NCERT Solutions for Class 11 Chemistry Hydrogen [Old Book]

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Chemistry
Chapter : 9
Chapter Name : Hydrogen

Q9.1 Justify the position of hydrogen in the periodic table on the basis of its electronic
con guration.

Answer. Hydrogen is the rst element of the periodic table. Its electronic con guration is [1s 1

]. Due to the presence of only one electron in its 1s shell, hydrogen exhibits a dual behaviour,
i.e., it resembles both alkali metals and halogens.
Resemblance with alkali metals:
1. Like alkali metals, hydrogen contains one valence electron in its valence shell.
1
H : 1s

1
Li : [He]2s

1
Na : [Ne]3s

Hence, it can lose one electron to form a unipositive ion.
2. Like alkali metals, hydrogen combines with electronegative elements to form oxides,
halides, and sulphides.
Resemblance with halogens:
1. Both hydrogen and halogens require one electron to complete their octets.
1
H : 1s

2 2 5
F : 1s 2s 2p

2 2 6 2 5
Cl : 1s 2s 2p 3s 3p

Hence, hydrogen can gain one electron to form a uninegative ion.
2. Like halogens, it forms a diatomic molecule and several covalent compounds.
Though hydrogen shows some similarity with both alkali metals and halogens, it differs from
them on some grounds. Unlike alkali metals, hydrogen does not possess metallic
characteristics. On the other hand, it possesses a high ionization enthalpy. Also, it is less
reactive than halogens.
Owing to these reasons, hydrogen cannot be placed with alkali metals (group I) or with
halogens (group VII). In addition, it was also established that H ions cannot exist freely as
+

they are extremely small. H ions are always associated with other atoms or molecules.
+

Hence, hydrogen is best placed separately in the periodic table.

Page : 296 , Block Name : Exercise

Q9.2 Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes?

1
1. Protium, H

Answer. Hydrogen has three isotopes. They are:
2
2. Deuterium, H or D, and

3
3. Tritium, H or T

The mass ratio of protium, deuterium and tritium is 1:2:3.

Page : 296 , Block Name : Exercise

Q9.3 Why does hydrogen occur in a diatomic form rather than in a mono-atomic form under
normal conditions?

Answer. The ionization enthalpy of hydrogen atom is very high (1312kJmol ). Hence, it is
−1

very hard to remove its only electron. As a result, its tendency to exist in the mono-atomic
form is rather low. Instead, hydrogen forms a covalent bond with another hydrogen atom and
exists as a diatomic H molecule.
2

Page : 296 , Block Name : Exercise

Q9.4 How can the production of dihydrogen, obtained from ‘coal gasi cation’, be increased?

Answer. Dihydrogen is produced by coal gasi cation method as:
1270k

C(s) + H2 O(g) ⟶ CO(g) + H2(g)

(coal)

The yield of dihydrogen (obtained from coal gasi cation) can be increased by reacting carbon
monoxide (formed during the reaction) with steam in the presence of iron chromate as a
catalyst.
673K
CO(g) + H2 O(g) CO2(g) + H2(g)
Cualyst

This reaction is called the water-gas shift reaction. Carbon dioxide is removed by scrubbing it
with a solution of sodium arsenite.

Page : 296 , Block Name : Exercise

Q9.5 Describe the bulk preparation of dihydrogen by electrolytic method. What is the role of
an electrolyte in this process ?

Answer. Dihydrogen is prepared by the electrolysis of acidi ed or alkaline water using
platinum electrodes. Generally, 15 – 20% of an acid (H2SO4) or a base (NaOH) is used.
Reduction of water occurs at the cathode as:
− −
2H2 O + 2e ⟶ 2H2 + 2OH

At the anode, oxidation of OH– ions takes place as:
− 1 −
2OH ⟶ H2 O + O2 + 2e
2

Net reaction can be represented as:
1
H2 O(l) ⟶ H2(x) + O2(g)
2

Electrical conductivity of pure water is very low owing to the absence of ions in it. Therefore,
electrolysis of pure water also takes place at a low rate. If an electrolyte such as an acid or a
base is added to the process, the rate of electrolysis increases. The addition of the electrolyte
makes the ions available in the process for the conduction of electricity and for electrolysis to
take place.

Page : 296 , Block Name : Exercise

Q9.6 Complete the following reactions:
Δ
(i) H2 (g) + Mm Oo (s) ⟶

Δ
(ii) CO(g) + H2 (g)
catalyst

Δ
C3 H8 (g) + 3H2 O(g) −
catalyst

heat
Zn(s) + NaOH(aq)
heat

Δ

Answer. (i) H 2(g)
+ Mm Oo(s) ⟶ mM(s) + H2 O(l)

(ii) CO (g)
+ 2H2(g)

Δ

calalyst

CH3 OH(l)

(iii) C H
3 8(g)
+ 3H2 O(g)
Δ

calalyst

3CO(g) + 7H2(g)

(iv) Zn (s)
+ 2NaOH(aq)

bat
Na2 ZnO2(oq) + H2(g)
⟶

Page : 296 , Block Name : Exercise

Q9.7 Discuss the consequences of high enthalpy of H–H bond in terms of chemical reactivity
of dihydrogen.

Answer. The ionization enthalpy of H–H bond is very high (1312 kJ mol ).This indicates that
−1

hydrogen has a low tendency to form H ions. Its ionization enthalpy value is comparable to
+

that of halogens. Hence, it forms diatomic molecules (H ), hydrides with elements, and a large
2

number of covalent bonds.
Since ionization enthalpy is very high, hydrogen does not possess metallic characteristics
(lustre, ductility, etc.) like metals.

Q9.8 What do you understand by
(i) electron-de cient,
(ii) electron-precise, and
(iii) electron-rich compounds of hydrogen? Provide justi cation with suitable examples.

Answer. Molecular hydrides are classi ed on the basis of the presence of the total number of
electrons and bonds in their Lewis structures as:
1. Electron-de cient hydrides
2. Electron-precise hydrides
3. Electron-rich hydrides
An electron-de cient hydride has very few electrons, less than that required for representing
its conventional Lewis structure e.g. diborane (B H ). In B H , there are six bonds in all, out
2 6 2 6

of which only four bonds are regular two centered-two electron bonds. The remaining two
bonds are three centered-two electron bonds i.e., two electrons are shared by three atoms.
Hence, its conventional Lewis structure cannot be drawn.
An electron-precise hydride has a suf cient number of electrons to be represented by its
conventional Lewis structure e.g. CH4. The Lewis structure can be written as:

Four regular bonds are formed where two electrons are shared by two atoms.
An electron-rich hydride contains excess electrons as lone pairs e.g. N H .
3

There are three regular bonds in all with a lone pair of electrons on the nitrogen atom.

Q9.9 What characteristics do you expect from an electron-de cient hydride with respect to its
structure and chemical reactions?

Answer. An electron-de cient hydride does not have suf cient electrons to form a regular
bond in which two electrons are shared by two atoms e.g., B H , Al2H6 etc.
2 6

These hydrides cannot be represented by conventional Lewis structures. B H , for example,
2 6

contains four regular bonds and two three centered-two electron bond. Its structure can be
represented as:

Since these hydrides are electron-de cient, they have a tendency to accept electrons. Hence,
they act as Lewis acids.
B2 H6 + 2NMe ⟶ 2BH3 ⋅ NMe3

B2 H6 + 2CO ⟶ 2BH3 ⋅ CO

Q9.10 Do you expect the carbon hydrides of the type (C H n 2n + 2 ) to act as ‘Lewis’ acid or
base? Justify your answer.

Answer. For carbon hydrides of type C Hn 2n + 2 , the following hydrides are possible for
n = 1 ⇒ CH4

n = 2 ⇒ C 2 H6

n = 3 ⇒ C 3 H8

For a hydride to act as a Lewis acid i.e., electron accepting, it should be electron-de cient.
Also, for it to act as a Lewis base i.e., electron donating, it should be electron-rich.
Taking C H as an example, the total number of electrons are 14 and the total covalent bonds
2 6

are seven. Hence, the bonds are regular 2e–-2 centered bonds.

Hence, hydride C H has suf cient electrons to be represented by a conventional Lewis
2 6

structure. Therefore, it is an electron-precise hydride, having all atoms with complete octets.
Thus, it can neither donate nor accept electrons to act as a Lewis acid or Lewis base.

Q9.11 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.

Answer. Non-Stoichiometric hydrides are hydrogen-de cient compounds formed by the
reaction of dihydrogen with d-block and f-block elements. These hydrides do not follow the
law of constant composition. For example: LaH , YbH , TiH
2.87 2.55 etc.
1.5−1.8

Alkali metals form stoichiometric hydrides. These hydrides are ionic in nature. Hydride ions
have comparable sizes (208 pm) with alkali metal ions. Hence, strong binding forces exist
between the constituting metal and hydride ion. As a result, stoichiometric hydrides are
formed.
Alkali metals will not form non-stoichiometric hydrides.

Q9.12 How do you expect the metallic hydrides to be useful for hydrogen storage? Explain.

Answer. Metallic hydrides are hydrogen de cient, i.e., they do not hold the law of constant
composition. It has been established that in the hydrides of Ni, Pd, Ce, and Ac, hydrogen
occupies the interstitial position in lattices allowing further absorption of hydrogen on these
metals. Metals like Pd, Pt, etc. have the capacity to accommodate a large volume of hydrogen.
Therefore, they are used for the storage of hydrogen and serve as a source of energy.

Q9.13 How does the atomic hydrogen or oxy-hydrogen torch function for cutting and welding
purposes ? Explain.

Answer. Atomic hydrogen atoms are produced by the dissociation of dihydrogen with the help
of an electric arc. This releases a huge amount of energy . This energy can be used to generate
a temperature of 4000 K, which is ideal for welding and cutting metals. Hence, atomic
hydrogen or oxy-hydrogen torches are used for these purposes. For this reason, atomic
hydrogen is allowed to recombine on the surface to be welded to generate the desired
temperature.

Q9.14 Among NH , H O and HF, which would you expect to have highest magnitude of
3 2

hydrogen bonding and why?

Answer. The extent of hydrogen bonding depends upon electronegativity and the number of
hydrogen atoms available for bonding. Among nitrogen, uorine, and oxygen, the increasing
order of their electronegativities are N < O < F.
Hence, the expected order of the extent of hydrogen bonding is HF > H O > NH 2 3

But, the actual order is H O > HF > NH
2 3

Although uorine is more electronegative than oxygen, the extent of hydrogen bonding is
higher in water. There is a shortage of hydrogens in HF, whereas there are exactly the right
numbers of hydrogens in water. As a result, only straight chain bonding takes place. On the
other hand, oxygen forms a huge ring-like structure through its high ability of hydrogen

bonding.
In case of ammonia, the extent of hydrogen bonding is limited because nitrogen has only one
lone pair. Therefore, it cannot satisfy all hydrogens.

Q9.15 Saline hydrides are known to react with water violently producing re. Can CO , a well
2

known re extinguisher, be used in this case? Explain.

Answer. Saline hydrides (i.e., NaH, LiH, etc.) react with water to form a base and hydrogen gas.
The chemical equation used to represent the reaction can be written as:
MH(s) + H2 O(aq) ⟶ MOH(aq) + H2(g)

The reaction is violent and produces re.
CO is heavier than dioxygen. It is used as a re extinguisher because it covers the re as a
2

blanket and inhibits the supply of dioxygen, thereby dousing the re.
CO can be used in the present case as well. It is heavier than dihydrogen and will be effective
2

in isolating the burning surface from dihydrogen and dioxygen.

Q9.16 Arrange the following
(i) CaH , BeH and TiH in order of increasing electrical conductance.
2 2 2

(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, MgH and H Oin order of increasing reducing property.
2 2

Answer. (i) The electrical conductance of a molecule depends upon its ionic or covalent
nature. Ionic compounds conduct, whereas covalent compounds do not.
BeH is a covalent hydride. Hence, it does not conduct.
2

CaH is an ionic hydride, which conducts electricity in the molten state. Titanium hydride,
2

TiH is metallic in nature and conducts electricity at room temperature. Hence, the
2

increasing order of electrical conductance is as follows:
BeH2 < CaH2 < TiH2

(ii) The ionic character of a bond is dependent on the electronegativities of the atoms
involved. The higher the difference between the electronegativities of atoms, the smaller is
the ionic character.
Electronegativity decreases down the group from Lithium to Caesium. Hence, the ionic
character of their hydrides will increase (as shown below).
LiH < NaH < CsH
(iii) Bond dissociation energy depends upon the bond strength of a molecule, which in turn
depends upon the attractive and repulsive forces present in a molecule.
The bond pair in D–D bond is more strongly attracted by the nucleus than the bond pair in H–
H bond. This is because of the higher nuclear mass of D2. The stronger the attraction, the
greater will be the bond strength and the higher is the bond dissociation enthalpy. Hence, the
bond dissociation enthalpy of D–D is higher than H–H.
However, bond dissociation enthalpy is the minimum in the case of F–F. The bond pair
experiences strong repulsion from the lone pairs present on each F-centre.
Therefore, the increasing order of bond dissociation enthalpy is as follows:
F− F < H − H < D − D

(iv) Ionic hydrides are strong reducing agents. NaH can easily donate its electrons. Hence, it is
most reducing in nature.
Both, MgH and H O are covalent hydrides. H O is less reducing than MgH since the bond
2 2 2 2

dissociation energy of H O is higher than MgH
2 2

Hence, the increasing order of the reducing property is H O < MgH < NaH
2 2

Q9.17 Compare the structures of H O and H O .
2 2 2

Answer. In gaseous phase, water molecule has a bent form with a bond angle of 104.5 . The
∘

O–H bond length is 95.7 pm. The structure can be shown as:

Hydrogen peroxide has a non-planar structure both in gas and solid phase. The dihedral angle
in gas and solid phase is 111.5 and 90.2 respectively.
∘ ∘

Q9.18 What do you understand by the term ’autoprotolysis’ of water? What is its signi cance?

Answer. Auto-protolysis (self-ionization) of water is a chemical reaction in which two water
molecules react to produce a hydroxide ion and a hydronium ion .
The reaction involved can be represented as:
H2 O(l) + H2 O(i) ⟷
+ −
H3 O + OH
(α) (aq)

Hydronium ion Hydroxide ion

Auto-protolysis of water indicates its amphoteric nature i.e., its ability to act as an acid as well
as a base.
The acid-base reaction can be written as:
H O
2 + H O
(l)
-> H O2 (l)
+3
+
OH
(aq)
−

(aq)

Q9.19 Consider the reaction of water with F and suggest, in terms of oxidation and
2

reduction, which species are oxidised/reduced.

Answer. The reaction between uorine and water can be represented as:
2F2(g) + 2H2 O(t) ⟶
+
4H + 4F(aq) + O2(g)
(aq)

This is an example of a redox reaction as water is getting oxidized to oxygen, while uorine is
being reduced to uoride ion.
The oxidation numbers of various species can be represented as:

Fluorine is reduced from zero to (– 1) oxidation state. A decrease in oxidation state indicates
the reduction of uorine.
Water is oxidized from (– 2) to zero oxidation state. An increase in oxidation state indicates

oxidation of water.

Q9.20 Complete the following chemical reactions.
PbS(s) + H2 O2 (aq) →

−
MnO (aq) + H2 O2 (aq) →
4

CaO(s) + H2 O(g) →

AlCl3 (g) + H2 O(l) →

Caa N2 (s) + H2 O(l) →

Classify the above into
(a) hydrolysis,
(b) redox and
(c) hydration reactions.

Answer. (i) PbS (s)
+ H2 O2(αq) → PbSO4(s) + H2 O(l)

H2 O 2 is acting as an oxidizing agent in the reaction. Hence, it is a redox reaction.
(ii) 2MnO −

4(oq)
+ 5H2 O2(g) → 6H
+

(oq)

2+
→ 2Mn + 8H2 O(i) + 5O2(g)
(aq)

H2 O 2is acting as a reducing agent in the acidic medium,
thereby oxidizing MnO . Hence, the given reaction is a redox reaction.
−

4(aq)

(iii) CaO (s)
+ H2 O(g) → Ca(OH)2(αq)

The reactions in which a compound reacts with water to produce other compounds are called
hydrolysis reactions. The given reaction is hydrolysis.
(iv) 2AlCl + 3H O
3(x) → Al O 2 + 6HCl
(t) 2 3(s) (αq)

The reactions in which a compound reacts with water to produce other compounds are called
hydrolysis reactions. The given reaction represents hydrolysis of AlCl . 3

(v) Ca N 3 + 6H O
2(s) → 3Ca(OH)
2 (l) + 2NH 2(aH) 3(g)

The reactions in which a compound reacts with water to produce other compounds are called
hydrolysis reactions. The given reaction represents hydrolysis of Ca N . 3 2

Q9.21 Describe the structure of the common form of ice.

Answer. Ice is the crystalline form of water. It takes a hexagonal form if crystallized at
atmospheric pressure, but condenses to cubic form if the temperature is very low.
The three-dimensional structure of ice is represented as:

The structure is highly ordered and has hydrogen bonding. Each oxygen atom is surrounded
tetrahedrally by four other oxygen atoms at a distance of 276 pm. The structure also contains
wide holes that can hold molecules of appropriate sizes interstitially.

Q9.22 What causes the temporary and permanent hardness of water ?

Answer. Temporary hardness of water is due to the presence of soluble salts of magnesium and
calcium in the form of hydrogen carbonates (MHCO , where M = Mg, Ca) in water.
3

Permanent hardness of water is because of the presence of soluble salts of calcium and
magnesium in the form of chlorides in water.

Q9.23 Discuss the principle and method of softening of hard water by synthetic ion exchange
resins.

Answer. The process of treating permanent hardness of water using synthetic resins is based
on the exchange of cations (e.g., Na , Ca , Mg etc) and anions (e.g., Cl , SO , HCO
+ 2+ 2+ − 2− −

4 3

etc) present in water by H and OH ions respectively.
+ −

Synthetic resins are of two types:
1) Cation exchange resins

2) Anion exchange resins
Cation exchange resins are large organic molecules that contain the −SO H group. The resin 3

is rstly changed to RNa (from RSO H) by treating it with NaCl. This resin then exchanges
3

Na
+
ions with Ca and Mg
2+
ions, thereby making the water soft.
2+

2+ +
2RNa + M ⟶ R2 M(s) + 2Na
(aq) (aq)

There are cation exchange resins in H form. The resins exchange H ions for
+ +

Na
+
ions with Ca and Mg
2+
ions. 2+

2+ +
2RH + M ⇄ MR2(s) + 2H
(αq) (αq)

Anion exchange resins exchange OH ions for anions like Cl , HCO , and SO present in
− − − 2−

3 4

water.
+ −
RNH2(s) + H2 O(l) ⇄ RNH ⋅ OH
3 (s)

−
↓ +X
(αq)

+ − −
RNH ⋅ X + OH
3 (s) (αq)

During the complete process, water rst passes through the cation exchange process. The
water obtained after this process is free from mineral cations and is acidic in nature.
This acidic water is then passed through the anion exchange process where OH ions −

neutralize the H ions and de-ionize the water obtained.
+

Q9.24 Write chemical reactions to show the amphoteric nature of water.

Answer. The amphoteric nature of water can be described on the basis of the following
reactions:
1) Reaction with H S 2

The reaction takes place as:
+ −
H2 O(t) + H2 S(aq) ⟹ H3 O + HS
(aq) (αq)

In the forward reaction, H O 2 (l)
accepts a proton from H S
2 (aq)
. Hence, it acts as a Lewis base.
2) Reaction with NH 3

The reaction takes place as:

In the forward reaction, H O 2 (l) denotes its proton to NH . Hence, it acts as a Lewis acid.
3

3) Self-ionization of water
In the reaction, two water molecules react as:

Q9.25 Write chemical reactions to justify that hydrogen peroxide can function as an oxidising
as well as reducing agent.

Answer. Hydrogen peroxide, H O acts as an oxidizing as well as a reducing agent in both
2 2

acidic and alkaline media.
Reactions involving oxidizing actions are:
2+ + 3+
2Fe + 2H + H2 O2 ⟶ 2Fe + 2H2 O

2+ 4+ −
2) Mn + H2 O2 ⟶ Mn + 2OH

3) PbS + 4H2 O2 ⟶ PbSO4 + 4H2 O

2+ 3+ −
4) 2Fe + H2 O2 ⟶ 2Fe + 2OH

Reactions involving reduction actions are:
− ∗ 2+
2MnO + 6H + 5H2 O2 ⟶ 2Mn + 8H2 O + 5O2
4

− −
2) I2 + H2 O2 + 2OH ⟶ 2I + 2H2 O + O2

+ −
3) HOCl + H2 O2 ⟶ H3 O + Cl + O2
−
2MnO4 +3H2 O2 −
4) ⟶ 2MnO2 + 3O2 + 2H2 O + 2OH

Q9.26 What is meant by ‘demineralised’ water and how can it be obtained ?

Answer. Demineralised water is free from all soluble mineral salts. It does not contain any
anions or cations.
Demineralised water is obtained by passing water successively through a cation exchange (in
the H form) and an anion exchange (in the OH form) resin.
+ −

During the cation exchange process,H exchanges for Na , Mg , Ca , and other cations
+ + 2+ 2+

present in water.
2RH + M
(s)
⇄ MR
2+
+ 2H
(aq)
……(1) 2(s)
+

(aq)

In the anion exchange process, OH exchanges for anions such as CO − 2−

3
, SO
2−

4
, Cl
−
, HCO
−

3

etc. present in water.
+ −
RNH2(s) + H2 O(t) ⇄ RNH ⋅ OH
3 (s)

+ − −
RNH OH + X ⇄
3 (s) (aq)

RNH
+

3
⋅ X
−

(s)
+ OH
−

(aq)
........(2)
ions liberated in reaction (2) neutralize H ions liberated in reaction (1), thereby
− +
OH

forming water.
+ −
H + OH ⟶ H2 O(l)
(aq) (aq)

Page : 298 , Block Name : Exercise

Q9.27 Is demineralised or distilled water useful for drinking purposes? If not, how can it be
made useful?

Answer. Water is an important part of life. It contains several dissolved nutrients that are
required by human beings, plants, and animals for survival. Demineralised water is free of all
soluble minerals. Hence, it is not t for drinking.
It can be made useful only after the addition of desired minerals in speci c amounts, which
are important for growth.

Page : 298 , Block Name : Exercise

Q9.28 Describe the usefulness of water in biosphere and biological systems.

Answer. Water is essential for all forms of life. It constitutes around 65% of the human body
and 95% of plants. Water plays an important role in the biosphere owing to its high speci c
heat, thermal conductivity, surface tension, dipole moment, and dielectric constant.
The high heat of vapourization and heat of capacity of water helps in moderating the climate
and body temperature of all living beings.
It acts as a carrier of various nutrients required by plants and animals for various metabolic
reactions.

Page : 298 , Block Name : Exercise

Q9.29 What properties of water make it useful as a solvent? What types of compound can it
(i) dissolve, and
(ii) hydrolyse ?

Answer. A high value of dielectric constants (78.39 C2/Nm2) and dipole moment make water a
universal solvent.
Water is able to dissolve most ionic and covalent compounds. Ionic compounds dissolve in
water because of the ion-dipole interaction, whereas covalent compounds form hydrogen
bonding and dissolve in water.
Water can hydrolyze metallic and non-metallic oxides, hydrides, carbides, phosphides,
nitrides and various other salts. During hydrolysis, H+ and OH– ions of water interact with the
reacting molecule.
Some reactions are:
CaO + H2 O ⟶ Ca(OH)2

NaH + H2 O ⟶ NaOH + H2

CaC2 + H2 O ⟶ C2 H2 + Ca(OH)2

Page : 298 , Block Name : Exercise

Q9.30 Knowing the properties of H2O and D2O, do you think that D2O can be used for
drinking purposes?

Answer. Heavy water (D O) acts as a moderator, i.e., it slows the rate of a reaction. Due to this
2

property of D O, it cannot be used for drinking purposes because it will slow down anabolic
2

and catabolic reactions taking place in the body and lead to a casualty.

Page : 298 , Block Name : Exercise

Q9.31 What is the difference between the terms ‘hydrolysis’ and ‘hydration’ ?

Answer. Hydrolysis is de ned as a chemical reaction in which hydrogen and hydroxide ions
(H+ and OH– ions) of water molecule react with a compound to form products. For example:
NaH + H2 O ⟶ NaOH + H2

Hydration is de ned as the addition of one or more water molecules to ions or molecules to
form hydrated compounds. For example:
CuSO4 + 5H2 O ⟶ CuSO4 ⋅ 5H2 O

Page : 298 , Block Name : Exercise

Q9.32 How can saline hydrides remove traces of water from organic compounds?

Answer. Saline hydrides are ionic in nature. They react with water to form a metal hydroxide
along with the liberation of hydrogen gas. The reaction of saline hydrides with water can be
represented as:
AH(s) + H2 O(η) ⟶ AOH(αq) + H2(g)

(where, A = Na, Ca,……)
When added to an organic solvent, they react with water present in it. Hydrogen escapes into
the atmosphere leaving behind the metallic hydroxide. The dry organic solvent distills over.

Page : 298 , Block Name : Exercise

Q9.33 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 behaviour towards water.

Answer. The elements of atomic numbers 15, 19, 23, and 44 are phosphorus, potassium,
vanadium, and ruthenium respectively.
1) Hydride of phosphorus
Hydride of nitrogen (PH ) is a covalent molecule. It is an electron-rich hydride owing to the
3

presence of excess electrons as a lone pair on phosphorus.

(2) Hydride of potassium

Dihydrogen forms an ionic hydride with potassium owing to the high electropositive nature of
potassium. It is crystalline and non-volatile in nature.
3) Hydrides of Vanadium and Ruthenium
Both vanadium and ruthenium belong to the d–block of the periodic table. The metals of d–
block form metallic or non–stoichiometric hydrides. Hydrides of vanadium and ruthenium are
therefore, metallic in nature having a de ciency of hydrogen.
4) Behaviour of hydrides towards water
Potassium hydride reacts violently with water as: KH + H O ⟶ KOH + H
(s) 2 (αq) (αq) 2(g)

Phosphorus (PH ) is covalent hydride and slightly soluble in water.
3

Hydrides of vanadium and Ruthenium do not react with water. Hence, the increasing order of
reactivity of the hydrides is (V, Ru) H < NH < KH 3

Page : 298 , Block Name : Exercise

Q9.34 Do you expect different products in solution when aluminium(III) chloride and
potassium chloride treated separately with (i) normal water (ii) acidi ed water, and (iii)
alkaline water? Write equations wherever necessary.

Answer. Potassium chloride (KCl) is the salt of a strong acid (HCl) and strong base (KOH).
Hence, it is neutral in nature and does not undergo hydrolysis in normal water. It dissociates
into ions as follows:
KCl -> K
(s)
+ Cl
+

(aq)
−

(aq)

In acidi ed and alkaline water, the ions do not react and remain as such.
Aluminium (III) chloride is the salt of a strong acid (HCl) and weak base [Al(OH)3]. Hence, it
undergoes hydrolysis in normal water.
-> Al(OH) + −
AlCl + 3H O
3(s) 2 (l) + 3H + 3Cl
3(s) (aq) (aq)

In acidi ed water, H ions react with Al(OH) forming water and giving Al ions. Hence, in
+ 3+
3

acidi ed water, AlCl will exist as Al 3
3+
andCl ions.
−

Acidified
AlCl3(s)
Water
3+ −
Al + 3Cl
(aq) (aq)

In alkaline water, the following reaction takes place:
− −
Al(OH)3(x) + OH ⟶ [Al(OH)4 ] + 2H2 O(l)
(aq) (aq)

Page : 298 , Block Name : Exercise

Q9.35 How does H O behave as a bleaching agent?
2 2

Answer. H O or hydrogen peroxide acts as a strong oxidizing agent both in acidic and basic
2 2

media.
When added to a cloth, it breaks the chemical bonds of the chromophores (colour producing
agents). Hence, the visible light is not absorbed and the cloth gets whitened.

Page : 298 , Block Name : Exercise

Q9.36 What do you understand by the terms:
(i) hydrogen economy
(ii) hydrogenation
(iii) ‘syngas’
(iv) water-gas shift reaction
(v) fuel-cell ?

Answer. (i) Hydrogen economy
Hydrogen economy is a technique of using dihydrogen in an ef cient way. It involves
transportation and storage of dihydrogen in the form of liquid or gas.
Dihydrogen releases more energy than petrol and is more eco–friendly. Hence, it can be used
in fuel cells to generate electric power. Hydrogen economy is about the transmission of this
energy in the form of dihydrogen.
(ii) Hydrogenation
Hydrogenation is the addition of dihydrogen to another reactant. This process is used to
reduce a compound in the presence of a suitable catalyst. For example, hydrogenation of
vegetable oil using nickel as a catalyst gives edible fats such as vanaspati, ghee etc.
(iii) Syngas
Syngas is a mixture of carbon monoxide and dihydrogen. Since the mixture of the two gases is
used for the synthesis of methanol, it is called syngas, synthesis gas, or water gas.
Syngas is produced on the action of steam with hydrocarbons or coke at a high temperature in
the presence of a catalyst.
= nCO + (2n + 1)H For example :
1270K
C H
n 2n+2 + nH O
2 2
Ni
1270K
CH4(g) + H2 O(g) = CO(g) + 3H2(g)
Ni

Syn gas

(iv) Water shift reaction
It is a reaction of carbon monoxide of syngas mixture with steam in the presence of a catalyst
as:
673K
CO(g) + H2 O(g) CO2(g) + H2(g)
Catalyst

This reaction is used to increase the yield of dihydrogen obtained from the coal gasi cation
reaction as:
C(s) + H2 O(g) ⟶ CO(g) + H2(g)

(v) Fuel cells
Fuel cells are devices for producing electricity from fuel in the presence of an electrolyte.
Dihydrogen can be used as a fuel in these cells. It is preferred over other fuels because it is
eco-friendly and releases greater energy per unit mass of fuel as compared to gasoline and
other fuels.

Page : 298 , Block Name : Exercise