -A (Direct combination of two elements)
2Fe + 3Cl₂ 2FeCl₃
Lead chloride: -E (Reaction of two solutions of salts to form a precipitate)
Pb(NO₃)₂ +2HCl PbCl₂ +2HNO₃
Sodium sulphate: – D( Titration of dilute acid with a solution of soluble base)
2NaOH + H₂SO₄ Na₂SO₄+2H₂O
Coppersulphate:– C (reaction of dilute acid with an insoluble base)
Cu(OH)₂ +H₂SO₄ CuSO₄ + 2H₂O

Solution 5.
(a) Lead chloride
(b) Silver chloride
(c) Barium sulphate and lead sulphate
(d) Basic lead chloride
(e) Sodium hydrogen sulphate
(f) Sodium potassium carbonate
(g) Sodium argentocyanide
(h) Potash alum
(i) Potassium bromide and potassium chloride
(j) Calcium sulphate

Solution 6.
An acid is a compound which when dissolved in water forms hydronium ions as the only positively charged ions. A base is a compound which is soluble in water and contains hydroxide ions. A base reacts with an acid to form a salt and water only. This type of reaction is known as neutralisation.

Solution 7.
(a) Blue litmus will turn into red which will indicate the solution to be acidic.
(b) No change will be observed.
(c) Red litmus will turn into blue will indicate the solution to be basic.

Solution 8.
(a) Since sodium hydroxide and sulphuric acid are both soluble, an excess of either of them cannot be removed by filtration. Therefore it is necessary to find out on small scale, the ratio of solutions of the two reactants.
(b) As iron chloride is highly deliquescent, so it is kept dry with the help of fused calcium chloride.
(c) On heating the hydrate, HCl acid is released and basic salt (FeOCl) or ferric oxide remains. Hence, anhydrous ferric chloride
cannot be prepared by heating the hydrate.

Solution 9.
Zinc Sulphate – Displacement

Ferrous sulphide – synthesis

Barium sulphate – Precipitation

Ferric Sulphate- Oxidation

Sodium sulphate – Neutralisation

Solution 10.
(a) pH of pure water is 7 at 25^oC. No, the pH does not change when common salt is added.
(b) Acids: H₂SO₄ and HNO₃Bases: Ammonium hydroxide and sodium hydroxide.
Salts: Barium chloride and sodium chloride.

Solution 11.
Neutralization is the process by which H⁺ ions of an acid react completely with the [OH]^– ions of a base to give salt and water only.
(a)

(b) Neutralization is simply a reaction between H⁺ ions given by strong acid and OH^– ions given by strong base. In case of all
strong acids and strong bases, the number of H⁺ and OH^– ions produced by one mole of a strong acid or strong base is always same. Hence the heat of neutralization of a strong acid with strong base is always same.

Solution 12.
Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts img 12

Solution 13.
(a) Iron (III) Chloride: Iron chloride is formed by direct combination of elements.
Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts img 13

Solution 14.
(a) By neutralisation:
NaOH + HCl → NaCl + H₂O
(b) By precipitation:
Pb(NO₃)₂ + 2NaCl → PbCl₂ + 2NaNO₃(c)CuCO₃ + H₂SO₄→ CuSO₄ + H₂O + CO₂(d) Simple displacement:
Zn + H₂SO₄→ ZnSO₄ + H₂

Solution 15.
(a) Na₂CO₃ + H₂SO₄ (dil) → Na₂SO₄ + H₂O + CO₂(b) CuCO₃ + H₂SO₄ (dil) → CuSO₄ + H₂O + CO₂(c) Fe + H₂SO₄ (dil) → FeSO₄ + H₂(d) Zn + H₂SO₄ (dil) → ZnSO₄ + H₂ZnSO₄ + Na₂CO₃ → ZnCO₃ + Na₂SO₄

Solution 16.
(a) NaHSO₄(b) AgCl
(c) CuSO₄.5H₂O
(d) CuCO₃(d) Pb(NO₃)₂

Solution 17.
(a) acid salt
(b) NaOH+ HCl → NaCl + H₂O

Solution 18.
(a) Alkali
(b) Precipitate
(c) Weak acid

Solution 19.

Copper (II) chloride – B
Iron (II) chloride – A
Iron (III) chloride – C
Lead (II) chloride – E
Sodium chloride – D

Solution 20.
Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts img 14

Exercise Intext 4

Solution 1.
It is the amount of water molecules which enter into loose chemical combination with one molecule of the substance on crystallisation from its aqueous solution.
Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts img 15

Solution 2.
(a) Crystalline hydrated salts which on exposure to the atmosphere lose their water of crystallisation partly or completely and change into a powder. This phenomenon is called efflorescent and the salts are called efflorescent.
Examples: CuSO₄.5H₂O, MgSO₄.7H₂O, Na₂CO₃.10H₂O
(b) Water-soluble salts which on exposure to the atmosphere absorb moisture from the atmosphere and dissolve in the same and change into a solution. The phenomenon is called deliquescence and the salts are called deliquescent.
Examples: CaCl₂, MgCl₂, ZnCl₂

Solution 3.
Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts img 16

Solution 4.
Conc. sulphuric acid is hygroscopic in nature and can remove moisture from other substances; therefore, it is used as a drying agent. It is also used as a dehydrating agent because it has a strong affinity for water and thus absorbs water quickly from compounds.

Solution 5.

blue
red
hydrogen gas
basic, alkaline
graphite

Solution 6.
(a) Sodium hydrogensulphate [NaHSO₄] is an acid salt and is formed by the partial replacement of the replaceable hydrogen ion in a dibasic acid [H₂SO₄]. The [H] atom in NaHSO₄ makes it behave like an acid.
So, on dissolving in water, it gives hydrogen ions.
(b) Desiccating agentsare used to absorb moisture. Anhydrous calcium chloride (CaCl₂) has the capacity of absorbing moisture as it is hygroscopic in nature. So, it is used in a desiccator.

Solution 7.
(a) Increase
(b) Increase
(c) Decrease
(d) Increase
(e) Increase

Solution 8.
(a) Table salt turns moist and ultimately forms a solution on exposure to air especially during the rainy season. Although pure sodium chloride is not deliquescent, the commercial version of the salt contains impurities (such as magnesium chloride) which are deliquescent substances.
(b) The impurity can be removed by passing a current of dry hydrogen chloride gas through a saturated solution of the affected salt. Pure sodium chloride is produced as a precipitate which can be recovered by filtering and washing first with a little water and finally with alcohol.
(c) Conc. sulphuric acid
(d) Common salt and sugar

Solution 9.
(a) Water of crystallization
(b) White
(c) By heating with any dehydrating agent
(d) Anhydrous calcium chloride

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Selina Concise Chemistry Class 10 ICSE Solutions Chemical Bonding

August 11, 2022July 15, 2017 by Veerendra

Selina Concise Chemistry Class 10 ICSE Solutions Chemical Bonding

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 2 Chemical Bonding. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Chemistry Chapter 2 Chemical Bonding

Exercise Intext 1

Solution 1.

Atoms lose, gain or share electrons to attain noble gas configuration.

Solution 2.

(a) A chemical bond may be defined as the force of attraction between any two atoms, in a molecule, to maintain stability.
(b) The chemical bond formed between two atoms by transfer of one or more electrons from the atom of a metallic electropositive element to an atom of a non-metallic electronegative element.
(c) The chemical bond formed due to mutual sharing of electrons between the given pairs of atoms of non-metallic elements.

Solution 3.

Conditions for formation of Ionic bond are:

The atom which changes into cation should possess 1, 2 or 3 valency electrons. The other atom which changes into anion should possess 5, 6 or 7 electrons in the valence shell.
A high difference of electronegativity of the two atoms is necessary for the formation of an Ionic bond.
There must be an overall decrease in energy i.e., energy must be released.
For this an atom should have low value of Ionisation potential and the other atom should have high value of electron affinity.
Higher the lattice energy, greater will be the case of forming an ionic compound.

Solution 4.

It will form a cation: M³⁺M₂(SO₄)₃M(NO₃)₃M₃(PO₄)₃M₂(CO₃)₃M(OH)₃

Solution 5.

Atoms combine with other atoms to attain stable octet or noble gas configuration.

Solution 6.

Ionic compounds are generally formed between metals and non-metals as metals always lose electrons to form cations while non-metals gain electrons forming anions to complete their octet. These oppositely charged ions are held together by electrostatic force of attraction and hence results in an ionic compound.

Solution 7.

Selina Concise Chemistry Class 10 ICSE Solutions Chemical Bonding img 1

Solution 8.

(a) X has 7 electrons in its outermost shell and Y has only one electron in its outermost shell so Y loses its one electron and X gains that electron to form an ionic bond.
(b) The formula of the compound would be XY.

Solution 9.

Selina Concise Chemistry Class 10 ICSE Solutions Chemical Bonding img 2

Solution 10.

(a) Sodium atom and sodium ion

Sodium atom has one electron in M shell while sodium ion has 8 electrons in L shell.
Sodium atom is neutral while sodium ion is positively charged.
Sodium atom is highly reactive while its ion is inert.
Sodium atom is poisonous while sodium ion is non-poisonous.

(b) Chlorine atom and chlorine ion

Chlorine atom has 7 electrons in its M shell while Chloride ion has 8 electrons in the same shell.
Chlorine atom is neutral while chloride ion is negatively charged.
Chlorine atom is highly reactive while its ion is inert.
Chlorine gas is poisonous while chloride ion is non-poisonous.

Solution 11.

Fluoride ion is negatively charged while neon atom is neutral.

Solution 12.

(a) Transfer of electron(s) is involved in the formation of an electrovalent bond. The electropositive atom undergoes oxidation, while the electronegative atom undergoes reduction. This is known as a redox process.

Oxidation: In the electronic concept, oxidation is a process in which an atom or ion loses electron(s).
Zn → Zn²⁺+ 2e^–

Reduction: In the electronic concept, the reduction is a process in which an atom or ion accepts electron(s).
Cu²⁺ + 2e^–→ Cu

(b)

Zn → Zn²⁺+ 2e^– (Oxidation)
Pb²⁺+ 2e^– → Pb (Reduction)
Zn → Zn²⁺+ 2e^– (Oxidation)
Cu²⁺+ 2e^–→ Cu (Reduction)
Cl₂ + 2e^–→ 2Cl^– (Reduction)
2Br^–→ Br₂ + 2e^–(Oxidation)
Sn²⁺→ Sn⁴⁺ + 2e^– (Oxidation)
2Hg²⁺+ 2e^–→ Hg₂(Reduction)
Cu⁺→ Cu²⁺+ e^–(Oxidation)
Cu⁺+ e^–→ Cu (Reduction)

(c)

2K + Cl₂→2KCl

Oxidation: In the electronic concept, oxidation is a process in which an atom or ion loses electron(s).
K → K⁺+ e^–
Reduction: In the electronic concept, the reduction is a process in which an atom or ion accepts electron(s).
Cl₂ + 2e^–→ 2Cl^–
Oxidising agent
An oxidising agent oxidises other substances either by accepting electrons or by providing oxygen or an electronegative ion, or by removing hydrogen or an electropositive ion.
Cl₂ + 2e^–→ 2Cl^–
Reducing agent
A reducing agent reduces other substances either by providing electrons or by providing hydrogen or an electropositive ion, or by removing oxygen or an electronegative ion.
K → K⁺+ e^–

Exercise Intext 2

Solution 1.

(i) Both atoms should have four or more electrons in their outermost shells, i.e., non-metals.
(ii) Both the atoms should have high electronegativity.
(iii) Both the atoms should have high electron affinity and high ionisation potential.
(iv) Electronegativity difference between the two atoms should be zero or negligible.
(v) The approach of the atoms towards one another should be accompanied by decrease of energy.

Solution 2.

(a) A is a non-metal; B is a metal while C is a chemically inert element.
(b) BA

Solution 3.

(a) (i) E (ii) B
(b) C₂D
(c) A and C are metals while B, D and E are non -metals.

Solution 3(2017).

Selina Concise Chemistry Class 10 ICSE Solutions Chemical Bonding img 4

Solution 4.

(a) Ionic compounds are formed as a result of transfer of one or more electrons from the atom of a metallic electropositive element to an atom of a non-metallic electronegative element.
A polar covalent compound is the one in which there is an unequal distribution of electrons between the two atoms.

(b) Ionic compounds, made up of ions, are generally crystalline solids with high melting and boiling points.
They are soluble in water and good conductors of electricity in aqueous solution and molten state.
Covalent compounds, made up of molecules, can exist as soft solids or liquids or gases with low melting and boiling points. They are generally insoluble in water and poor conductors of electricity.

(c) Polar covalent compounds are formed between 2 non-metal atoms that have different electronegativities and therefore have unequal sharing of the bonded electron pair. Non-polar compounds are formed when two identical non-metals equally share electrons between them.

Solution 5.

(a) X⁺(b) X will be a strong reducing agent as it will have the tendency to donate its valence electron.

Solution 6.

Covalent compounds are said to be polar when shared pair of electrons are unequally distributed between the two atoms. For example in HCl, the high electronegativity of the chlorine atom attracts the shared electron pair towards itself. As a result, it develops a slight negative charge and hydrogen atom develops a slight positive charge. Hence, a polar covalent bond is formed.

Solution 7.

During the formation of a non-polar covalent bond between two similar atoms or dissimilar atoms, the atoms involved in sharing share the electrons equally. The molecule of methane has four carbon-hydrogen single covalent bonds. It is a non-polar covalent compound as the electrons are shared by the carbon and hydrogen atoms equally and hence the shared pair lies between the atoms at an equal distance from both carbon and hydrogen atom.

Solution 7.

Selina Concise Chemistry Class 10 ICSE Solutions Chemical Bonding img 6

b. Methane is a covalent compound and is non-polar in nature. This is because the shared pair of electrons is equally distributed between the two atoms. So, no charge separation takes place and the molecule is symmetrical and electrically neutral.

Solution 8.

(a) Properties of Ionic Compounds:

Ionic compounds usually exist in the form of crystalline solids.
Ionic compounds have high melting and boiling points.
Ionic compounds are generally soluble in water but insoluble in organic solvents.
They are good conductors of electricity in the fused or in aqueous solution state.

(b) Properties of Covalent Compounds:

The covalent compounds exist as gases or liquids or soft solids.
The melting and boiling points of covalent compounds are generally low.
Covalent compounds are insoluble in water but dissolve in organic solvents.
They are non-conductors of electricity in solid, molten or aqueous state.

Solution 9.

(a)

A reaction in which oxidation and reduction occur simultaneously is called an oxidation-reduction, or simply, a redox reaction.
Redox reactions involve the transfer of electrons between two chemical species.
The reaction in which electron is gained is called a reduction reaction and the reaction in which electron is lost is called oxidation reaction.
The compound that loses an electron is said to be oxidized, the one that gains an electron is said to be reduced.

Selina Concise Chemistry Class 10 ICSE Solutions Chemical Bonding img 7

(c)
(i) Potassium undergoes oxidation as it loses an electron and forms a cation.
(ii) Chlorine undergoes reduction as it gains an electron and forms chloride anion.
(iii) Potassium acts a reducing agent and gets oxidised.
(iv) Chlorine acts an oxidizing agent and gets reduced.

Solution 9.

Selina Concise Chemistry Class 10 ICSE Solutions Chemical Bonding img 8

Solution 10.

(a) Electrovalent compounds in the solid state do not conduct electricity because movement of ions in the solid state is not possible due to their rigid structure. But these compounds conduct electricity in the molten state. This is possible in the molten state since the electrostatic forces of attraction between the oppositely charged ions become weak. Thus, the ions move freely and conduct electricity.

(b) The atoms of covalent compounds are bound tightly to each other in stable molecules, but the molecules are generally not very strongly attracted to other molecules in the compound. On the other hand, the atoms (ions) in electrovalent compounds show strong attractions to other ions in their vicinity. This generally leads to low melting points for covalent solids, and high melting points for electrovalent solids.

(c) Electrovalent compounds dissolve in polar solvents like water because the forces of attraction between positive and negative charges become weak in water. But since covalent compound are made up of molecules, they do not ionize in water and hence do not dissolve in water.

(d) Since it takes a lot of energy to break the positive and negative charges apart from each other, the ionic compounds are so hard. But on applying stress, Ions of the same charge are brought side-by-side and so the opposite ions repel each other and crystal breaks into pieces.

(e) Since polar covalent compounds are made up of charged particles, they conduct electricity in aqueous solution.

Solution 10.

Dipole molecule is a molecule that has both, slight positive and slight negative charge.
For example, in HCl hydrogen has a slight positive charge and chlorine has a slight negative charge. The dipole moment of HCl molecule is 1.03 D and may be represented as:

Solution 11.

i. Y = 9
ii. Z = 12

b. Ionic bond with molecular formula ZY₂.

Solution 12.

MgCl₂ – Electrovalent compound	CCl₄ – Covalent compound
They are hard crystalline solids consisting of ions.	These are gases or liquids or soft solids.
They have high melting and boiling points.	They have low melting and boiling points.
They conduct electricity in the fused or aqueous state.	They do not conduct electricity in the solid, molten or aqueous state.
These are soluble in inorganic solvents but insoluble in organic solvents.	These are insoluble in water but dissolve in organic solvents.

Solution 13.

Potassium chloride is an electrovalent compound and conducts electricity in the molten or aqueous state because the electrostatic forces of attraction weaken in the fused state or in aqueous solution.

Polar covalent compounds like hydrogen chloride ionise in their solutions and can act as an electrolyte. So, both can conduct electricity in their aqueous solutions.

Solution 14.

a. HCland NH₃

b. HCl + H₂O → H₃O⁺ + Cl^–NH₃ + H₂O →NH₄⁺ + OH^–

Solution 15.

Formula of compound when combined with sulphur – MSFormula of compound when combined with chlorine –MCl₂

Solution 16.

Selina Concise Chemistry Class 10 ICSE Solutions Chemical Bonding img 10
(c) If the compound formed between A and B is melted and an electric current is passed through the molten compound, then element A will be obtained at the cathode and B at the anode of the electrolytic cell.

Exercise 1

Solution 1.

The bond formed between two atoms by sharing a pair of electrons, provided entirely by one of the combining atoms but shared by both is called a coordinate bond. It is represented by an arrow starting from the donor atoms and ending in the acceptor atom.

Conditions:

One of the two atoms must have at least one lone pair of electrons.
Another atom should be short of at least a lone pair of electrons.

The two lone pair of electrons in the oxygen atom of water is used to form coordinate bond with the hydrogen ion which is short of an electron resulting in the formation of the hydronium ion.

H₂O + H⁺ H₃O⁺ Over here the hydrogen ion accepts one lone pair of electrons of the oxygen atom of water molecule leading to the formation of a coordinate covalent bond.

Solution 2.

A pair of electrons which is not shared with any other atom is known as a lone pair of electrons. It is provided to the other atom for the formation of a coordinate bond.

A pair of electrons which is shared between two atoms resulting in the formation of a covalent bond is called a shared pair.

Solution 3.

a. Polar covalent bond
b. Ionic bond
c. O and H are bonded with a single covalentbond and oxygen possesses a single negative charge in the hydroxyl ion.
d. Covalent bond
e. Coordinate bond
f. Electrovalentbond, dative bond (or coordinate bond) and covalent bond

Solution 4.

Solution 5.

Solution 6.

	Sodium	Phosphorus	Carbon
Formula of chloride	NaCl	PCl₅	CCl₄
Nature of bonding	Ionic	Covalent	Covalent
Physical state of chloride	Solid	Solid	Liquid

Solution 7.

a.
CaO- 1 calcium atom + 1 oxygen atom
Cl₂ – 2 chlorine atoms
H₂O – 2 hydrogen atoms + 1 oxygen atom
CCl₄ – 1 carbon atom + 4 chlorine atoms

Ca – will donate two electrons
O – will accept two electrons
Cl – will accept one electron, so two Cl atoms will share an electron pair.
C – will accept four electrons by sharing electrons pairs with hydrogen forming covalent bonds.
H – will donate one electron by sharing an electron pair with carbon.

Solution 8.

(a) Unequal, polar
(b) Middle, equally
(c) Electrovalent, electrostatic

Solution 9.

Y is getting reduced.
Y is positive and it will migrate towards negative electrode that is cathode.

Solution 10.

Selina Concise Chemistry Class 10 ICSE Solutions Chemical Bonding img 12

Solution 1 (2004).

Solution 1 (2005).

(a) (i) C (ii) C (iii) D
(b) (i)reduced (ii) negative
(c) (i) H₃O⁺ions
Selina Concise Chemistry Class 10 ICSE Solutions Chemical Bonding img 14
(ii) Like dissolves like. Since carbon tetrachloride is non-polar and water is polar compound, carbon tetrachloride does not dissolve in water.
(iii) Solid
(iv) No as ionic bonds can only be made by transfer of electrons from a metal to non metal.

Solution 1 (2006).

(a) (i) B (ii) A
(b) (i) Reduction
(ii) Oxidation
(iii) Reduction

Solution 1 (2007).

(i) Ions
(ii) Electrons are shared between the atoms of two or more elements
(iii) Two
(iv) Magnesium is oxidized and chlorine is reduced

Solution 1 (2008).
(a)
(i) D
(b)
(i) Covalent bond
(ii) Coordinate bond.

Solution 1 (2009).
Selina Concise Chemistry Class 10 ICSE Solutions Chemical Bonding img 15

Solution 1 (2010).

a. Oxidation

b.
i. ionic bond
ii. covalent and oordinate bond
iii. covalent bond

Solution 1 (2011).

c. HCl is a covalent compound formed by sharing one electron between chlorine and hydrogen. Because chlorine is more electronegative than hydrogen, the shared pair of electrons shifts towards the chlorine atom. So, a partial negative charge (δ^–) develops on chlorine and a partial positive charge (δ⁺) develops on hydrogen. Hence, the covalent bond is polar in nature.

Solution 1 (2012).

Selina Concise Chemistry Class 10 ICSE Solutions Chemical Bonding img 17

Solution 1 (2013).

a. Dative or coordinate bond
b. B Ammonium chloride
c. C Are insoluble in water
d.

Carbon tetrachloride	Sodium chloride
It is insoluble in water but dissolves in organic solvents.	It is soluble in water but insoluble in organic solvents.
It is a non-conductor of electricity due to the absence of ions.	It does not conduct electricity in the solid state but conducts electricity in the fused or aqueous state.

Solution 1 (2014).

a. B
b. D
c. Ionisation
d. Their constituent particles are molecules. These exist as gases or liquids or soft solids because they have weak forces of attraction between their molecules.

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Selina Concise Chemistry Class 10 ICSE Solutions Periodic Table, Periodic Properties and Variations of Properties

August 11, 2022July 14, 2017 by Veerendra

Selina Concise Chemistry Class 10 ICSE Solutions Periodic Table, Periodic Properties and Variations of Properties

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 1 Periodic Table, Periodic Properties and Variations of Properties. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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ICSE Solutions Selina ICSE Solutions

Selina ICSE Solutions for Class 10 Chemistry Chapter 1 Periodic Table, Periodic Properties and Variations of Properties

Exercise Intext 1

Solution 1.

(a) The modern periodic law states that “The properties of elements are the periodic functions of their atomic number.” Henry Moseley put forward the modern periodic law.
(b) A tabular arrangement of the elements in groups (vertical columns) and periods (horizontal rows) highlighting the regular trends in properties of elements is called a Periodic Table. Modern Periodic table has 7 periods and 18 groups.

Solution 2.

Valency is the combining capacity of the atom of an element. It is equal to the number of electrons an atom can donate or accept or share. It is just a number and does not have a positive or negative sign.
Group 1elements have 1 electron in their outermost orbital, while Group 7 elements have 7 electrons in their outermost orbital.
Valency depends on the number of electrons in the outermost shell (i.e. valence shell).
If the number of electrons present in the outermost shell is 1, then it can donate one electron while combining with other elements to obtain a stable electronic configuration.
If the number of electrons present in the outermost shell is 7, then its valency is again 1 (8 – 7 = 1) as it can accept 1 electron from the combining atom.
In a given period, the number of electrons in the valence (outermost) shell increases from left to right. But the valency increases only up to Group 14, where it becomes 4, and then it decreases, that is, it becomes 1 in Group 17.

Solution 3.

The horizontal rows are known as periods and vertical columns in the periodic table are known as groups.

Solution 4.

Periodicity is observed due to the similar electronic configuration..

Solution 5.

(i) Though the number of shells remain the same, number of valence electrons increases by one, as we move across any given period from left to right.
(ii) While going from top to bottom in a group, the number of shells increases successively i.e. one by one but the number of valence electrons remains the same.

Solution 6.

(a) Elements in the same group have equal valency.
(b) Valency depends upon the number of valence electrons in an atom.
(c) Copper and zinc are transition elements.
(d) Noble gases are placed at the extreme right of the periodic table.

Solution 7.

(a) Sodium and Potassium
(b) Calcium and Magnesium
(c) Chlorine and Bromine
(d) Neon and Argon
(e) Iron and Cobalt
(f) Cerium and Europium
(g) Uranium and Neptunium

Solution 8.

(a) The properties that reappear at regular intervals, or in which there is a gradual variation at regular intervals, are called periodic properties and the phenomenon is known as the periodicity of elements.
(b) The third period elements, Na, Mg, Al, Si, P and Cl summarize the properties of their respective groups and are called typical elements.
(c) The elements of the second period show resemblance in properties with the elements of the next group of the third period leading to a diagonal relationship. Such elements are called bridge elements.

Solution 9.

Beryllium and magnesium will show similar chemical reactions as calcium. Since these elements belong to same group 2 and also have two electrons in their outermost shell like calcium.

Solution 10.

Metals: Lithium, Beryllium, Sodium, Magnesium, Aluminium, Potassium, Calcium
Metalloids: Boron, Silicon
Non-metals: Hydrogen, Helium, Carbon, Nitrogen, Oxygen, Fluorine, Neon, Phosphorous, Sulphur, Chlorine, Argon

Solution 11.

(i) Properties: Non-metallic, highest electronegativity in the respective periods, highest ionisation potentials in the respective periods, highest electron affinity in the respective periods
(ii) Salt-forming; hence, the common name is halogens.

Solution 12.

The main characteristic of the last element in each period of the periodic table is they are inert or chemically unreactive.
The general name of such elements is ‘Noble gases’.

Solution 13.

According to atomic structure, the number of valence electrons determines the first and the last element in a period.

Solution 14.

Elements	Valency	Formula of oxides
Na	1	Na₂O
Mg	2	MgO
Al	3	Al₂O₃
Si	4	SiO₂
P	5	P₂O₅
S	2	SO₂
Cl	1	Cl₂O

Solution 15.

(i) Noble gases
(ii) Representative elements
(iii) Transition elements
(iv) Halogens
(v) Alkaline Earth metals

Solution 16.

(i) 30
(ii) It belongs to group 12 and fourth period.
(iii) It is a metal.
(iv) The name assigned to this group is IIB.

Solution 17.

(i) Electronic configuration of P: 2,8,5
(ii) 15^th Group and 3^rd Period.
(iii) Valency of P = 8 – 5 = 3
(iv) Phosphorus is a non-metal.
(v) It is an oxidizing agent.
(vi) Formula with chlorine = PCl₃

Exercise Intext 2

Solution 1.

Atomic size is the distance between the centre of the nucleus of an atom and its outermost shell.
It’s measured in Angstrom and Picometre.

Solution 2.

(i) The atomic size of an atom increases when we go down a group from top to bottom.
(ii) It increases as we move from right to left in a period.

Solution 3.

Second Period: Fluorine <Neon< Oxygen< Nitrogen < Carbon < Boron< Beryllium < Lithium.
Third Period: Chlorine < Argon < Sulphur < Phosphorus < Silicon < Aluminum < Magnesium < Sodium.

Solution 4.

(i) The size of Neon is bigger compared to fluorine because the outer shell of neon is complete(octet).As a result, the effect of nuclear pull over the valence shell electrons cannot be seen. Hence the size of Neon is greater than fluorine.
(ii) Since atomic number of magnesium is more than sodium but the numbers of shells are same, the nuclear pull is more in case of Mg atom. Hence its size is smaller than sodium.

Solution 5.

(i) An atom is always bigger than cation since cation is formed by the loss of electrons; hence protons are more than electrons in a cation. So the electrons are strongly attracted by the nucleus and are pulled inward.
(ii) An anion is bigger than an atom since it is formed by gain of electrons and so the number of electrons are more than protons. The effective positive charge in the nucleus is less, so less inward pull is experienced. Hence the size expands.
(iii) Fe ²⁺ is bigger than Fe³⁺since Fe ²⁺ has more number of electrons than Fe³⁺and hence the inner pull by nucleus is less strong on it as compared to the pull on Fe³⁺.

Solution 6.

In increasing metallic character: F < O < N < C < B < Be < Li
In decreasing non-metallic character: Cl > S > P > Si > Al > Mg > Na

Solution 7.

(i) Across a period, the chemical reactivity of elements first decreases and then increases.
(ii) Down the group, chemical reactivity increases as the tendency to lose electrons increases down the group.

Solution 7.

The periodic variation in electronic configuration as one move sequentially in increasing order of atomic number produces a periodic variation in properties.
As the elements are arranged in increasing order of atomic number, the metals with tendency to lose electrons are placed on the left and the metallic character decreases from left to right and increases down a group and non-metals with tendency to gain electrons are placed automatically on the right and the non-metallic character increase across a period and decreases down a group.

Solution 8.

(i) The metallic character decreases as we go from left to right in a period.
(ii) It increases as we go down a group.

Solution 9.

(i) The element from the 17^th group has 7 electrons in its outermost shell.
(ii) The name of the element is chlorine.
(iii) Chlorine belongs to the halogen family.
(iv) The element has ₁₃Y²⁷three electrons in its outermost shell which it can donate; hence, its valency is three. While the valency of chlorine is 1. Thus, ₁₃Y²⁷ which is Aluminium can donate three electrons, and chlorine can accept 1 electron to get the stable electronic configuration.
Therefore, the formula of the compound is AlCl₃.

Solution 10.

(i) Yes, these elements belong to the same group but are not from the same period.
(ii) We know that m.p. decreases on going down the group. Hence, from the above table, the elements can be ordered according to their period as follows:

Elements	B	C	A
m.p.	180.0	97.0	63.0

The metallic character increases as one moves down the group.
Hence, the order of the given elements with increasing metallic character is as follows:
B

Solution 10.

The melting and boiling points of metals decrease on going down the group.
Example: Observe the trend in group 1 elements given in the following table:
Selina Concise Chemistry Class 10 ICSE Solutions Periodic Table, Periodic Properties and Variations of Properties img 1

Solution 11.

Correct option: (ii) Potassium

Solution 12.

Correct option: (iii) I^–

Solution 13.

(i) Barium will form ions most readily as the outermost valence electron which experiences the least force of attraction by positively charged nucleus can be given away readily to form cations.
(ii) All Group II elements have two valence electrons.

Solution 14.

Selina Concise Chemistry Class 10 ICSE Solutions Periodic Table, Periodic Properties and Variations of Properties img 2

Solution 15.

Group = 1
Period = 4
Valence electrons = 1
Valency = 1
Metal

Solution 16.

(i) It belongs to group II and has 2 valence electrons, so it is a metal.
(ii) Barium is placed below calcium in the group. Since the reactivity increases below the group, barium is more reactive than calcium.
(iii) It needs to lose 2 valence electrons to complete its octet configuration, so its valency is 2.
(iv) The formula of its phosphate will be Ba₃(PO₄)2.
(v) As we move from left to right in a period, the size decreases, so it will be smaller than caesium.

Solution 17.

Sincethe size of the atom increases down the group, the ionic radii will also increase. Hence, the order of increasing atomic numbers in the group is Z < Y < X.

Solution 18.

(i) All groups do not contain both metals and non-metals. Group I and II contain only metals.
(ii) Atoms of elements in the same group have same number of valence electrons. They have same number of electrons present in their outermost shell.
(iii) The non-metallic character increases across a period with increase in atomic number. This is because across the period, the size of atom decreases and the valence shell electrons are held more tightly.
(iv) On moving from left to right in a period, the reactivity of elements first decreases and then increases, while in groups, chemical reactivity of metals increases going down the group whereas reactivity of non-metals is decreases down the group.

Solution 19.

(i) A metal of valency one = 19
(ii) A solid non-metal of period 3 = 15
(iii) A rare gas = 2
(iv) A gaseous element with valency 2 = 8
(v) An element of group 2 = 4

Solution 20.

(i) The properties of the elements are a periodic function of their atomic number.
(ii) Moving across a period of the periodic table, the elements show increasing non-metallic character.
(iii) The elements at the bottom of a group would be expected to show more metallic character than the elements at the top
(iv) The similarities in the properties of a group of elements are because they have the same number of outer electrons.

Solution 21(i).

An anion is formed by the gain of electrons. In the chloride ion, the number of electrons is more than the number of protons. The effective positive charge in the nucleus is less, so the less inward pull is experienced. Hence, the size expands.
Selina Concise Chemistry Class 10 ICSE Solutions Periodic Table, Periodic Properties and Variations of Properties img 3

Solution 21(ii).

The inert gas argon is the next element after chlorine in the third period.

In a period, the size of an atom decreases from left to right due to an increase in nuclear charge with an increase in the atomic number. However, the size of the atoms of inert gases is bigger than the previous atom of halogen in the respective period. This is because the outer shell of inert gases is complete. They have the maximum number of electrons in their outermost orbit; thus, electronic repulsions are maximum. Hence, the size of the atom of an inert gas is bigger.

Solution 21(iii).

Ionisation potential of the element increases across a period because the atomic size decreases due to an increase in the nuclear charge, and thus, more energy is required to remove the electron(s).

Solution 21(iv).

Alkali metals are strong reducing agents because they lose electrons easily to complete their octet.

Solution 22(i).

Neon (Atomic number = 10)
Electronic configuration = 1s²2s²2p⁶

Solution 22(ii).

Electronic configuration = 2, 8, 3
Hence, atomic number = 13
The element having atomic number 13 is Aluminium.

Solution 22(iii).

The element has a total of three shells; hence, the element belongs to the third period.
Five valence electrons indicate that the element belongs to the fifth group (VA).
Hence, the element is phosphorus.

Solution 22(iv).

The element has a total of four shells; hence, the element belongs to the fourth period.
Two valence electrons indicate that the element belongs to the second group (IIA).
Hence, the element is calcium.

Solution 22(v).

Twice as many electrons in its second shell as in its first shell indicates electronic configuration 1s²2s².
From the electronic configuration, the total number of electrons is 4.
We know that
Number of electrons = Number of protons = Atomic number
The element with atomic number 4 is beryllium.

Solution 23(i).

Period 1:
Number of elements = 2
Hydrogen, helium

Period 2:
Number of elements = 8
Lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, neon

Period 3:
Number of elements = 8
Sodium, magnesium, aluminium, silicon, phosphorus, sulphur, chlorine, argon

Solution 23(ii).

A common feature of the electronic configuration of the elements at the end of Period 2 and Period 3 is that the atoms have 8 electrons in their outermost shell.

Solution 23(iii).

If an element is in Group 17, it is likely to be non-metallic in character, while with one electron in its outermost energy level (shell), then it is likely to be metallic.

Solution 1.

(i) Electron affinity
(ii) Atomic size
(iii) Metallic character
(iv) Non-metallic character
(v) Ionization energy

Exercise Intext 3

Solution 1.

(a) The energy required to remove an electron from a neutral isolated gaseous atom and convert it into a positively charged gaseous ion is called Ionization energy or ionization potential.

(b) M(g)+ I.E → M⁺(g)+ e^–M can be any element
It is measured in electron volts per atom. Its S.I unit kJmol^-1.

Solution 2.

Ionisation potential values depend on

Atomic size: The greater the atomic size, the lesser the force of attraction. Electrons of the outermost shell lie further away from the nucleus, so their removal is easier and the ionisation energy required is less.
Nuclear charge: The greater the nuclear charge, greater is the attraction for the electrons of the outermost shell. Therefore, the electrons in the outermost shell are more firmly held because of which greater energy is required to remove them.

Solution 3.

(a) Ionization energy increases as we move from left to right across a period as the atomic size decreases.
(b) Ionization energy decreases down a group as the atomic size increases.

Solution 4.

Helium has the highest ionization energy of all the elements while cesium has the lowest ionization energy.

Solution 5.

Second period: Lithium

Second period: Neon > Fluorine > Oxygen > Nitrogen > Carbon > Boron > Beryllium > Lithium
Third Period: Argon> Chlorine > Sulphur > Phosphorus > Silicon > Aluminum >Magnesium > Sodium

Solution 6(2010).

(a) Electron affinity is the energy released when a neutral gaseous atom acquires an electron to form an anion.
(b) Second period: Lithium
Second period: Lithium<Boron < Carbon < Oxygen < Fluorine
Neon, Nitrogen and Beryllium do not follow the trend.

Solution 7.

Electron affinity values generally increase across the period left to right and decrease down the group top to bottom.

Solution 8(a).

Electronegativity is the tendency of an atom in a molecule to attract the shared pair of electrons towards itself.
Electronegativity is a dimensionless property; hence, it has no unit.

Solution 8(b).

Correct option – (i).
The element with least electronegativity is lithium.

Solution 9.

(a) On moving across a period, nuclear pull increases because of the increase in atomic number, and thus, the atomic size decreases. Hence, elements cannot lose electrons easily. Hence, Group 17 elements are strong non-metals, while Group 1 elements are strong metals.

(b) On moving across a period, nuclear pull increases because of the increase in atomic number, and thus, the atomic size decreases. Hence, elements cannot lose electrons easily. Hence, Group 17 elements are strong non-metals, while Group 1 elements are strong metals. Down a group, the atomic size increases and the nuclear charge also increases. The effect of an increased atomic size is greater as compared to the increased nuclear charge. Therefore, metallic nature increases as one moves down a group, i.e. they can lose electrons easily.

(c) The atomic size of halogens is very small. The smaller the atomic size, the greater the electron affinity, because the effective attractive force between the nucleus and the valence electrons is greater in smaller atoms, and so the electrons are held firmly.

(d) The reducing property depends on the ionisation potential and electron affinity of the elements. In a period, from left to right in a horizontal row of the periodic table, the atomic size decreases and the nuclear charge increases, so the electron affinity and ionisation energy both increase. Hence, the tendency to lose electrons decreases across the period from left to right and thus the reducing property also decreases across the period from left to right.

The electron affinity and ionisation potential decreases along the group from top to bottom. Hence, the tendency to lose electrons increases, and thus, the reducing property also increases along the group from top to bottom.

(e) In a period, the size of an atom decreases from left to right. This is because the nuclear charge, i.e. the atomic number increases from left to right in the same period, thereby bringing the outermost shell closer to the nucleus. Therefore, considering the third period given above, it has been found that sodium is the largest in size, while chlorine is the smallest.

Solution 10.

(i) Ionization energy
(ii) Metallic character
(iii) Electronegativity

Solution 11.

(a) G (due to the smallest atomic size).
(b) G and O as both have outermost electronic configuration np⁵.
(c) A and I as both have outermost electronic configuration ns¹.
(d) D (1s²2s²2p²)
(e) I as alkali metals have least ionisation energy. Also, ionisation energy decreases with an increase in the atomic size that decreases on moving down the group.
(f) O, as halogens have the least atomic size.

Solution 12.

(a) Thallium. Because the metallic character increases down the group, thallium will have the most metallic character.
(b) Boron. Electronegativity decreases down the group as the size increases; hence, boron will be the most electronegative atom.
(c) Three. The number of electrons present in the valence shell is the same for each group. Hence, all these elements and thallium will have 3 valence electrons.
(d) BCl₃(e) Since metallic character decreases from left to right and non-metallic character increases from left to right, elements in the group to the right of this boron group will be less metallic in character.

Solution 9.

(a) As we move from left to right the increase in atomic number and decrease in size results in a greater nuclear pull. As a result, the ability to attract the electrons increases, and so does the electron affinity.
But noble gases have complete stable octet configuration, hence their electron affinity is lower than halogens.
Hence halogens on extreme right have highest electron affinity in a period.

(b) Chlorine is smaller than sulphur with a bigger atomic number. Since its nuclear pull is more, hence its electron affinity is higher than sulphur.

Solution 10.

Since size of chlorine is bigger than fluorine hence the electrons being farther away from the nucleus experience a lesser force of attraction, hence electron negativity of chlorine is less than fluorine.

Solution 11.

Electronegativity measures an atom’s tendency to attract shared pair of electrons towards itself.
Its S.I unit is Pauling unit.

Solution 12.

(i) The element fluorine has the highest electronegativity and Caesium has the lowest electronegativity.
(ii) The nature of oxides changes from basic to acidic as we move from left to right in third period. Hence sodium forms most basic oxide while oxide of Aluminum is amphoteric and oxides of phosphorus, sulphur and chlorine are progressively acidic.

Exercise 1

Solution a(2015)

(i) Lithium
Reason: Electronegativity increases from left to right. Lithium is present on the left side of the periodic table; hence, it will be the least electronegative element.

Solution b(2015)

(i) Ba metal will form ions readily because the ionisationenergy decreases down the group as the size increases.
(ii) On moving down the group, the number of electrons in the outermost shell, i.e.valence electrons remain the same. So, the valency in a group remains the same, i.e. 2.

Solution a(2009)

Correct option is A. Lithium
In a period from left to right, electron affinity decreases as the non-metallic character increases.

Solution a(2009)

Correct option is A. Lithium

In a period from left to right, electron affinity decreases as the non-metallic character increases.

Solution b(2009)

The most electronegative is J.
Valence electrons present in G are 5.
B contains 1 valence electron and H contains 6 valence electrons. So, the valency of B is +1 and the valency of H is – 2.
In the compound between F and J, the type of bond formed will be covalent.
The electron dot structure for the compound formed between C and K is

Solution a(2010)

The number of electrons in the valence shell of a halogen is 7.
Correct option: D

Solution b(2010)

Electronegativity across the period increases.

Solution c(2010)

Non-metallic character down the group decreases.

Solution d(2010)

Atomic number of an element is 16.

It belongs to Period 3.
The number of valence electrons in the element is 6.
The element is a non-metal.

Solution a(2011)

The oxidising power of elements depends on the tendency to gain electrons which increases from left to right along a period due to increase in nuclear pull.

Solution (b)2011

Across a period, the ionisation potential increases.
Down the group, electron affinity decreases.

Solution c(i)(2011)

In the periodic table, alkali metals are placed in Group I. So, the correct option is A.

Solution c(ii)(2011)

The correct option is C.

The elements of halogen family are non-metallic in nature.

Solution (d)2011

Three shells indicate that the element belongs to the third period.

Three valence electrons indicate that the element belongs to the third group.

Solution a(2012)

Correct option: (D) Argon

Solution b(2012)

Because the atomic radius decreases across a period. Due to this, attraction between the nucleus and the electron increases. This results in an increase in the ionisation potential.
Alkali metals are good reducing agents because they have a greater tendency to lose electrons.

Solution c(2012)

Electronic configuration of E with atomic number 19 = 1s²2s²2p⁶3s²3p⁶4s¹E is a metal.

Electronic configuration of F with atomic number 8 =
1s²2s²2p⁴F is a non-metal.

Electronic configuration of G with atomic number 17 =
1s²2s²2p⁶3s²3p⁵G is a non-metal.

Solution d(2012)

A metal present in Period 3, Group I of the periodic table is sodium.

Solution a(2013)

Correct option: (D) Fluorine

Solution b(2013)

I
R
M
5
T
Y
Ionic bond will be formed and the molecular formula is A₂H.

Solution c(2013)

The element which has the highest ionisation potential is helium (He).

Solution a (2014)

Correct option: D (atomic radius decreases and nuclear charge increases)
Correct option: A (3 shells and 2 valence electrons)

Solution b (2014)

(a) An element Z having atomic number 16 is Sulphur.

(i) Sulphur belongs to Period 3 and Group 16.
(ii) Sulphur is a non-metal.

Solution e (2014)

Electron affinity

Solution f (2014)

A: (ii)
B: (i)

Solution d (2014)

Ionic bond exists between M and O.
1 electron is present in the outermost shell of M.
M belongs to Group 1 in the modern periodic table.

Solution 1(2008)

Solution 2

A covalent oxide of a metalloid. – SiO₂ (Si is a metalloid)
An oxide which when dissolved in water forms acid. – SO₂ (SO₂ + H₂O → H₂SO₃)
A basic oxide. – Na₂O (Na₂O + H₂O → 2NaOH)
An amphoteric oxide. – Al₂O₃ (shows both acidic and basic properties)

Solution 3 (2015)

a. Mg,Cl, Na, S, Si (increasing order of atomic size) –
Cl < S < Si < Mg < Na
99 pm < 104 pm < 117 pm < 160 pm < 186 pm

b. Cs, Na, Li, K, Rb (increasing metallic character)
Li < Na < K < Rb < Cs

c. Na, K,Cl, S, Si (increasing ionisation potential) –
Cl < S < Si < Na < K
1256 < 999 < 786 < 496 < 419

d. Cl, F, Br, I (increasing electron affinity) –
I < Br < F < Cl
-295 KJ mol^-1 < -324 KJ mol^-1 < -327.9 KJ mol^-1 < -349 KJ mol^-1

e. Cs, Na, Li, K,Rb (decreasing electronegativity) –
Li > Na > K = Rb > Cs
1.0 > 0.9 > 0.8 = 0.8 > 0.7

Solution 4

(a) An element with atomic number 9 and 35
(b) An element with atomic number 9.

Solution 5

The ionisation energy is the minimum energy required to remove the outermost electron from a gaseous neutral atom to form a cation.
Position in a group: X will be above Y ( because of ionisation energy decreases down the group )
Position in a period: X will be the right side of Y ( because ionisation energy increases from left to right)

Solution 6

(a) Cl < Cl¯
(b) Mg²⁺ < Mg⁺ < Mg
(c) O < N < P

Solution 7

(a) Cl
Metals have low ionisation energy and non-metals have high ionisation energy. Also, across the period, ionisationenergy tends to increase. The elements P, Na and Cl belong to the third period. Na – Group 1, P – Group 15 and Cl – Group 17.

(b) Ne
Inert gases have zero electron affinity because of their stable electronic configuration.

(c) He
Ionisation energy decreases with an increase in the atomic size, i.e. it decreases as one moves down a group. Ne, He and Ar are inert gases. He – Period 1, Ne – Period 2 and Ar – Period 3.

Solution 8

Na, Mg, Al, Si, P, S, Cl

Solution 9 (2016)

(a) The element below sodium in the same group would be expected to have a lower electro-negativity than sodium, and the element above chlorine would be expected to have a higher ionisation potential than chlorine.
(b) On moving from left to right in a given period, the number of shells remains the same.
(c) On moving down a group, the number of valence electrons remains the same.
(d) Metals are good reducing agents because they are electron donors.

Solution 10

(a) Increases
(b) Increases
(c) Increases
(d) Decreases
(e) Increases

Solution 11

(a) Period 2
(b) Nitrogen (N), between carbon and oxygen
(c) Be< N< F
(d) Fluorine

Solution 1

(a) Na and F
(b) Argon
(c) C, N, O and F are non-metals present in period 2 while Na, Mg and Al are metals in period 3.
(d) Silicon
(e) Argon
(f) Mg
(g) Fluorine
(h) K

Solution 1.

(a) The total number of electron shells in an atom determines the period to which the element belongs, and the valence electrons determine the group to which it will belong. So with the help of electronic configuration we can figure out the period and group number of an element.

Elements with one and two valence electrons belong to group 1 and 2 respectively, while to determine the group number of elements with 3 to 8 valence electrons, we add 10 to their valence electrons.

For example an element X has atomic number 15
Its configuration will be:
K shell has 2 electrons, L will have 8, and the remaining 5 will be placed in M shell
Since it has three shells it belongs to period 3 and with 5 valence electrons the element will be placed in five plus ten, that is the 15^th group
So with the help of electronic configuration we can figure out the period and group number of an element.

(b) Atomic number = Number of protons
Hence, number of protons in K atom = 19
Number of neutrons = Mass number – Atomic number
Hence, number of neutrons in K atom = 39-19 = 20
Number of electrons = Number of protons
Hence, number of electrons= 19
And electronic configuration of K atom = 2, 8, 8, 1
Since K atom has 4 shells, hence it belongs to fourth period.
With one valence electron, it belongs to group 1
Number of protons in P atom = 15
Number of neutrons in P atom = 31-15 = 16
Number of electrons in P atom = 15
And electronic configuration of P atom = 2, 8, 5
Since it has three shells, it belongs to period 3 and with 5 valence electrons Phosphorus is found in five plus ten that is 15^th group.

Solution 2.

(a) Fluorine, chlorine and bromine are non-metals with seven valence electrons. They are highly electronegative elements with valency of one. They exist as diatomic molecules. They form ionic compounds with alkali metals.
(b) They are known as halogens. The term means salt forming and therefore compounds containing these elements are called salts.

Solution 3.

The last element in each period of the periodic table is a gaseous element with its valence shell completely filled. Except for helium with complete duplet configuration, rest all the 5 gases have complete octet configuration.
These group 18 elements are commonly referred to as noble gases.

Solution 4.

The electronic configuration of an element determines its position in Modern Periodic table.
The element with one valence electron is the first while the element with 8 valence electrons is placed in the 18th group of a period.

Solution 5.

(i) The number of valence electrons increases by one as we move from left to right in a period.
The group number 1 and 2 have 1 and 2 valence electrons respectively while group 13 to 18 have group number minus 10 = valence electrons. So,group 13 to 18 have 3, 4, 5, 6, 7 and 8 valence electrons respectively.

(ii) Valency is determined by the number of valence electrons. For elements belonging to group 1, 2 and 13, the valency is equal to the number of valence electrons, so their valency is 1, 2 and 3 respectively.
Since the elements in group 14 to 17 needs to gain electrons to complete their octet configuration. Their valency is 8 minus the number of valence electrons. So their valencies are 4, 3, 2 and 1 respectively.

Solution 6.

(a) Periods
(b) Increases
(c) Decreases

Solution 7.

(a) Since it belongs to group II, it has 2 valence electrons and hence it is a metal.
(b) Barium is placed below calcium in the group. Since, it has more number of shells; it is easier for it to lose its valence electrons to complete its octet configuration. Hence it is more reactive than calcium.
(c) It needs to lose its 2 valence electrons to complete its octet configuration; therefore its valency is also 2.
(d) The formula of its phosphate will be (Ba)₃(PO₄)₂(e) As we move from left to right in a period, the size decreases, therefore, it will be smaller than Cesium.

Solution 8.

(a) The number of valence electrons increases by one as we move across any given period.
Therefore as we move from Lithium to Neon in period 2, the valence electrons will increase from 1 to 7.

(b) The metallic character decreases as we move from left to right while the non-metallic character increases.
On going from left to right in a period, the chemical reactivity of elements first decreases and then increases.
For example in period 3, Sodium is the most reactive metal and Chlorine is the most reactive non-metal and Silicon is least reactive.

(c) The oxides of metals are basic and that of non-metals are acidic in general. Therefore since metallic strength decreases and non-metallic strength increases on moving from left to right across a period, the strength of basic oxides decreases, while the strength of acidic oxides increases.
For example, sodium forms a basic oxide, while sulphur and phosphorus form acidic oxides.

Solution 9.

(a) Noble gases- H and P
(b) Halogens- G and O
(c) Alkali metals – A and I
(d) D and L have valency of 4
(e) I with atomic number 11.
(f) Cl has the least atomic size in period 3 with atomic number 17.

Solution 10.

As we move down a group, the numbers of shells increases and hence the atomic size increases.
Therefore, Z will have the smallest atomic number followed by Y, while X will have the largest atomic number.
So the elements in order of increasing atomic number will be Z<Y<X.

Solution 11.

(a) Since, the distance of the valence electrons from the nucleus keeps on increasing down the group, therefore, the ionization energy keeps on decreasing. Hence the reactivity of alkali metals increases from lithium to francium.

(b) As we move down a group, the size keeps on increasing, so it becomes more difficult for atoms to attract electrons. Thus reactivity of halogens decreases from Fluorine to Astatine.

Solution 12.

(a) Since it belongs to period 3 it has 3 shells, K, L and M. The outermost M shell will have 2 valence electrons as it is placed in group II.
(b) With 2 valence electrons, its valency will be 2.
(c) Since it has electronic configuration of 2, 8, 2, its atomic number is 12 and hence X is Magnesium.
(d) It is a metal.

Solution 13.

(a) Group 1since the valence electrons is 1.
(b) With 4 shells T belong to period 4.
(c) Number of electrons = 2+8+8+1=19
(d) T needs to lose one electron to complete its octet hence its valency is 1.
(e) Since it has one valence electron, it is a metal.

Solution 14.

(a) Group 1: Lithium< Sodium< Potassium< Rubidium < Caesium< Francium
Group 17: Fluorine < Chlorine < Bromine< Iodine < Astatine

(b) Group 1: Francium
Group 17: Astatine< Iodine< Bromine< Chlorine< Fluorine

(c) Group 1: Francium< Cesium< Rubidium< Potassium< Sodium< Lithium
Group 17: Astatine< Iodine< Bromine< Chlorine< Fluorine

(d) Group 1: Francium
Group 17: Astatine

(e) Group 1: Lithium>Sodium> Potassium> Rubidium> Cesium> Francium
Group 17: Fluorine > Chlorine> Bromine > Iodine > Astatine

Solution 15.

(a) atomic number
(b) period, non-metallic character
(c) more
(d) number of outer electrons

Solution 16.

(a) Anion is formed by the gain of electrons. Thus the numbers of electrons are more than protons. The effective positive charge in the nucleus is less, so less inward pull is experienced. Hence the size expands. So the size of an atom is greater than the size of parent atom.

(b) Since Argon has stable octet configuration, so due to the inter- electronic repulsions the effect of nuclear pull over the valence shell electrons cannot be seen which results in the bigger size.

(c) Since size of Bromine is bigger than chlorine, so it becomes more difficult for Br atoms to attract electrons. Thus, Cl is more reactive than Br.

Solution 17.

(a) Neon
(b) Aluminum
(c) Phosphorus
(d) Calcium
(e) Carbon

Solution 18.

(a) Na and F
(b) Argon
(c) C, N, O and F are non-metals present in period 2 while Na, Mg and Al are metals in period 3.
(d) Silicon
(e) Argon
(f) Mg
(g) Fluorine
(h) K

Solution 19.

(a) Element with atomic number 9 and 35
(b) Element with atomic number 9.

Solution 20.

(a) Period 1 has 2 elements while period 2 and period 3 have 8 elements each.
(b) Hydrogen and helium
(c) The elements at the end of period 2 and Period 3 have 8 electrons in its outermost shell.
(d) Non metallic, metallic.

Solution 21.

Position in a group: X and Y
Position in a period: Y and X

Solution 22.

Period no. = no. of shells, so n=3
From the formula M₂O₃its valency is 3.
Since it is a metal, its valence shell has 3 electrons.
So its electronic configuration is 2,8,3
Atomic number=13
Hence the metal is Aluminum with valency 3.

Solution 23.

(a) Since the elements in a group have same number of valence electrons, they can either contain metals or non-metals like alkali and alkaline metals have only metals whereas halogens are non-metals.

(b) No two elements have the same number of electrons instead atoms of the same elements in the same group have the same number of valence electrons.

(c) Non-metals have the tendency to gain electrons to attain stable configuration and therefore are said to be electronegative. As we move from left to right the increase in atomic number and decrease in size results in a greater nuclear pull. As a result the non-metallic character increases across a period.

(d) On moving from left to right in a period, the reactivity first decreases and then increases since the tendency to lose electrons first decreases on going from left to right and then from P to Cl, tendency to gain electrons increases, so reactivity increases then. In case of a group, reactivity increases on going down since the tendency to lose electrons increases but for non-metals, reactivity decreases on going down the group as the tendency to gain electrons decreases down the group.

Solution 24.

(a) Cl < Cl¯
(b) Mg²⁺ < Mg⁺ < Mg
(c) O < N < P

Solution 25.

(a) Cl
Metals have low ionisation energy and non-metals have high ionisation energy. Also, across the period, ionisation energy tends to increase. The elements P, Na and Cl belong to the third period. Na – Group 1, P – Group 15 and Cl – Group 17.

(b) Ne
Inert gases have zero electron affinity because of their stable electronic configuration.

Solution 26.

(a) (iv) Argon
(b) (iii) Calcium
(c) (iii) Helium

Solution 1 (2003).

(a) (Al)₂(SO₄)₃(b) Covalent bonding
(c) Same number of valence electrons
(d) Helium
(e) 8
(f) Electron affinity
(g) Decreases, atomic number, number of shells

Solution 1 (2004).

(a) Na, Mg, Al, Si, P, S, Cl
(b)
(i) Lower, higher
(ii) remains the same
(iii) remains the same

Solution 1 (2005).

(a) Increases
(b) Increases
(c) Increases
(d) Decreases
(e) Increases

Solution 1 (2006).

(a) Period 2
(b) Nitrogen (N), between carbon and oxygen
(c) Carbon
(d) Be < N < F
(e) Fluorine

Solution 1 (2007).

(a) Thallium has the most metallic character since metallic character increases down the group.
(b) Boron has the highest electronegativity since it has the smallest size in the group.
(c) 3. Since all the elements in a group have same number of valence electrons.
(d) BCl₃(e) The elements in the group to the right of boron group would be less metallic as with the decrease in size and increase in atomic number, it will be more difficult for them to lose electrons.

Solution 1 (2008).

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