NCERT Solutions for Class 11 Chemistry Chapter 9

Q.1 Justify the position of the hydrogen in the periodic table on the basis of its electronic configuration.

Ans.

The electronic configuration of hydrogen is 1s¹. On the basis of its electronic configuration it is placed in the alkali metal group in the periodic table. Alkali metals have only one electron in the outermost shell (ns¹) and their position in the periodic table is group 1 or 1A. So on the basis of electronic configuration, hydrogen is placed in group 1or 1A in the periodic table.

Q.2 Write the name of the isotopes of hydrogen. What is the mass ratio of these isotopes?

Ans.

Hydrogen exists in the form of three isotopes.

1. Protium or ordinary hydrogen
   \left({}_{\text{1}}^{\text{1}}\text{H}\right)
2. Deuterium or heavy hydrogen
   \left({}_{\text{1}}^{\text{2}}\text{H}\right)
3. Tritium
   \left({}_{\text{1}}^{\text{3}}\text{H}\right)

The mass ratio of these three isotopes is as follows:

H: D: T = 1: 2: 3

Q.3 Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions?

Ans.

The electronic configuration of hydrogen atom is 1s¹.To get a stable inert gas configuration of helium, hydrogen requires one extra electron. Thus it shares its one electron with another hydrogen atom to form a stable diatomic molecule. So, hydrogen exists in diatomic rather than monoatomic form in normal conditions.

Q.4 How can the production of dihydrogen, obtained from ‘coal gasification’, be increased?

Ans.

The production of syngas (mixture of CO and H₂) from coal is called coal gasification.

\text{C}\left(\text{s}\right)\text{+}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\stackrel{\text{1270K}}{\to }\text{CO}\left(\text{g}\right)\text{+}{\text{H}}_{\text{2}}\left(\text{g}\right)

The production of the hydrogen gas can be increased if CO in the gas mixture can be removed from the medium. For this purpose, CO is reacted with steam in the presence of iron chromate catalyst.

${\displaystyle \mathrm{C}\mathrm{O}(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}(\mathrm{g})\underset{\mathrm{F}\mathrm{e}\mathrm{C}\mathrm{r}{\mathrm{O}}_4}{\overset{673\mathrm{K}}{\to }}\mathrm{C}{\mathrm{O}}_2(\mathrm{g})+{\mathrm{H}}_2(\mathrm{g})}$

The CO₂ can be removed by scrubbing with a solution of sodium arsenite leaving behind H₂ which can be collected.

Q.5 Describe the bulk preparation of dihydrogen by electrolytic method. What is the role of an electrolyte in this process?

Ans.

Electrolysis of water produces dihydrogen. A small quantity of acid or base is added to water to make it conductive.
In the cell, iron sheet is used as cathode and nickel plated iron sheet as anode. These are separated from each other by an asbestos diaphragm.
On passing electric current, dihydrogen is collected at cathode while dioxygen at anode.

{\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right)\underset{\text{Acid/base}}{\overset{\text{Electrolysis}}{\to }}{\text{2H}}_{\text{2}}\text{O}\left(\text{g}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)

The catalyst used in this process increases the conductivity of water as in neutral conditions water is non-conductor of electricity.

Q.6

Complete the following reactions:

\begin{array}{l}\text{i)}{\text{H}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{M}}_{\text{m}}{\text{O}}_{\text{o}}\left(\text{s}\right)\stackrel{\text{Δ}}{\to }\\ \text{ii)}\text{CO}\left(\text{g}\right)\text{+}{\text{H}}_{\text{2}}\left(\text{g}\right)\stackrel{\text{Δ}}{\to }\\ \text{iii)}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right)\text{+}{\text{3H}}_{\text{2}}\text{O}\left(\text{g}\right)\stackrel{\text{Δ}}{\to }\\ \text{iv)}\text{Zn}\left(\text{s}\right)\text{+}\text{NaOH}\left(\text{aq}\right)\stackrel{\text{Δ}}{\to }\end{array}

Ans.

\begin{array}{l}\text{i)}{\text{oH}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{M}}_{\text{m}}{\text{O}}_{\text{o}}\left(\text{s}\right)\stackrel{\text{Δ}}{\to }\text{mM}\left(\text{s}\right)\text{+}{\text{oH}}_{\text{2}}\text{O}\left(\text{l}\right)\\ \text{ii)}\text{CO}\left(\text{g}\right)\text{+}{\text{2H}}_{\text{2}}\left(\text{g}\right)\underset{\text{Catalyst}}{\overset{\text{Δ}}{\to }}{\text{CH}}_{\text{3}}\text{OH}\left(\text{l}\right)\\ \text{iii)}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right)\text{+}{\text{3H}}_{\text{2}}\text{O}\left(\text{g}\right)\stackrel{\text{Ni,}\text{1270K}}{\to }\text{3CO}\left(\text{g}\right){\text{+ 7H}}_{\text{2}}\left(\text{g}\right)\\ \text{iv)}\text{Zn}\left(\text{s}\right)\text{+}\text{2NaOH}\left(\text{aq}\right)\stackrel{\text{Δ}}{\to }\underset{\text{Sod}\text{.}\text{zincate}}{{\text{Na}}_{\text{2}}{\text{ZnO}}_{\text{2}}\left(\text{aq}\right)}{\text{+H}}_{\text{2}}\left(\text{g}\right)\end{array}

Q.7 Discuss the consequences of high enthalpy of H–H bond in terms of chemical reactivity of dihydrogen.

Ans.

Hydrogen is unreactive at room temperature due to high enthalpy of H–H bond. But at high temperature or in presence of catalyst, hydrogen reacts with metals and non-metals to form hydrides.

Q.8 What do you understand by (i) electron-deficient, (ii) electron-precise, and (iii) electron-rich compounds of hydrogen? Provide justification with suitable examples.

Ans.

i) Hydrides of group 13 (BH₃, AlH₃) have incomplete octet; they do not have sufficient number of electrons to form covalent bonds. So they are called electron-deficient hydrides. For that reason these compounds exist in polymeric forms.

ii) Hydrides of group 14 (CH₄, SiH₄, GeH₄) have exact number electrons to form covalent compounds. These are called electron-precise hydrides.

iii) Hydrides of group 15,16 and 17 (NH₃, PH₃, H₂O, H₂S,HCl, etc.) have excess electrons than required to form the covalent bonds. These are called electron-rich hydrides. The extra electrons are present as lone pairs of electrons.

Q.9 What characteristics do you expect from an electron-deficient hydride with respect to its structure and chemical reactions?

Ans.

The electron-deficient hydrides have insufficient number of electrons to form covalent bonds, so they exist in polymeric forms. These hydrides cannot be represented by conventional Lewis structures. B₂H₆, for example, contains four regular bonds and two three centre-two electron bonds. Its structure can be represented as:

Further to make-up the electron deficiency, they react with other metals, non-metals and their compounds. So these compounds are supposed to be very reactive.
B₂H₆(g) + 3O₂(g) → B₂O₃(s) + 3H₂O(g)
As they are electron deficient, they act as Lewis acids, thus form complexes with Lewis bases.

{\text{B}}_{\text{2}}{\text{H}}_{\text{6}}\text{+}\text{2}\stackrel{\text{••}}{\text{N}}{\text{Me}}_{\text{3}}\stackrel{}{\to }\text{2H}\underset{\text{Complex}}{{}_{\text{3}}\text{B}\stackrel{}{\leftarrow }{\text{NMe}}_{\text{3}}}

Q.10 Do you expect the carbon hydrides of the type (C_nH_2n+2) to act as ‘Lewis’ acid or base? Justify your answer.

Ans.

The compounds of the type (C_nH_2n+2) are electron-precise hydrides. They have exact number of electrons that are required to form covalent bonds. They don’t have any tendency to gain or lose electrons, so they can’t act as Lewis acid or Lewis base.

Q.11 What do you understand by the term “non-stoichiometric hydrides”? Do you expect this type of the hydrides to be formed by alkali metals? Justify your answer.

Ans.

The hydrides which are deficient in hydrogen and in which ratio of metal to hydrogen is fractional are known as non- stoichiometric hydrides. Mainly d and f- block elements can form this type of hydrides. The fractional ratio also varies with change in temperature and pressure, i.e. it is not fixed. The hydrogen atoms in this type of hydrides occupy the lattice holes.

Alkali metals are highly reducing, they transfer their lone electron to the hydrogen atom to form hydride ion (H^–). This is formed by complete transfer of an electron, so the ratio of metal to hydrogen is fixed.Thus, alkali metals form only stoichiometric hydrides.In other words,alkali metals cannot form the non-stoichiometric hydrides.

Q.12 How do you expect the metallic hydrides to be useful for hydrogen storage? Explain.

Ans.

In the metallic hydrides, the hydrogen is absorbed as H-atoms in the lattices. Due to the inclusion of hydrogen atoms, metal lattices are unstable. When heat is supplied to the metallic hydride,it decomposes to form hydrogen and finely divided metal.

So the hydrides of the transition metals can be used to store and transport hydrogen as hydrogen produced by these metal hydrides is used as fuels. This is known as hydrogen economy.

Q.13 How does the atomic hydrogen or oxy-hydrogen torch function for cutting and welding purposes? Explain.

Ans.

When an electric arc is passed through molecular hydrogen, atomic hydrogen is produced between two electrodes. The lifetime of atomic hydrogen is very less(0.3 sec.). It immediately gets converted to molecular form, releasing a large amount of energy which is used for cutting and welding purposes in the form of atomic hydrogen torch. Atomic hydrogen torch can produce temperature about 3400K.

{\text{H}}_{\text{2}}\underset{\text{3773-4273}\text{K}}{\overset{\text{Electric}\text{arc}}{\to }}\text{2H}

Q.14 Among NH₃, H₂O and HF, which would you expect to have highest magnitude of hydrogen bonding and why?

Ans.

N, O, F are highly electronegative. All the given compounds undergo intermolecular hydrogen bonding. The electronegativity order among N, O, F atoms, is as follows F > O > N. Since, in the H-F molecule, the difference in the electronegativities of H and F is highest, therefore, hydrogen bonding is the strongest in H-F.

Q.15 Saline hydrides are known to react with water violently producing fire. Can CO₂, a well known fire extinguisher, be used in this case? Explain.

Ans.

Saline hydrides such as NaH, CaH₂, are mainly hydrides of s block elements. They react violently with water to form corresponding metal hydroxides with the evolution of dihydrogen.

NaH(s) + H₂O(l) → NaOH(aq) + H₂(g)

The reaction is highly exothermic, so produced hydrogen catches fire. CO₂ can’t be used as fire extinguisher for these purposes as CO₂ gets reduced by the metal hydride to form sodium formate. Free CO₂ gas is not available for these purposes.

NaH + O₂ → HCOONa

Q.16 Arrange the following

(i) CaH₂, BeH₂ and TiH₂ in order of increasing electrical conductance.

(ii) LiH, NaH and CsH in order of increasing ionic character.

(iii) H–H, D–D and F–F in order of increasing bond dissociation enthalpy.

(iv) NaH, MgH₂ and H₂O in order of increasing reducing property.

Ans.

i) BeH₂ is a covalent compound, so it does not conduct electricity. CaH₂ is an ionic compound, so it conducts electricity in fused state while TiH₂ conducts electricity under normal conditions. The order of increasing conductivity is as follows:

BeH₂ < CaH₂ < TiH₂

ii) Li, Na, Cs belong to same group of the periodic table. The electronegativity decreases from Li to Cs. So the ionic character increases down the group. The order of the increasing ionic character of the hydrides is as follows:

LiH < NaH < CsH.

iii) Due to the greater molecular mass, the bond pair in D-D bond is attracted strongly than that in H-H bond. Hence, bond dissociation energy of D-D bond is greater than that of H-H bond. In F-F bond, due to repulsion between lone pair and bond pair, the energy required for bond dissociation is less. Thus the order of bond dissociation enthalpy is as follows:

F-F < H-H < D-D

iv)The ionic hydrides are powerful reducing agents. MgH₂ and H₂O are covalent compounds, but the bond dissociation energy of H₂O is much higher than that of MgH₂. NaH is ionic hydride. So the order of increasing reducing character is as follows-

H₂O < MgH₂ < NaH

Q.17 Compare the structures of H₂O and H₂O₂.

Ans.

In water, the oxygen atom is sp³ hybridised. H₂O molecule has two lone pairs and two bond pairs of electrons. Due to the greater repulsion between lp–lp as compared to bp–bp, the bond angle decreases from 109.5° to 104.5°. Hence, water has bent structure.

H₂O₂ has a non-planar structure. In H₂O₂, the two O-atoms are linked to each other with single covalent bond. The O-atoms are further attached to the H-atoms with single covalent bond.The two O-H bonds are situated in different planes and the dihedral angle between the two planes is 111.5° in gas phase. Its structure is just like that of an open book.

Q.18 What do you understand by the term ‘auto-protolysis’ of water? What is its significance?

Ans.

The term ‘auto-protolysis’ means self ionization of water. The reaction can be represented as follows :

\underset{aci{d}_1}{{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)}\text{+}\underset{bas{e}_2}{{\text{H}}_{\text{2}}\text{O}}\underset{}{⇌}\underset{aci{d}_2}{{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}}\left(\text{aq}\right)\text{+}\underset{bas{e}_1}{{\text{OH}}^{\text{–}}\text{(aq)}}

For this reason, water is amphoteric in nature, it reacts with both acids and bases. For example,

H₂O(l) + NH₃(aq) → NH₄⁺(aq) + OH^–(aq)

Here H₂O acts as acid towards base NH₃.

H₂O(l) + H₂S(aq) → H₃O+(aq) ⁺ HS^– (aq

Here H₂O acts as base towards acid H₂S.

Q.19 Consider the reaction of water with F₂ and suggest, in terms of oxidation and reduction, which species are oxidised/reduced.

Ans.

Water acts as reducing agent when it reacts with fluorine. It gets oxidized to O₂. F₂ acts as an oxidising agent and gets reduced to F^–.

Fluorine is reduced from zero to (– 1) oxidation state.

Water is oxidized from (– 2) to zero oxidation state.

Q.20

\begin{array}{l}\text{Complete the following chemical reactions}\text{.}\\ \text{i) PbS}\left(\text{s}\right){\text{+ H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\stackrel{}{\to }\\ {\text{ii) MnO}}_{\text{4}}^{\text{–}}\left(\text{aq}\right){\text{+ H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\stackrel{}{\to }\\ \text{iii) CaO}\left(\text{s}\right){\text{+ H}}_{\text{2}}\text{O}\left(\text{g}\right)\stackrel{}{\to }\\ {\text{iv) AlCl}}_{\text{3}}\left(\text{g}\right){\text{+ H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }\\ {\text{v) Ca}}_{\text{3}}{\text{N}}_{\text{2}}\left(\text{s}\right){\text{+ 6H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }\\ \text{Classify the above into}\left(\text{a}\right)\text{hydrolysis,}\left(\text{b}\right)\text{redox and}\left(\text{c}\right)\text{hydration reactions}\text{.}\end{array}

Ans.

\begin{array}{l}\text{i) PbS}\left(\text{s}\right){\text{+ 4H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\stackrel{}{\to }\underset{\text{White}}{{\text{PbSO}}_{\text{4}}}\left(\text{s}\right)\text{+}{\text{4H}}_{\text{2}}\text{O}\\ {\text{ii) 2MnO}}_{\text{4}}^{\text{–}}\left(\text{aq}\right){\text{+ 5H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\text{+}{\text{6H}}^{\text{+}}\text{(aq)}\stackrel{}{\to }{\text{2Mn}}^{\text{2+}}\text{(aq)}\text{+}{\text{8H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{+}{\text{5O}}_{\text{2}}\left(\text{g}\right)\\ \text{iii) CaO}\left(\text{s}\right){\text{+ H}}_{\text{2}}\text{O}\left(\text{g}\right)\stackrel{}{\to }\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{aq}\right)\\ {\text{iv) AlCl}}_{\text{3}}\left(\text{g}\right){\text{+ 3H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }\text{Al}{\left(\text{OH}\right)}_{\text{3}}\left(\text{s}\right)\text{+}\text{3HCl}\left(\text{aq}\right)\\ {\text{v) Ca}}_{\text{3}}{\text{N}}_{\text{2}}\left(\text{s}\right){\text{+ 6H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }\text{3Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{aq}\right)\text{+}{\text{2NH}}_{\text{3}}\left(\text{aq}\right)\end{array}

(a) Hydrolysis reactions: (iv), and (v)
(b) Redox reactions: (i) and (ii)
(c) Hydration reaction (iii)

Q.21 Describe the structure of the common form of ice.

Ans.

In ice form of water, each oxygen atom is surrounded by another four oxygen atoms and there is one hydrogen atom in between each pair of oxygen atom. Each oxygen atom is bonded with four hydrogen atoms, two of them are covalently bonded with oxygen atom and rest two by weak hydrogen bonds.

Q.22 What causes the temporary and permanent hardness of water?

Ans.

The temporary hardness is due to the presence of bicarbonates of calcium and magnesium, i.e. Ca(HCO₃)₂, Mg(HCO₃)₂.

The permanent hardness is due to the presence of chlorides and sulphates of sodium and magnesium, i.e.,CaCl₂, CaSO₄, MgCl₂, and MgSO₄.

Q.23 Discuss the principle and method of softening of hard water by synthetic ion exchange resins.

Ans.

Ion exchange resins are giant organic molecule of high molecular weight. The working principle of ion exchange resins are discussed below:

Step1: The hard water is passed through a tank packed with cation exchange resin which is supported over gravel. All the cations present in hard water will exchange with the H⁺ ions present in the resin.

2R — COO^–H⁺ + CaCl₂ → (RCOO)₂Ca + 2H⁺ + 2Cl^–

(Cation exchange resin) (From hard water)

2R — COO^–H⁺ + MgSO₄ → (RCOO)₂Mg + 2H⁺ + SO₄^2–

(Cation exchange resin) (From hard water)

Step2. The water comes out from the bottom of the tank is richer in H⁺ ions. This will pass through another tank which is packed with the anion exchange resins supported over gravel. The Cl^– and SO₄^2– ions present in the hard water exchange with OH^– ions of the resin.

\begin{array}{l}\text{R}\text{—}\stackrel{\text{+}}{\text{N}}{\text{H}}_{\text{3}}{\text{OH}}^{\text{–}}\text{+}{\text{Cl}}^{\text{–}}\stackrel{}{\to }\text{R}\text{—}\stackrel{\text{+}}{\text{N}}{\text{H}}_{\text{3}}{\text{Cl}}^{\text{–}}{\text{+ OH}}^{\text{+}}\\ \text{(anion}\text{exchange}\text{(From}\text{hard}\text{(Exhausted}\text{resin)}\\ \text{resin)}\text{water)}\end{array}

R — COO^–H⁺ + CaCl₂ → (RCOO)₂Ca + 2H⁺ + 2Cl^–

(Cation exchange resin) (From hard water)

\begin{array}{l}\text{2R}\text{—}\stackrel{\text{+}}{\text{N}}{\text{H}}_{\text{3}}{\text{OH}}^{\text{–}}\text{+}{\text{SO}}_4^{2–}\stackrel{}{\to }{\stackrel{+}{\left(\text{R}\text{—}{\text{NH}}_{\text{3}}\right)}}_2{\text{SO}}_4^{2–}{\text{+ 2OH}}^{–}\\ \text{(anion}\text{exchange}\text{(From}\text{hard}\text{(Exhausted}\text{resin)}\\ \text{resin)}\text{water)}\end{array}

The H⁺ ions coming from the first tank and OH^– from the second tank combine to form water.

H⁺ + OH^– → H₂O

Thus, the water obtained from this method is free from all types of cations and anions.

Regenaration of resins:

The exhausted resin in the first tank can be regenerated by treatment with the concentrated H₂SO₄ or HCl.

Q.24 Write chemical reactions to show the amphoteric nature of water.

Ans.

Due to the amphoteric nature of water, it acts both as an acid as well as a base. Towards strong base(e.g. NH₃), it behaves as a strong acid and towards strong acid (e.g. H₂S), it behaves as a strong base.

${\displaystyle \begin{array}{l}\underset{\mathrm{A}\mathrm{c}\mathrm{i}{\mathrm{d}}_1}{{\mathrm{H}}_2\mathrm{O}(\mathrm{l})}+\underset{\mathrm{B}\mathrm{a}\mathrm{s}{\mathrm{e}}_2}{\mathrm{N}{\mathrm{H}}_3(\mathrm{a}\mathrm{q})}\to <!--\; \to \; -->\underset{\mathrm{A}\mathrm{c}\mathrm{i}{\mathrm{d}}_2}{{\mathrm{N}\mathrm{H}}_4^{+}(\mathrm{a}\mathrm{q})}+\underset{\mathrm{B}\mathrm{a}\mathrm{s}\mathrm{e}{}_1}{\mathrm{O}{\mathrm{H}}^{-<!--\; -\; -->}(\mathrm{a}\mathrm{q})}\\ \underset{\mathrm{B}\mathrm{a}\mathrm{s}\mathrm{e}{}_1}{{\mathrm{H}}_2\mathrm{O}(\mathrm{l})}+\underset{\mathrm{A}\mathrm{c}\mathrm{i}{\mathrm{d}}_2}{{\mathrm{H}}_2\mathrm{S}(\mathrm{a}\mathrm{q})}\to <!--\; \to \; -->\underset{\mathrm{A}\mathrm{c}\mathrm{i}{\mathrm{d}}_2}{{\mathrm{H}}_3{\mathrm{O}}^{+}(\mathrm{a}\mathrm{q})}+\underset{\mathrm{B}\mathrm{a}\mathrm{s}\mathrm{e}{}_2}{\mathrm{H}{\mathrm{S}}^{-<!--\; -\; -->}(\mathrm{a}\mathrm{q})}\end{array}}$

Q.25 Write chemical reactions to justify that hydrogen peroxide can function as an oxidising as well as reducing agent.

Ans.

H₂O₂ can act as an oxidising and reducing agent in both acidic and basic medium.

(i) Reducing agent in acidic medium

{\text{2MnO}}_{\text{4}}^{\text{–}}\left(\text{aq}\right)\text{+}{\text{5H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\text{+}{\text{6H}}^{\text{+}}\left(\text{aq}\right)\stackrel{}{\to }{\text{2Mn}}^{\text{2+}}\left(\text{aq}\right){\text{+ 8H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{+}{\text{5O}}_{\text{2}}\left(\text{g}\right)

(ii) Reducing agent in basic medium

{\text{I}}_{\text{2}}\left(\text{s}\right){\text{+ H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right){\text{+ 2OH}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }{\text{2I}}^{\text{–}}\left(\text{aq}\right)\text{+}{\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)

(iii) Oxidising agent in acidic medium

{\text{2Fe}}^{\text{2+}}\left(\text{aq}\right)\text{+}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\text{+}{\text{2H}}^{\text{+}}\left(\text{aq}\right)\stackrel{}{\to }{\text{2Fe}}^{\text{3+}}\left(\text{aq}\right)\text{+}{\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right)

(iv) Oxidising agent in basic medium

Mn²⁺(aq) + H₂O₂(aq) + 2OH^–(aq) → MnO₂(s) + 2H₂O(l)

Q.26 What is meant by ‘demineralised’ water and how can it be obtained?

Ans.

The water which is free from any sort of cation and anion is called demineralised water. It can be produced by passing hard water first through cation exchange resin and then through anion exchange resin. The steps occurred are as follows:

Step1: The hard water is passed through a tank packed with cation exchange resins which is supported over gravel. All the cations present in hard water will exchange with the H⁺ ions present in the resin.

2R — COO^–H⁺ + CaCl₂ → (RCOO)₂Ca + 2H⁺ + 2Cl^–

(Caption exchange resin) (From hard water)

2R — COO^–H⁺ + MgSO₄ → (RCOO)₂Mg + 2H⁺ + SO₄^2–

(Caption exchange resin) (From hard water)

Step2. The water comes out from the bottom of the tank is richer in H⁺ ions. This will pass through another tank which is packed with the anion exchange resins supported over gravel. The Cl^– and SO₄^2– ions present in the hard water exchange with OH^– ions of the resin.

\begin{array}{l}\text{R}\text{—}\stackrel{\text{+}}{\text{N}}{\text{H}}_{\text{3}}{\text{OH}}^{\text{–}}\text{+}{\text{Cl}}^{\text{–}}\stackrel{}{\to }\text{R}\text{—}\stackrel{\text{+}}{\text{N}}{\text{H}}_{\text{3}}{\text{Cl}}^{\text{–}}\text{+}{\text{OH}}^{\text{–}}\\ \text{(anion}\text{exchange}\text{resin)}\text{(From}\text{hard}\text{water)}\text{(Exhausted}\text{resin)}\end{array} \begin{array}{l}\text{2R}\text{—}\stackrel{\text{+}}{\text{N}}{\text{H}}_{\text{3}}{\text{OH}}^{\text{–}}\text{+}{\text{SO}}_{\text{4}}^{\text{2–}}\stackrel{}{\to }{\stackrel{\text{+}}{\left(\text{R}\text{—}{\text{NH}}_{\text{3}}\right)}}_{\text{2}}{\text{SO}}_{\text{4}}^{\text{2–}}\text{+}{\text{2OH}}^{\text{–}}\\ \text{(anion}\text{exchange}\text{resin)}\text{(From}\text{hard}\text{water)}\text{(Exhausted}\text{resin)}\end{array}

The H⁺ ions coming from the first tank and OH^– ions from the second tank combine to form water.

H⁺ + OH^– → H₂O

The resulting water is free from both cations and anions.

Q.27 Is demineralised or distilled water useful for drinking purposes? If not, how can it be made useful?

Ans.

No, demineralised or distilled water is not useful for drinking purposes because it does not contain minerals. To make it drinkable, proper amount of minerals should be added to the demineralised water.

Q.28 Describe the usefulness of water in biosphere and biological systems.

Ans.

Water is essential for the survival of all living beings. It is the main constituent of human body. Water has high specific heat, thermal conductivity, surface tension, dipole moment and dielectric constant etc. as compared to other liquids. It is also called universal solvent, used for transportation of minerals and nutrients in plants and animals for the metabolism. Water is main reactant of photosynthesis of plants which releases oxygen into the atmosphere.

Q.29 What properties of water make it useful as a solvent? What types of compound can it (i) dissolve, and (ii) hydrolyse?

Ans.

(i) Due to the following two properties, water can dissolve most of the ionic and covalent compounds

• High dielectric constant
• High dipole moment

Ionic compounds dissolves in water due to the ion-dipole interaction or solvation of the ions. The covalent compounds such as alcohol, amines, urea, glucose etc. dissolve in water due to H-bonding. So water is called universal solvent.

(ii) Water can hydrolyse many oxides, hydrides, carbides, nitrides and other salts. H⁺ and OH^– ions of water react with anions and cations of the compounds respectively to form acid and base or both as follows:

CaO(s) + H₂O(l) → Ca(OH)₂(aq)

CaH₂(s) + 2H₂O(l) → Ca(OH)₂(aq) + 2H₂(g)

CaC₂(s) + 2H₂O(l) → Ca(OH)₂(aq) + C₂H₂(g)

Acetylene

Q.30 Knowing the properties of H₂O and D₂O, do you think that D₂O can be used for drinking purposes?

Ans.

Deuterium has higher molar mass than hydrogen and it slows down all the chemical reactions happening within human body. Thus, heavy water does not support life, so normal water is used for drinking purposes.

Q.31 What is the difference between the terms ‘hydrolysis’ and ‘hydration’?

Ans.

The Reaction between salt and H₂O which produces original acid and base is called hydrolysis. For example,

Na₂CO₃ + 2H₂O → 2NaOH + H₂CO₃

Salt Base Acid

On the other hand, addition of water molecule to ions or salts to form hydrated ions or salts is called hydration.

Q.32 How can saline hydrides remove traces of water from organic compounds?

Ans.

Saline hydrides react with water and produce corresponding metal hydroxides with liberation of hydrogen gas. Thus, traces of water present in organic solvents can be removed by distilling them over saline hydrides. H₂ escapes into the air and metal hydroxide is left in the flask, while dry organic solvent distills over.

Q.33 What do you expect the nature of hydrides is, if formed by elements of atomic numbers 15, 19, 23 and 44 with dihydrogen? Compare their behavior towards water.

Ans.

The hydride formed by element of atomic number 15 (element with Z= 15, is non-metal, i.e., P) should be covalent hydride (i.e. PH₃).

The hydride formed by element of atomic number 19 (element with Z=19, is alkali metal, i.e. K), should be ionic or saline hydride (i.e., KH).

Element with atomic number 23 is a transition metal of group 3 ( i.e., V) and hence forms a metallic or interstitial hydride (i.e., VH_0.56).

The element with atomic number 44 is a transition metal of group 8 (i.e., Ru),hence, does not form any hydride.

Behaviour towards water: Only saline hydrides react with water to form H₂ gas.

2KH(s) + 2H₂O(l) → 2KOH(aq) + H₂(g)

Q.34 Do you expect different products in solution when aluminum (III) chloride and potassium chloride treated separately with (i) normal water (ii) acidified water, and (iii) alkaline water? Write equations wherever necessary.

Ans.

KCl is salt of strong acid and strong base. It does not undergo hydrolysis in normal water. It remains as hydrated ions in water.

\text{KCl(s)}\stackrel{\text{Water}}{\to }{\text{K}}^{\text{+}}\text{(aq)}{\text{+ Cl}}^{\text{–}}\text{(aq)}

Since the aqueous solution of KCl is neutral, so in acidic and basic medium, it remains same.
AlCl₃ is salt of weak base, Al(OH)₃ and strong acid, HCl. In normal water, it hydrolyses to form Al(OH)₃, H⁺ and Cl^– ions.
AlCl₃(s) + 3H₂O(l) → Al(OH)₃(s) +3H⁺(aq) + 3Cl^–(aq)
In acidic medium, H⁺ ions react with Al(OH)₃ to generate Al³⁺ ions and H₂O. Thus, in acidic medium, AlCl₃ exists as Al³⁺ (aq) and Cl^– (aq) ions.

{\text{AlCl}}_{\text{3}}\text{(s)}\stackrel{\text{Acidified}\text{water}}{\to }{\text{Al}}^{\text{3+}}\text{(aq)}{\text{+ 3Cl}}^{\text{–}}\text{(aq)}

Al(OH)₃ in alkaline water forms soluble tetrahydroxoaluminate complex or meta-aluminate ion.
Al(OH)₃(s) + OH^–(aq) → [Al(OH)₄]^–(aq) or AlO₂^–(aq) + 2H₂O(l)

Tetrahydroxoaluminate Meta-aluminate ion
The complete reaction may be written as:

Q.35 How does H₂O₂ behave as a bleaching agent?

Ans.

H₂O₂ decomposes to water and nascent oxygen. The nascent oxygen reacts with the dyes or pigments used as colouring matters, which in turn get oxidized. The colour fades. It is used for the bleaching of the delicate materials like ivory, silk, wool and feather.

{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\stackrel{}{\to }\underset{\begin{array}{l}\text{Nascent}\\ \text{Oxygen}\end{array}}{{\text{H}}_{\text{2}}\text{O+}\left[\text{O}\right]}

Q.36 What do you understand by the terms?

(i) hydrogen economy (ii) hydrogenation (iii) ‘syngas’ (iv) water-gas shift reaction

(v) fuel-cell?

Ans.

(i) Hydrogen Economy: Coal and petroleum are conventional energy sources but their reserve are limited in the world. In search of new energy sources, one proposed way is to burn hydrogen as a fuel in industries, power plants, motor vehicles. This is called hydrogen economy.

(ii)Hydrogenation: The addition of hydrogen across the double bond and triple bond to form saturated compounds is called hydrogenation. Since, vegetable oils contain many double bonds, they are called polyunsaturated oils. When these oils are exposed to the air, they get oxidized (C=C bond undergoes oxidation), thus develop unpleasant taste. To avoid this, double bonds are hydrogenated. The vegetable oils react with hydrogen in the presence of nickel catalyst and get converted into solid fats. This process is called hydrogenation or hardening of oils. This process is used to manufacture vegetable ghee from vegetable oils.

(iii)Syngas: It is the mixture of CO and H₂ gases. It can be produced by the reaction of steam on hydrocarbon or coke at high temperature in the presence of nickel as catalyst.

{\text{CH}}_{\text{4}}\left(\text{g}\right){\text{+ H}}_{\text{2}}\text{O}\left(\text{g}\right)\underset{\text{Ni}}{\overset{\text{1270K}}{\to }}\text{CO}\left(\text{g}\right)\text{+}{\text{H}}_{\text{2}}\left(\text{g}\right)

When the syngas is produced from coal, the process is called ‘coal gasification’.

\text{C}\left(\text{s}\right){\text{+ H}}_{\text{2}}\text{O}\left(\text{g}\right)\underset{\text{Ni}}{\overset{\text{1270K}}{\to }}\text{CO}\left(\text{g}\right){\text{+ H}}_{\text{2}}\left(\text{g}\right)

(iv)Water-gas shift reaction: The amount of hydrogen in the water gas can be further increased by oxidizing CO to CO₂ by passing the mixture over FeCrO₄ in the presence of steam.

\underset{\text{Water}\text{gas}}{\text{CO}\left(\text{g}\right)\text{+}{\text{H}}_{\text{2}}\left(\text{g}\right)}\text{+}\underset{\text{Steam}}{{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)}\underset{\text{673K}}{\overset{{\text{FeCrO}}_{\text{4}}}{\to }}\underset{\text{Syngas}}{{\text{CO}}_{\text{2}}\left(\text{g}\right){\text{+ 2H}}_{\text{2}}\left(\text{g}\right)}

This reaction is called water-gas shift reaction.

(v)Fuel-cell: Fuel cell is a device which converts the energy produced during the combustion of fuel directly into the electrical energy. For example, Hydrogen-oxygen fuel cells. These are used for generating electrical energy. Dihydrogen is used as fuel. The conventional fossil fuels cause environmental pollution, but fuel cells do not cause such type of pollution and may release more amount of energy.