Std. XI Sci. Success Chemistry - II
09
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HYDROGEN
HORIZON Publications
Contents
9.0
Prominent scientist
9.1
Introduction, position of hydrogen in periodic table
9.2
Occurrence of hydrogen (dihydrogen)
9.3
Isotopes of hydrogen
9.4
Preparation of dihydrogen
9.5
Properties of dihydrogen
9.6
Uses of dihydrogen
9.7
Hydrides (ionic, covalent, interstitial)
9.8
Water (physical and chemical properties of water)
9.9
Heavy water
9.10 Hydrogen peroxide (preparation properties and structure)
9.11 Hydrogen as a fuel
Formulae
Solved Problems
Multiple Choice Questions
9.0 Prominent Scientists
i.
ii.
iii.
Name of the Scientist
Aristotle (Greek Philosopher)
Alchemist
Henry Cavendish
iv.
Antoine Lavoisier
Contributions
Water is one of the four elementary components of nature.
Regarded water as an element.
Considered water as combustion product of hydrogen,
Discovered hydrogen
Discovered water as a compound of hydrogen and oxygen
9.1 Introduction
*Q.1. What is another name of hydrogen?
Ans. Hydrogen, in the elemental form exists as a diatomic molecule (H2) and hence, it is called
dihydrogen.
*Q.2. Explain the position of hydrogen in the periodic table on the basis of its electronic
configuration.
Ans. i.
Hydrogen has atomic number 1 and electronic configuration 1s1, similar to the outermost
electronic configuration of alkali metals i.e. ns1.
ii. It forms univalent ion similar to alkali metal ions.
iii. It forms oxides, halides and sulphides like alkali metals.
iv. Hydrogen resembles halogen of group 17 in the properties like, high ionization enthalpy,
formation of uninegative ion (H–), formation of hydrides with many elements, formation of a
large number of covalent compounds, existence as a diatomic molecules (H2).
v.
Loss of e– from hydrogen results into the nucleus having very small size of – 1.5 × 10–3 pm and
H+ ions never exists freely and are always associated with other atoms or molecules.
vi. Hence, due to unique behaviour of hydrogen, it is placed separately in the periodic table.
Hydrogen
9.1
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Note :
9.2 Occurrence
*Q.3. How does hydrogen occur in nature ?
Ans. i.
Hydrogen is the main element of the solar system.
ii. In the combined form it constitutes about 15.4% of earth’s crust and oceans.
iii. The earth’s atmosphere contains about 0.15% of hydrogen by mass.
iv. It occurs in plant and animal tissues as carbohydrates and proteins.
v.
It is present in hydrides, hydrocarbons and many other compounds.
vi. It is the third abundant element in the earth’s crust.
*Q.4. Why does hydrogen occur in a diatomic form rather than monoatomic form under normal
conditions?
Ans. i.
The atomic number of hydrogen is one and it has one electron in its valence shell having
electronic configuration 1s1.
ii. It can acquire stable configuration of helium (i.e. atomic number 2 electronic configuration 1s2)
by sharing one e– with another H atom.
iii. Thus, two hydrogen atoms are bonded covalently forming a diatomic, H2 molecule.
iv. The monoatomic hydrogen is not stable under normal conditions.
Hydrogen
9.2
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HORIZON Publications
9.3 Isotopes of Hydrogen
Note :
Property
Active abundance %
Relative atomic mass (g mol–1)
Melting point/K
Boiling point/K
Density/g L–1
Enthalpy of fusion/kJ mol–1
Enthalpy of vapourization/kJ mol–1
Enthalpy of bond dissociation/kJ mol–1 at 298.3 K
Internuclear distance/pm
Ionization enthalpy/kJ mol–1
Electron gain enthalpy/kJ mol–1
Covalent radius/pm
Ionic radius (H–)/pm
Hydrogen
99.985
1.008
13.96
20.39
0.09
0.117
0.904
435.88
74.14
1312
– 73
37
208
Deuterium
0.0156
2.014
18.73
23.67
0.18
0.197
1.226
443.35
74.14
-
Tritium
10–15
3.016
20.62
25.0
0.27
-
*Q.5. Mention the isotopes of hydrogen.
Ans. Hydrogen has three isotopes.
i.
Protium, 1H1
ii. Deuterium, 1H2 or D
iii. Tritium 1H3 or T
*Q.6. Write a brief note on isotopes of hydrogen.
Ans. i.
Hydrogen has three isotopes.
a.
Protium or Hydrogen, 1H1, which is a predominant form of isotope. It has no neutron in
its nucleus.
b.
Deuterium, 1H2 or D, which is known as heavy hydrogen. It has one neutron in its
nucleus.
c.
Tritium, 1H3 or T which is a radioactive isotope with half life 12.33 years. It has two
neutrons in its nucleus.
ii. Since these isotopes have identical electronic configuration their chemical properties are
identical. They have different physical properties due to different masses. This effect is known
as isotopic effect.
iii. The heavy isotopes, deuterium and tritium are used for nuclear fusion, and as tracers in the
study of reaction mechanisms.
9.4 Preparation of Dihydrogen, H2
*Q.7. Explain laboratory methods for the preparation of dihydrogen.
OR
Explain briefly the laboratory preparation for pure hydrogen.
Ans. Laboratory method for preparation of hydrogen:
i.
It is usually prepared by the reaction of zinc with aqueous alkali.
Zn + 2NaOH 
→ Na2ZnO2 + H2↑
ii. It can also be prepared by the reaction of granulated zinc with dilute hydrochloric acid.
Zn + 2HCl 
→ ZnCl2 + H2↑
Hydrogen
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Std. XI Sci. Success Chemistry - II
iii.
iv.
v.
Q.8.
Ans.
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Very pure hydrogen gas is prepared by the action of pure dil. H2SO4 on magnesium ribbon.
Mg + H2SO4(dil.) 
→ MgSO4 + H2↑
Pure hydrogen gas is also prepared by action of water on sodium hydride.
NaH + H2O 
→ NaOH + H2↑
It is also prepared by action of KOH on scrap aluminium or silicon.
2Al + 2KOH + 2H2O 
→ 2KAlO2 + 3H2↑
This method gives very pure hydrogen and is known as Uyeno’s method.
How is pure hydrogen obtained by Uyeno’s method?
Pure hydrogen can be prepared by the action of potassium hydroxide (KOH) on scrap aluminium or
silicon.
2Al + 2KOH + 2H2O 
→ 2KAlO2 + 3H2↑
This method gives very pure hydrogen and is known as Uyeno’s method.
*Q.9. Describe the bulk preparation of dihydrogen by electrolytic method. What is the role of an
electrolyte in the process?
OR
How is dihydrogen prepared by electrolysis of acidified water?
Ans. i.
Hydrogen can be obtained by electrolysis of acidified water using platinum electrodes.
ii. Dihydrogen is liberated at cathode while dioxygen (O2) is liberated at anode.
iii. Since, pure water contains covalent bond, it cannot conduct electricity.
iv. Hence, acid (Sulphuric acid) helps in conduction of electricity by ionizing water molecules.
Thus electrolyte helps in conduction of electricity.
+
2−
Acidic medium :
H2SO4  
 2H + SO 4
v.
+
–
H2O  
 H + OH
At cathode :
2H+ + 2e– 
→ 2H; H + H 
→ H2↑
–
–
At anode :
4OH 
→ 4OH + 4e ; 4OH 
→ 2H2O + O2↑
Since, the discharge potential of sulphate ions is much higher than that of hydrogen ions,
sulphate ions are not discharged at the anode.
Role of an electrolyte : The electrolyte used is acidified water. Since pure water is a poor
conductor of electricity, it is acidified with dilute sulphuric acid to increase its electrical
conductivity.
Q.10. How can pure H2 be obtained from barium hydroxide?
Ans. Electrolysis of warm aqueous solution of barium hydroxide between nickel electrodes gives high
purity H2.
Q.11. Describe Bosch process of preparation of H2.
OR
How water gas can be used to prepare H2 ?
Ans. i.
In Bosch process, water gas (CO + H2) is mixed with twice the volume of steam and the mixture
is passed over heated catalyst (Fe2O3) at 773 K in the presence of a promoter (Cr2O3) or (ThO2),
here, CO is oxidized to CO2.
Fe2O3 + Cr2O3
→ CO2↑ + 2H2
CO + H2 + H2O 
773 K
ii.
Hydrogen
Water gas
Steam
Carbon dioxide is removed by dissolving it in water under pressure (20-25 atmospheres) and
hydrogen left behind is collected.
9.4
Std. XI Sci. Success Chemistry - II
HORIZON Publications
Q.12. Explain how is dihydrogen obtained by the electrolysis of brine solution ?
Ans. i.
Dihydrogen is obtained as a by-product in the manufacture of sodium hydroxide and chlorine by
electrolysis of brine solution (NaCl solution).
ii. Preferentially H+ ions are reduced at cathode liberating hydrogen gas at cathode, while Cl– ions
are oxidised at anode liberating Cl2 gas.
iii. On the electrolysis of brine (sodium chloride) solution; the following reactions take place.
NaCl 
→ Na+ + Cl–
+
–
H2O  
 H + OH
Reactions at cathode
Reactions at anode
+
–
2H + 2e 
2Cl– 
→ 2H;
→ 2Cl + 2e–;
2H 
2Cl 
→ H2(g);
→ Cl2(g)
+
–
iv. Na and OH ions remain in the solution which give sodium hydroxide (NaOH).
Q.13. Give preparation of H2 using coke or hydrocarbon.
OR
How is H2 prepared from coke or hydrocarbons ?
Ans. Reaction of steam on hydrocarbons or coke at high temperatures in the presence of catalyst yields
hydrogen.
1270 K
CnH2n + 2 + nH2O 
→ nCO(g) + (2n + 1)H2↑
Ni
1270 K
Eg. CH4(g) + H2O(g) 
→ CO(g) + 3H2(g) ↑
Ni
Q.14. Explain Lane’s process.
OR
What is Lane’s process ?
OR
How is hydrogen prepared from steam ?
Ans. Reaction of steam on coke or hydrocarbons in the presence of nickel catalyst, at high temperature
1270 K yields hydrogen.
∆
C(s) + H2O 
→ CO(g) + H2(g)
Ni
1270 K
CH4(g) + H2O 
→ CO(g) + 3H2(g)
Ni
*Q.15. Explain the terms :
i.
Syngas
ii. Water gas shift reaction.
Ans. i.
Syngas :
a.
Syngas is the mixture of CO and H2. It is also called water gas.
b.
It is used for the synthesis of methanol and many hydrocarbons, hence, called syngas or
synthesis gas.
c.
It is produced from saw dust or scrap wood.
d.
Methanol can be prepared from syngas.
∆
CO(g) + 2H2(g) →
CH3 – OH
Catalyst
e.
It is used to prepare hydrogen.
Fe2O3 + Cr2O3
CO + H2 + H2O 
→ CO2↑ + 2H2(g)
773 K
Water
Hydrogen
gas
Steam
9.5
Std. XI Sci. Success Chemistry - II
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Water gas shift reaction :
a.
Production of dihydrogen can be increased by reacting carbon monoxide of syngas
mixtures with steam in presence of iron chromate as catalyst.
673 K
CO(g) + H2O(g) →
CO2(g) + H2(g)
Catalyst
b.
This is “water-gas shift reaction”.
Carbon dioxide thus formed is removed by scrubbing with sodium arsenite solution.
Q.16. Complete the following reactions.
*i.
∆
?
CO(g) + H2O(g) →
Catalyst
∆
*ii. Zn(s) + 2NaOH 
→ ?
1270 K
*iii. CH4(g) + H2O 
→ ?
Ans.
iv.
NaH + H2O 
→ ?
v.
∆
3Fe + 4H2O 
→?
Fe
vi.
Mg + H2SO4(dil.) 
→ ?
i.
∆
CO(g) + H2O(g) →
CO2(g) + H2(g)
Catalyst
ii.
Steam
∆
Zn(s) + 2NaOH 
→ Na2ZnO2 + H2(g)
iii.
iv.
1270 K
CH4(g) + H2O 
→ CO(g) + 3H2(g)
NaH + H2O 
→ NaOH + H2↑
v.
∆
3Fe + 4H2O 
→ Fe3O4 + 4H2(g)
Fe
i.
Mg + H2SO4(dil.) 
→ MgSO4 + H2↑
9.5 Properties of Dihydrogen
*Q.17. What are the physical properties of dihydrogen?
Ans. i.
Chemical reactivity of dihydrogen involves the breaking of H – H bond.
ii. Dihydrogen is a colourless, odourless, tasteless, non–poisonous gas.
iii. It is combustible but non supportive of combustion.
iv. It is lighter than air.
v.
It is insoluble in water.
vi. It is neutral to litmus.
vii. The bond dissociation enthalpy of H – H bond is very high (435.8 kJ mol–1) so very high
energy is required to break the H – H bond.
viii. Therefore, dihydrogen is not very reactive under normal conditions.
*Q.18. What is the effect of high enthalpy of H – H bond in terms of chemical reactivity of
dihydrogen?
Ans. Chemical reactivity of dihydrogen involves the breaking of H – H bond.
i.
The bond dissociation enthalpy of H – H bond is very high (435.88 KJ mol–1 at 298 K).
ii. Due to high bond enthalpy, it is not very reactive at room temperature.
iii. However, at high temperature or in the presence of catalysts, hydrogen combines with many
metals and non-metals to form hydrides.
Hydrogen
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Std. XI Sci. Success Chemistry - II
HORIZON Publications
Q.19. Describe the chemical behaviour of dihydrogen.
OR
Give reason atomic hydrogen is produced at a high temperature.
Ans. i.
Due to high bond dissociation enthalpy (435.88 kJ mol–1 at 298 K) hydrogen is not very active. It
is relatively inert at room temperature.
ii. Thus, the atomic hydrogen is produced at a high temperature in an electric arc or under
ultraviolet radiations.
iii. Since, its orbital is incomplete with 1s1 electronic configuration, it does combine with almost
all the elements.
iv. Many reactions of hydrogen are due to
a.
loss of electron to form H+ ion,
b.
gain of an electron to form H– ion and
c.
sharing of electrons to form single covalent bond.
Q.20. What is the reaction of H2 with the following ?
i.
Halogen
ii. O2(g)
iii. N2(g)
iv. Na(s)
v.
Pd2+
vi. Metal oxide
Ans. i.
Reaction with halogen :
H2 + X2 
→ 2HX (X = F, Cl, Br, I)
Fluorine reacts violently with hydrogen even in the dark and at very low temperature to form
hydrogen fluoride.
63 K
H2 + F2 
→ 2HF
Chlorine reacts with H2 gas in the presence of U.V. light forming HCl.
U.V.
H2(g) + Cl2(g) 
→ 2HCl(g)
Iodine requires presence of catalyst.
ii.
Reaction with dioxygen (O2) : Hydrogen reacts with dioxygen to form water and the reaction
is highly exothermic.
Catalyst
2H2(g) + O2(g) →
2H2O()
iii.
Reaction with dinitrogen (N2) : Hydrogen reacts with dinitrogen in presence of Fe catalyst at
673 K and 200 atmospheric pressure to form ammonia.
Fe, 673 K
3H2(g) + N2(g) 
∆H = – 92.6 kJ mol–1
→ 2NH3(g);
200 atm.
This is the Haber’s process for the manufacture of ammonia.
iv.
Reaction with sodium metal (Na) : Hydrogen combines with metallic sodium to form sodium
hydride.
2Na + H2 
→ 2NaH
Sodium hydride
v.
Reaction with Pd2+ : Hydrogen reduces Pd2+ ions in aqueous solution to palladium metal.
Pd (2aq+ ) + H2(g) 
→ Pd(s) + 2H (+aq )
vi.
Reaction with metal oxides : Hydrogen reduces certain metal oxides to corresponding metals.
Example :
CuO + H2 
→ Cu + H2O
ZnO + H2 
→ Zn + H2O
Fe3O4 + 4H2 
→ 3Fe + 4H2O
Hydrogen
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Q.21. “H2 reacts with fluorine even at low temperature”. Explain.
Ans. i.
Hydrogen reacts with halogens (X2) to give hydrogen halides (HX).
H2(g) + X2(g) 
(X = F, Cl, Br, I)
→ 2HX(g)
Halogen halide
ii.
iii.
iv.
The reaction with fluorine occurs even in dark.
The reaction with iodine requires a catalyst.
Fluorine has great affinity for hydrogen and combines at very low temperature (63 K) forming
hydrogen fluoride.
H2 + F2 
→ 2HF
Q.22. Fill in the blanks to complete the following reactions.
dark
i.
H2 + F2 
→ ……………
63 K
Ans. 2HF
ii.
Catalyst
2H2 + O2 →
………..
or ∆
Ans. 2H2O
iii. …….. + H2 
→ 2NaH
Ans. 2Na
iv. H2 + ……… 
→ Cu + H2O
Ans. CuO
*Q.23. Explain the term Hydrogenation.
Ans. i.
The reaction in which hydrogen gas reacts with unsaturated organic compounds in the presence
of a catalyst to form hydrogenated (saturated) compounds is called hydrogenation.
ii. Hydrogenation gives some useful products of commercial importance.
Example :
i.
Hydrogenation of vegetable oils form fats (where unsaturated triglycerides are converted to
saturated triglycerides).
Finely divided Ni
fat
Vegetable oil + H2 →
450, 8−10 atm
ii.
iii.
Vanaspati ghee and margarine are obtained by hydrogenation.
Hydrogenation of olefins gives aldehydes which on further reduction give alcohols.
H2 + CO + CH3 – CH = CH2 
→ CH3 – CH2 – CH2 – CHO
Olefin (propylene)
n-butyraldehyde
H2 + CH3 – CH2 – CH2 – CHO 
→ CH3 – CH2 – CH2 – CH2 – OH
n-butyl alcohol
9.6 Uses of Dihydrogen
*Q.24. What are the uses of H2 ?
Ans. Following are the uses of dihydrogen :
i.
In the manufacture of vanaspati Ghee (fat) by hydrogenation of polyunsaturated vegetable oils
like soyabean, cotton seed, etc.
ii. In the synthesis of ammonia by Haber’s process.
iii. In the manufacture of methanol, hydrogen chloride and metal hydrides.
iv. In fuel cells for generating electrical energy.
v.
In metallurgy to reduce oxides of heavy metals.
vi. In atomic hydrogen and oxyhydrogen torches.
vii. Liquid hydrogen mixed with liquid oxygen is used as a rocket fuel to generate very high
temperatures of about 4000 K by allowing H atoms to combine forming H2 gas.
Hydrogen
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Std. XI Sci. Success Chemistry - II
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*Q.25. How does atomic hydrogen or oxyhydrogen torch function for cutting and welding ?
Ans. i.
For cutting and welding of metals very high temperature is required.
ii. By dissociation of dihydrogen with the help of an electric arc, atomic hydrogen is produced.
iii. Atoms of hydrogen are then allowed to recombine on the surface to be welded which generate
very high temperatures (around 4000 K), since the formation of dihydrogen from hydrogen
atoms is an exothermic reaction.
2H(g) 
→ H2(g); ∆H = – 435.8 kJ mol–1
Q.26. Mention the benefits of using H2 in fuel cell for generating electrical energy.
Ans. i.
Using dihydrogen in fuel cell has many advantages over conventional fossil fuel and electric
power.
ii. It does not produce pollution and releases greater energy per unit mass of fuel in comparison to
gasoline and other fuels.
9.7 Hydrides
*Q.27. What are hydrides and Mention different types of hydrides.
Ans. Hydrides : The binary compounds of hydrogen and other elements are called hydrides.
Eg. : LiH, MgH2, B2H6
Types of hydrides :
i.
Ionic or saline or salt like hydrides. Eg. LiH, MgH2
ii. Covalent or molecular hydrides. Eg. H2O, NH3
iii. Metallic or non stoichiometric or interstitial hydrides. Eg. TiH1.8 – 2.0, LaH2.87
*Q.28. Explain ionic hydrides.
Ans. i.
Ionic hydrides are stoichiometric compounds with highly electropositive character formed by the
reaction with dihydrogen and s-block elements.
Eg. : LiH, MgH2
ii. They are also called saline or salt like hydrides.
iii. BeH2 and MgH2 are polymeric in structure.
iv. In the solid state, they do not conduct electricity.
However, in the molten state they conduct electricity and liberate dihydrogen at anode,
indicating the existence of H– ion.
anode
2H– 
→ H2(g) + 2e–
v.
They react violently with water producing dihydrogen gas.
KH(s) + H2O(aq) 
→ KOH(aq) + H2(g)
vi. LiH is used in the synthesis of hydrides like Lithium aluminium hydride (LiAlH4) or Lithium
borohydride (LiBH4) which are good reducing agents.
8LiH + Al2Cl6 
→ 2LiAlH4 + 6LiCl
2LiH + B2H6 
→ 2LiBH4
*Q.29. Explain covalent hydrides.
Ans. Covalent or molecular hydrides are formed by a reaction of dihydrogen with many of p-block
elements. Eg. : H2O, CH4, NH3, HF.
According to the number of electrons in their Lewis structures, they are classified into (i) electron
deficient (ii) electron precise and (iii) electron rich hydrides.
i.
Electron deficient hydrides : These hydrides have very few electrons for writing their
conventional Lewis structure. They are incomplete octet molecules. As they can accept
electron pair, they are Lewis acids.
Eg. : Diborane (B2H6).
Hydrogen
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iii.
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Electron precise hydrides : They have the required number of electrons to write their Lewis
structure.
Eg. : Methane (CH4).
Electron rich hydrides : They have excess electrons which are present as lone pairs. They are
electron pair donors and therefore behave as Lewis bases.
Eg. : NH3, H2O, HF.
The presence of lone pair of electrons on electronegative atoms like N (1 lone pair), O (2 lone
pairs), F (3 lone pairs) in these hydrides, result in the formation of hydrogen bond between the
molecules. This leads to association of molecules and high boiling points.
Q.30. Explain non-stoichiometric or interstitial hydrides
Ans. i.
Non-stoichiometric hydrides or interstitial hydrides are hydrides of many d and f block metals.
Eg. : TiH1.8 – 2.0, LaH2.87
ii. They are deficient in hydrogen as they do not bear the exact stoichiometric ratio. Law of
constant proportion does not hold good for these hydrides.
iii. They are also called interstitial hydrides or metal hydrides as hydrogen occupies interstices in
the metal lattice.
*Q.31. Explain the following terms with suitable examples.
i.
Electron deficient hydrides
ii. Electron precise hydrides
iii. Electron rich compounds of hydrogen
Ans. i.
Electron deficient hydrides : These hydrides have very few electrons for writing their
conventional Lewis structure. They are incomplete octet molecules. As they can accept
electron pair, they are Lewis acids.
Eg. : Diborane (B2H6).
ii. Electron precise hydrides : They have the required number of electrons to write their Lewis
structure.
Eg. : Methane (CH4).
iii. Electron rich hydrides : They have excess electrons which are present as lone pairs. They are
electron pair donors and therefore behave as Lewis bases.
Eg. : NH3, H2O, HF.
Q.32. Among NH3, H2O and HF which would you expect to have the higher magnitude or hydrogen
bonding and why ?
[Intext Question text book page no. 207]
Ans. HF will have the highest magnitude of hydrogen bonding, which can be explained as follows:
i.
Hydrogen bond is formed when difference in the Electronegativity of hydrogen atom and the
other atom is large.
ii. F is the most electronegative atom among N, O and F.
iii. Hence among the hydrides NH3, H2O and HF, the Electronegativity difference between H and
F atoms is the greatest.
iv. Hence, HF has the highest magnitude of hydrogen bonding.
+δ
–δ
+δ
–δ
+δ
–δ
…H–F…H–F…H–F…
Q.33. “Metallic hydrides are also called interstitial hydrides”. Give reason.
Ans. i.
In metallic hydrides, hydrogen occupies interstices in the metal lattice producing distortion
without any change in its type.
ii. Due to this, interstitial solids are formed.
iii. Hence metallic hydrides are termed as interstitial hydrides.
Hydrogen
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Q.34. Distinguish between Ionic and Covalent hydrides.
Ans.
No.
Ionic hydrides
Covalent hydrides
i.
Ionic hydrides are formed by s-block Covalent hydrides are formed by p-block
elements.
elements.
ii.
Ionic hydrides are crystalline, non-volatile Covalent hydrides are amorphous, volatile
and non-conducting in solid state.
compounds.
iii.
Formed generally by metals.
Formed by non-metals.
9.8 Water
Q.35. “Chemistry of world is chemistry of water”. Justify.
Ans. i.
Water is the most common abundant and easily obtainable of all chemical compounds. It is a
crucial compound for man’s survival.
ii. Water is the nature’s most precious and wonderful gift to mankind.
iii. It is an universal solvent and can exist in solid, liquid and gaseous form.
iv. Water is the most essential commodity for the existence of plants, animals and man.
v.
A major part of all living organisms is made up of water.
vi. Human body has about 65% and some plants have 95% water.
vii. It is used as cleansing agent in speeding up chemical action and the basic material needed by
almost all industries.
viii. Hence, life on earth would be totally impossible in the absence of water.
ix. Thus, it is rightly said that the “Chemistry of world is the chemistry of water”.
Q.36. Write the estimated source of H2O supply of the world.
Ans. The distribution of water over the earth’s surface is not uniform.
The estimated source of water supply is given in table below.
Source
Oceans
Saline lakes and inland seas
Polar ice and glaciers
Under ground water
Lakes
Soil moisture
Water vapour in atmosphere
Rivers
% of Total
97.33
0.008
2.04
0.61
0.009
0.005
0.001
0.0001
Q.37. Mention the important physical properties of water.
Ans. Physical properties of water :
i.
Water is a colourless, tasteless liquid.
ii. Pure water is a poor conductor of electricity.
iii. It has high dielectric constant (78.39).
iv. Its boiling point is 373 K and freezing point 273 K.
v.
Its density is 1.00 g cm–3.
vi. There is extensive hydrogen bonding between water molecules.
Q.38. Why is water regarded as a universal solvent ?
Ans. i.
Water has a high dielectric constant.
ii. Hence, it has high ability to dissolve most of the inorganic (ionic) compounds.
Hydrogen
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Std. XI Sci. Success Chemistry - II
iii.
iv.
v.
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It dissolves some organic covalent compounds like alcohols, carboxylic acids due to the
formation of hydrogen bonding.
It is available in huge amounts.
Thus, water is regarded as a universal solvent.
Q.39. How does unique properties of water help to maintain this biosphere?
Ans. i.
The high heat of vapourization and high heat of fusion capacity of water is responsible for
moderation of climate and body temperature of living organisms.
ii. In comparison with other liquids, water has a high specific heat, thermal conductivity, surface
tension, dipole moment and dielectric constant.
iii. Water is excellent solvent for transportation of ions and molecules, needed for plant and animal
metabolism.
iv. Even covalent organic compounds like alcohol, glucose, sugar, urea have high solubility in
water because of their ability to form H-bonds with water while the insolubility of some ionic
compounds like AgCl, BaSO4, CaF2, AlF3 in water is due to their exceptionally high attractive
forces in the lattice.
v.
Due to these unique properties, water plays a key role in the biosphere.
Q.40. Why does water shows unusual properties in the condensed phase ?
Ans. i.
The unusual properties of water in the condensed phases like liquid and solid phase are due to
the presence of extensive hydrogen bonding between water molecules.
ii. The freezing point, boiling point, heat of vapourization and heat of fusion are abnormally
higher than those of hydrides of other elements of group 16 (H2S, H2Se).
iii. This is due to the presence of hydrogen bonding in H2O molecule.
Q.41. Describe hydrogen bonding in water.
Ans. i.
Due to the polar nature of water molecule, the water molecules are held together by
intermolecular hydrogen bonds in water.
ii. As a result, the water molecules are joined together in an extensive three dimensional network as
shown in the figure below.
iii. In this arrangement each oxygen is tetrahedrally surrounded by four hydrogen atoms, two by
covalent bonds and two by hydrogen bonds.
H
O H
H
H
O
H
O
H
H
H
O
O
H
Hydrogen bonds
H
H
O
H
O H
H
Hydrogen
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Std. XI Sci. Success Chemistry - II
HORIZON Publications
Q.42. Describe the structure of water with suitable diagram.
Ans. i.
Water molecule has two hydrogen atoms and one oxygen atom.
ii. The oxygen atom undergoes sp3 hybridisation and two O – H bonds are formed by sp-s overlapping.
iii. There are two lone pairs on oxygen atom which gives a bent or angular structure to H2O
molecule.
iv. The H – O – H bond angle is 104°35′ and the O – H bond length is 95.7 pm.
O
H
2δ–
O
95.7 pm
104°35′
δ+
H
H
δ+
H
(b)
(a)
v.
H
(c)
H
In the gaseous state water exists as discrete molecules but in the liquid state they are held
together by inter molecular hydrogen bonds. Hence water has higher boiling point (373 K).
H
O H
H
H
O
O H
H
H
H
O
O
H
Hydrogen bonds
H
H
O H
H
O H
Q.43. Explain the structure of ice.
Ans. i.
The crystalline form of water is ice.
ii. Ice has a highly ordered three dimensional hydrogen bonded structure.
iii. X-ray study of ice shows that each oxygen atom is surrounded tetrahedrally by four hydrogen
atoms at a distance of 276 pm.
iv. Hydrogen bonding gives ice an open cage type structure with wide holes.
v.
These holes can hold some other molecules of appropriate size interstitially.
=O
=H
The structure of ice
Hydrogen
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Std. XI Sci. Success Chemistry - II
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Q.44. Give reason ‘Ice floats on water’.
OR
Why does ice floats on water ?
Ans. i.
Liquid water and ice consists of aggregates of varying number of water molecules held together
by hydrogen bonds.
ii. The structure of ice is open structure having a number of vacant spaces.
iii. When the ice melts, some of the hydrogen bonds are broken and the water molecules go in
between the vacant spaces in the structure.
iv. As a result, the structure of liquid water is less open than structure of ice.
v.
Therefore, the density of water is more than that of ice. In other words, the density of ice is less
than that of water and ice floats over water.
*Q.45. What is auto-protolysis of water ? What is its significance?
Ans. Autoprotolysis of water is the self-ionisation of water molecules.
+
−
H2O() + H2O()  
 H3O ( aq ) + OH ( aq )
(acid-1)
(base-2)
(acid-2)
(base-1)
Significance of autoprotolysis : Autoprotolysis of water leads to its amhoteric nature.
Hence, water behaves as an acid as well as a base.
−
+
H2O() + NH3(aq)  
 OH ( aq ) + NH 4( aq )
acid
base
(water as an acid)
+
−
H2O() + HCl(aq)  
 H3O ( aq ) + Cl ( aq )
base
acid
(water as a base)
Q.46. Write a note on amphoteric nature of water.
Ans. i.
Water acts as an acid while reacting with a base by donating a proton.
−
+
H2O() + NH3(aq)  
 OH ( aq ) + NH 4( aq )
acid
ii.
base
iii.
base
Water acts as a base while reacting with a strong acid by accepting a proton.
+
−
H2O() + HCl(aq)  
 H3O ( aq ) + Cl ( aq )
acid
Thus, water has the ability to act both as a base and as an acid. Hence it has an amphoteric
nature. Therefore, pure water contains H3O+ and OH– in equal amount.
+
−
H2O() + H2O()  
 H3O ( aq ) + OH ( aq )
(acid-1)
(base-2)
(acid-2)
(base-1)
Q.47. Write or explain redox reactions involving water.
Ans. i.
Water is reduced to dihydrogen by highly electropositive sodium metal.
2H2O() + 2Na(s) 
→ 2NaOH(aq) + H2(g)
ii. Water can be oxidised to oxygen by fluorine.
2F2(g) + 2H2O() 
→ 4H (+aq ) + 4F (−aq ) + O2(g)
During photosynthesis, water is oxidised to oxygen.
Photosynthesis
6CO(g) + 12H2O() 
→ C6H12O6(aq) + 6H2O() + 3O2(g)
Q.48. What is hydrolysis ? Explain the hydrolysis of covalent and ionic compounds.
Ans. i.
Decomposition of a compound in presence of water is called as hydrolysis.
ii. Due to high dielectric constant, water has a very strong solvating tendency.
Hydrogen
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iii.
HORIZON Publications
It dissolves many ionic compounds. However, certain covalent and some ionic compounds are
hydrolysed in water.
P4O10(s) + 6H2O() 
→ 4H3PO4(aq)
SiCl4() + 2H2O() 
→ SiO2(s) + 4HCl(aq)
CaO + H2O 
→ Ca(OH)2
CaC2(s) + 2H2O() 
→ C2H2(s) + Ca(OH)2(aq)
Mg3N2(s) + 6H2O() 
→ 3Mg(OH)2(aq) + 2NH3(g)
Na2CO3(s) + 2H2O() 
→ 2NaOH(aq) + H2CO3(s)
*Q.49. What is the difference between the terms hydrolysis and hydration ?
Ans. Hydrolysis : It is a chemical process in which water molecules react with the molecules of certain
compounds and form new compounds which produce acidity or basicity.
Eg.
P4O10(s) + 6H2O() 
→ 4H3PO4(aq)
CaC2 + 2H2O 
→ C2H2 + Ca(OH)2
Na2CO3 + 2H2O 
→ 2NaOH + H2CO3
Hydration : The process of crystallisation from aqueous solutions forming hydrated salts in which
water molecules are associated with the molecules of the salt is called hydration.
The association of water in hydration is of the following types:
i.
Coordinated water : For example aqua complex salt : [Cr(H2O)6]3+3Cl–
ii. Interstitial water : BaCl2 ⋅ 2H2O
iii. Hydrogen bonded water : CuSO4 ⋅ 5H2O
which exists as [Cu(H2O)4]2+SO 24− ⋅ H2O
9.9 Heavy water
Q.50. What is heavy water and what are its uses ?
Ans. Heavy water is deuterium oxide (D2O) in which hydrogen atoms ( 11 H) of water molecule are replaced
by its heavy isotope ( 12 D).
As deuterium has twice the mass of an ordinary hydrogen. D2O is called heavy water. It is 11%
heavier than ordinary water.
Uses of heavy water :
i.
heavy water is used in nuclear reactors as moderator for neutrons.
ii. as tracer compound in the study of reaction mechanism and kinetics of reaction.
iii. in the preparation of other deuterium compounds.
Eg.
CaC2 + 2D2O 
→ C2D2 + Ca(OD)2
SO3 + D2O 
→ D2SO4
Al4C3 + 12D2O 
→ 3CD4 + 4Al(OD)3
Q.51. What is super heavy water ?
Ans. Super heavy water is T2O in which hydrogen atoms ( 11 H) of water molecules are replaced by its
heavy isotope tritium ( 13 T or 13 H).
Tritium is about three times as heavy as ordinary hydrogen, hence T2O is called super heavy water.
Q.52. Water is a mixture of different varieties of isotopic compounds. Explain.
Hydrogen
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Std. XI Sci. Success Chemistry - II
Ans.
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Water consists of two hydrogen atoms and one oxygen atom. Hydrogen and oxygen both have three
isotopes. Accordingly there are 18 different varieties of isotopic compounds as follows:
The isotopic varieties of compounds:
H216O
(18)
H217O
(19)
H218O
(20)
D216O
(20)
D217O
(21)
D218O
(22)
T216O
(22)
T217O
(23)
T218O
(24)
HD16O
(19)
HD17O
(20)
HD18O
(21)
HT16O
(20)
HT17O
(21)
HT18O
(22)
DT16O
(21)
DT17O
(22)
DT18O
(23)
Q.53. What are the physical properties of D2O ?
Ans. The physical properties of D2O are tabulated below.
Property
Molecular mass g mol–1
Melting point(K)
Boiling point(K)
Enthalpy of formation kJ mol–1
Enthalpy of vapourization at 373 K(kJ mol–1)
Temperature of maximum density(K)
Density 298 K(g cm–3)
Viscosity/centipoises
Dielectric constant(C2/Nm2)
Electrical conductivity 293 K(ohm–1 cm–1)
D2O
20.0276
276.8
374.4
–294.4
41.61
284.2
1.1059
1.107
78.06
-
*Q.54. Complete the following chemical reactions.
i.
CaO + H2O 
→
ii. AlCl3 + H2O 
→
iii. Ca3N2 + H2O 
→
iv. PbS + H2O2 
→
Ans. i.
CaO + H2O 
→ Ca(OH)2
ii. AlCl3 + 3H2O 
→ Al(OH)3 + 3HCl
iii. Ca3N2 + 6H2O 
→ 3Ca(OH)2 + 2NH3
iv. PbS(s) + 4H2O2(aq) 
→ PbSO4(s) + 4H2O()
Q.55. Knowing the properties of H2O and D2O do you think that D2O can be used for drinking
purpose ? [Intext Question text book page no. 210]
Ans. D2O cannot be used for drinking due to the following reasons:
i.
D2O is about 11% heavier than H2O.
ii. It is denser than ordinary water.
iii. It boils at higher temperature.
iv. It is mildly toxic and can interfere with normal functioning of our body.
v.
It influences biochemical processes in the body.
Hence, D2O is not suitable for drinking purpose.
Q.56. What are the uses of varieties of water ?
Ans. i.
Water is a mixture of 18 different varieties of isotopic compounds, the lightest being H216O and
the heaviest T218O whose molecular weights are 18 and 24 respectively.
Hydrogen
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Std. XI Sci. Success Chemistry - II
ii.
iii.
iv.
v.
HORIZON Publications
Among the different kinds of water, protonium water or the ordinary water is vital for life
process.
The other isotopic forms are important in reaction kinetics and have a significant role in
deciding the pathway of reactions.
Heavy water D2O is prepared by exhaustive electrolysis of water or as a by-product in some
fertilizer industries.
It is used for the preparation of other deuterium compounds.
Eg. CaC2 + 2D2O 
→ C2D2 + Ca(OD)2
SO3 + D2O 
→ D2SO4
Al4C3 + 12D2O 
→ 3CD4 + 4Al(OD)3
9.10 Hydrogen Perioxide
Q.57. Describe the laboratory method of preparing hydrogen peroxide.
Ans. i.
When hydrated barium peroxide, BaO2 ⋅ 8H2O is acidified with dilute sulphuric acid, hydrogen
peroxide is obtained along with a white precipitate of barium sulphate.
BaO2 ⋅ 8H2O(s) + H2SO4(dil) 
→ BaSO4(s) + H2O2(aq) + 8H2O()
ii. Insoluble barium sulphate is filtered off.
iii. Excess water is removed by evaporation.
iv. Anhydrous BaO2 cannot be used in this method since BaSO4 forms a protective layer around
unreacted BaO2 and prevents further reaction.
Q.58. How is hydrogen peroxide is prepared from sodium peroxide ?
OR
Explain Merck’s method of preparation of hydrogen peroxide.
Ans. Action of dilute sulphuric acid on sodium peroxide (Na2O2) gives hydrogen peroxide.
Na2O2 + H2SO4 
→ Na2SO4 + H2O2
This method is called Merck’s method.
Q.59. What are the different methods of preparation of hydrogen peroxide ?
Ans. Methods of preparation of H2O2 :
i.
By the action of dilute acids on sodium peroxide (Merck’s method) :
Na2O2 + H2SO4 
→ Na2SO4 + H2O2
Sodium peroxide
ii. By bubbling CO2 through a paste of BaO2 :
BaO2 + H2O + CO2 
→ BaCO3↓ + H2O2
Thin paste in ice
ppt.
cold water
iii. By the action of phosphoric acid on BaO2 :
3BaO2 + 2H3PO4 
→ Ba3(PO4)2↓ + 3H2O2
iv.
By the electrolysis of 50% H2SO4 :
Manufacture of 3H2O2 is done by electrolysis of 50% solution of H2SO4. Electrolysis of 50%
H2SO4 is done by using Pt as anode and graphite as cathode.
The reactions taking place are :
2H2SO4 
→ 2H+ + 2HSO 4−
At cathode : 2H+ + 2e– 
→ H2↑
−
At athode : 2HSO 4 
→ H2S2O8 + 2e–
Peroxodisulphuric acid
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Peroxodisulphuric acid us distilled with water under reduced pressure when low boiling H2O2
distills over along with water leaving behind boiling H2SO4.
H2S2O8 + H2O 
→ H2SO5 + H2SO4
H2SO5
+ H2O 
→ H2SO4 + H2O2
Peroxymonosulphuric
acid
This method is now used for the laboratory preparation of D2O2.
K2S2O8(s) + 2D2O(l) 
→ 2KDSO4(aq) + D2O2(l)
Q.60. How is phosphoric acid used to prepare hydrogen peroxide ?
Ans. Phosphoric acid on reaction with barium peroxide forms hydrogen peroxide and insoluble barium
phosphate which is removed by filteration.
3BaO2 + 2H3PO4 → Ba3(PO4)2↓ + 3H2O2
Q.61. Describe the industrial preparation of hydrogen peroxide.
OR
How is hydrogen peroxide obtained from 2-Ethylanthraquinol ?
Ans. i.
Industrially hydrogen peroxide is prepared by the auto oxidation of 2-Ethylanthraquinol.
ii. Air is bubbled through a solution of 2-Ethylanthraquinol (10% solution in benzene and
cyclohexane) to get hydrogen peroxide and the oxidised product (2-Ethylanthraquinone). It is a
reversible reaction.
2-Ethylanthraquinone is further reduced by H2 gas in presence of Pd catalyst to give back
2-Ethylanthraquinol.
OH
O
C2H5
C2H5
O2
 H2/Pd

OH
iii.
+ H2O2
O
Concentration of hydrogen peroxide : 1% hydrogen peroxide is obtained by this method. It
is extracted with water and distilled under reduced pressure. Pure H2O2 is obtained by freezing
out the remaining water.
Q.62. Explain how is hydrogen peroxide manufactured by electrolysis ?
Ans. i.
On a large scale, hydrogen peroxide is obtained by electrolysis of 50% solution of H2SO4 using
platinum anode and graphite cathode.
ii. The following reactions take place during electrolysis,
2H2SO4 
→ 2H+ + 2HSO −4
iii.
Reaction at cathode : 2H+ + 2e– 
→ H2 ↑
Reaction at anode : 2HSO 4− 
→
H2S2O8
+
2e–
Peroxodisulphuric acid
iv.
During electrolysis, H+ ions are discharged at cathode and reduced to hydrogen gas. At anode,
HSO −4 ions are oxidised to H2S2O8, (peroxodisulphuric acid).
When peroxodisulphuric acid is distilled with water under reduced pressure, low boiling
hydrogen peroxide (H2O2) distils over along with water leaving behind H2SO4.
H2S2O8 +
H2O 
+
H2SO5
→ H2SO4
Peroxodisulphuric acid
Peroxomonosulphuric acid
Now, H2SO5 oxidises H2O to H2O2.
+
H2O 
H2SO5
→ H2SO4 + H2O2
Peroxomonosulphuric acid
Hydrogen
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Std. XI Sci. Success Chemistry - II
HORIZON Publications
Q.63. How is hydrogen peroxide stored ?
OR
“Glass bottles should not be used to store hydrogen peroxide”. Explain.
Ans. i.
H2O2 is not stored in glass bottle since the alkali oxides present in glass catalyses its
decomposition.
ii. It is therefore, stored in paraffin wax coated plastic or Teflon bottles.
iii. Small amount of acid, glycerol, alcohol, acetamide and H3PO4 are often used as stabilizers to
check the decomposition.
Q.64. What is the equivalent weight of H2O2 ?
Ans. Hydrogen peroxide on decomposition liberates oxygen. The equivalent weight of hydrogen peroxide is
the weight which gives 8 gram oxygen.
H2O2 
→ H2O + [O]
34 g
16 g
Since, 16 g oxygen is obtained from 34 g H2O2, 8 g oxygen will be obtained from,
8× 34
= 17 g, H2O2
16
Hence, the equivalent weight of H2O2 is 17.
Q.65. What is the weight of H2O2 required to prepare one litre of 1 N H2O2 solution ?
Ans. 1 litre of 1 normal H2O2 solution will contain 1 gram equivalent of H2O2, which is equal to
equivalent weight of H2O2 i.e., 17 gram H2O2.
Q.66. What will be the volume of oxygen gas liberated at STP by the decomposition of 1 litre of 1 N
H2O2 ?
Ans. 1 litre of 1 normal H2O2 solution contains 17 gram of H2O2.
H2O2 
→ H2O +
1
O2
2
34 g
16 g
Now, 34 gram H2O2 on decomposition liberates 16 gram O2, hence 17 gram H2O2 will liberate,
16 × 17
= 8 gram O2.
34
Now, from gas Laws,
22.4 × 8
1 mol O2 ≡ 32 gram O2 occupies 22.4 L at STP, hence 8 gram O2 will occupy
= 5.6 L.
32
i.e., one litre of one normal H2O2 liberates 5.6 litres of O2 at STP on decomposition. OR
1 mL of 1 N H2O2 will liberate 5.6 mL of O2 at STP.
Hence, it can be concluded that 1 N H2O2 solution has volume strength 5.6. OR
H2O2 solution having volume strength 5.6 has normality one.
Q.67. How is volume strength of H2O2 solution related to (i) Normality (ii) Molarity of solution?
Ans. i.
1 N H2O2 ≡ 5.6 volume strength of H2O2
∴ volume strength of H2O2 = Normality × 5.6
ii. 1 mol H2O2 ≡ 34 g H2O2
1 gram equivalent of H2O2 ≡ 17 gram H2O2
∴ 1 mol H2O2 = 2 × gram equivalent of H2O2
∴ volume strength of H2O2 = 2 × molarity × 5.6
∴ volume strength of H2O2 = 11.2 × molarity.
Hydrogen
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Std. XI Sci. Success Chemistry - II
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Q.68. What is the relation between percentage strength and volume strength of H2O2 solution ?
Ans. Percentage strength indicates weight of H2O2 in 100 ml solution. For example, if H2O2 solution has
percentage strength x, it means 100 ml H2O2 solution contains x g H2O2
∴
1000 ml H2O2 solution contains
x ×1000
= x × 10 g H2O2
100
x ×10
∴
Gram equivalent of H2O2 in 1 litre solution =
Eq. wt. of H 2O 2
x ×10
∴
Normality of H2O2 =
Eq. wt. of H 2O 2
Since, equivalent weight of H2O2 = 17 g
Percentage strength × 10
Normality of H2O2 =
17
Q.69. Write physical properties of H2O2.
Ans. Physical properties of H2O2 :
i.
Pure hydrogen peroxide is a pale blue liquid.
ii. It is miscible with water in all proportions and forms a hydrate, H2O2. H2O having melting
point 221 K.
iii. It boils at 423 K.
iv. Its density in liquid form is 1.44 g cm–3 at 298 K.
v.
Its important physical properties are tabulated as follows:
Property
Melting point (K)
Boiling point (K) (extrapolated)
Vapour pressure 298 K (mm Hg)
Density: solid at 268.5 K (g cm–3)
Density: liquid at 298 K (g cm–3)
Viscosity at 290 K (centipoises)
Dielectric constant at 298 K (c2/Nm2)
Electrical conductivity at 298 K (m–1 cm–1)
H2O2
272.4
423
1.9
1.64
1.44
1.25
70.7
5.1 × 10–8
vi. A 30% solution of H2O2 is marketed as 100 volume hydrogen peroxide.
vii. It means that one millilitre of 30% H2O2 solution will give 100 V of oxygen at STP.
Commercially, it is marketed as 10 V, which means it contains 3% H2O2.
Q.70. Explain the structure of H2O2.
Ans. i.
Hydrogen peroxide has an open book (skew) structure as shown in the figure.
H
H
95.0 pm
147.5 pm
111.5°
H
(a) Gas phase
Hydrogen
90.2°
101.9°
94.8°
ii.
98.8 pm
145.8 pm
H
(b) Solid phase
In the gaseous phase, the O – H bond length is 95 pm whereas O – O bond length is 147.5 pm.
9.20
Std. XI Sci. Success Chemistry - II
iii.
HORIZON Publications
The H – O – O bond angle is 94.8°. The dihedral angle or angle between the planes containing
H – O – O – H is 111.5°. However in the crystalline state the angle between two planes is 90.2°.
*Q.71. Compare the structures of H2O and H2O2.
Ans.
Structure of H2O
i.
Structure of H2O is angular and V shaped.
ii.
iii.
iv.
HOH bond angle is 104°35′.
The O – H bond length is 95.7 pm.
O
H
Structure of H2O2
Structure of H2O2 is skew structure like
an open book and non-planar.
H – O – O bond angle is 94.8°.
The O – H bond length is 95 pm.
95.7 pm
104°35′
H
(a)
H
95.0 pm
147.5 pm
111.5°
94.8°
H
(a) Gas phase
H
H
(b)
H
98.8 pm
145.8 pm
90.2°
101.9°
H
(b) Solid phase
Q.72. Write the important uses of hydrogen peroxide.
Ans. Hydrogen peroxide is used:
i.
In environmental (green) chemistry : (a) to control pollution of domestic and industrial effluents,
(b) in restoration of aerobic condition to sewage waste and (c) in oxidation of cyanide.
ii. As an antiseptic and sold as perhydrol.
iii. As a hair-bleach.
iv. As a mild disinfectant.
v.
As an industrial bleaching agent for textiles, paper pulp, oils, fats and leather.
vi. In the manufacture of sodium perborate and percarbonate which are used as detergents.
vii. In the synthesis of hydroquinone, tartaric acid and certain pharmaceuticals.
*Q.73. How H2O2 behaves as a bleaching agent ?
Ans. i.
The bleaching action of hydrogen peroxide is due to the oxidation of colouring matter.
ii. H2O2 decomposes to give nascent oxygen which oxidises the colouring matter to colourless
matter.
H2O2 
→ H2O + [O]
Coloured matter + [O] 
→ colourless matter
*Q.74. Write chemical reactions to prove that H2O2 can function as an oxidising agent as well as
reducing agent.
Ans. Hydrogen peroxide acts as an oxidising as well as reducing agent.
Hydrogen
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Std. XI Sci. Success Chemistry - II
i.
ii.
iii.
www.horizonpublications.in
H2O2 as an oxidising agent :
In acidic medium :
a.
2Fe (2aq+ ) + 2H (+aq ) + H2O2(aq) 
→ 2Fe 3( aq+ ) + 2H2O()
b.
PbS(s) + 4H2O2(aq) 
→ PbSO4(s) + 4H2O()
In basic medium :
a.
2Fe2+ + H2O2 
→ 2Fe3+ + 2OH–
b.
Mn2+ + H2O2 
→ Mn4+ + 2OH–
c.
I2 + H2O2 + 2OH– 
→ 2I– + 2H2O + O2
d.
2MnO −4 + 3H2O2 
→ 2MnO2 + 3O2 + 2H2O + 2OH–
H2O2 as a reducing agent in acidic medium :
a.
2MnO −4 + 6H+ + 5H2O2 
→ 2Mn2+ + 8H2O + 5O2
b.
HOCl + H2O2 
→ H3O+ + Cl– + O2
H2O2 reduces Cl2 to Cl– forming HCl :
H2O2 + Cl2 
→ 2HCl + O2
Q.75. Give preparation of enthylene glycol from H2O2.
Ans. Ethylene reacts with hydrogen peroxide to form ethylene glycol.
CH2
CH2OH
||
|
CH2 + H2O2 
→ CH2OH
Ethylene
Ethylene glycol
Q.76. Explain auto oxidation-reduction of H2O2.
Ans. Hydrogen peroxide decomposes rapidly in light, dust, or in the presence of some finely divided metals
like Fe, Cu, Au, Ag, Pt.
H2O2 + H2O2 
→ 2H2O + O2
One molecule of H2O2 is oxidised to O2 i.e., O1– to O0 and the other molecule is reduced to H2O i.e.,
O1– to O2– and thus shows autooxidation reduction reaction.
Q.77. What is the action of H2O2 on NaOH?
Ans. Hydrogen peroxide forms two types of salts with sodium hydroxide:
i.
Sodium hydroperoxide (an acidic salt).
NaOH + H2O2 
→ NaHO2 + H2O
Sodium hydroperoxide
ii. Sodium peroxide (a normal salt).
2NaOH + H2O2 
→ Na2O2 + 2H2O
Sodium peroxide
9.11 Dihydrogen As a Fuel
*Q.78. Explain the term fuel cell.
Ans. i.
Fuel cell is a device which converts the energy produced during the combustion of a fuel directly
into electrical energy.
ii. Dihydrogen is used in hydrogen-oxygen fuel cells for generating electrical energy.
iii. It has many advantages over the conventional fossil fuels.
iv. It does not cause any pollution and releases more amount of energy per unit mass of fuel as
compared to gasoline and other fuels.
Hydrogen
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Std. XI Sci. Success Chemistry - II
HORIZON Publications
*Q.79. Explain the role of dihydrogen as a fuel.
OR
What are the advantages of using dihydrogen as fuel ?
Ans. i.
Dihydrogen is a clean fuel, as it produces much less air pollution.
ii. When burnt, hydrogen does not produce fossil fuels.
iii. Hydrogen can be easily transported.
iv. As generation of hydrogen is possible from sea water, it is limitless, renewable source of
energy.
v.
Hydrogen can serve as a storage energy from hydroelectric, solar and other renewable sources.
*Q.80. Explain the concept of hydrogen economy.
Ans. The concept of hydrogen economy is developed as a result of various properties of hydrogen.
Local power station
Industrial fuels and
reducing gas
Large electric power and
water electrolysis plant
Synthetic chemicals
and liquid fuel
Underground H2
transport pipeline
Domestic fuel
The Hydrogen Economy
i.
ii.
iii.
Hydrogen may serve as storage of energy from hydroelectric, solar and other sources.
The transport of H2 gas by pipeline is less expensive and more efficient.
H2 generated in plants in remote areas can be piped to urban centres.
Q.81. What are the difficulties in using hydrogen a fuel on large scale?
Ans. i.
Considerable energy is required for the production of hydrogen.
ii. So it is very expensive form of energy.
Q.82. Explain storage and production of electricity with hydrogen .
Ans. i.
Chemical storage scheme in the form of H2 is an attractive energy storage scheme.
ii. This gas is generated directly by electrolysis of H2O as shown in figure (a). Electricity is
passed between electrodes immersed in a conducting solution.
iii. H2 is generated at the cathode and O2 at the anode.
Air O2-deficient
Electron
Electron H2 for
motor
flow recycling
Air
or O2
Electron flow
O2
H2
H2
Pt wire
KOH
solution
OH
Pt wire
Porous carbon
electrode
KOH
solution
OH–
OH
(a) Production of Hydrogen
(b) Production of electricity from Hydrogen
Storage and production of electricity with hydrogen
Hydrogen
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Std. XI Sci. Success Chemistry - II
iv.
v.
www.horizonpublications.in
The energy stored in H2 can be converted into electricity using the reverse of the electrolytic
cell, called the fuel cell. It is shown in figure (b).
H2 is oxidised at the cathode where electrons are produced and passed through the circuit to the
anode, where O2 is reduced.
Note :
i.
Ionisation enthalpy: Is the energy needed to remove one or more electrons from neutral atom to
form a positively charged ion.
ii.
Discharge potential: It is the ability of ions to get discharged at the electrode.
iii.
Exothermic reaction: It is the chemical reaction that gives out energy or releases heat to the
surroundings.
iv.
Hydrogenation: Is the reduction reaction that results in an addition of hydrogen.
v.
Binary compounds: Are the compounds formed of two elements.
vi.
Electropositive: Is the tendency of an atom to release electrons to acquire positive charge.
vii.
Neutron moderator: Is a species used in the nuclear reactor to reduce the speed of neutrons making
them thermal neutrons capable of sustaining nuclear chain reaction.
viii. Oxidation: Process of adding oxygen to form an oxide.
ix.
Decomposition: Is the process of breaking down organic substance into simpler form of matter.
x.
Reduction: Is the reaction involving gain of electrons.
Hydrogen
9.24
Std. XI Sci. Success Chemistry - II
HORIZON Publications
Quick Review
1.
Isotopes of hydrogen :
Isotopes of Hydrogen
Protium
(1 H1)
Neutrons
(1)
2.
Deuterium
(1 H2 or D)
Neutrons
(1)
Tritium
(1 H3 or T)
Neutrons
Radioactive
(β particle emitter)
Half life is 12.33 years
Chemical properties of Dihydrogen :
i.
Name of the reaction
Reaction with halogens
(X= F,Cl, Br, I)
ii.
With dioxygen
iii. With dinitrogen
Conditions
F → dark, 63 K
I → catalyst
Catalyst and highly
exothermic
∆H = – 285.9 kJ/mol
Fe as catalyst
200 atm. Pressure
673 K temperature
∆H = – 92.6 kJ/mol
High temperature
iv.
With metals
v.
With metal ions and oxides H2 is
agent.
a
Chemical equations
H2 + X2 
→ 2HX
Catalyst
2H2(g) + O2(g) →
2H2O()
or ∆
673 K, Fe catalyst
3H2 + N2 
→ 2NH3(g)
200 atm
2Na + H2 
→ 2NaH
reducing H2 + Pd2+ 
→ Pd + 2H+
H2 + CuO 
→ Cu + H2O
ZnO + H2 
→ Zn + H2O
Fe3O4 + 4H2 
→ 3Fe + 4H2O
vi.
With organic compounds
Ni, 450 K
Vegetable oil + H2 
→
8−10 atm
a. Hydrogenation of
vegetable oil
b. Hydrogenation of
olefins
Hydrogen
Ni catalyst
Solid fat
(Vanaspati ghee)
H2 + CO + CH3CH = CH2 
→
CH3–CH2–CH2–CHO
Aldehyde
H2 + CH3–CH2–CH2–CHO 
→
CH3–CH2–CH2–CH2–OH
Alcohol
9.25
Std. XI Sci. Success Chemistry - II
3.
www.horizonpublications.in
Uses of Dihydrogen :
Manufacture of
vanaspati ghee
Mixture of liquid O2 and H2 is
used as a rocket fuel
Fuel cell for generating
electrical energy
Synthesis of
NH3
Manufacture of
CH3OH (Methanol)
Uses of H2
As a reducing
agent
Manufacture of
metal hydrides
Atomic hydrogen and
oxyhydrogen is used for cutting
and welding purpose
4.
Preparation of
HCl
Preparation of H2O2 :
Merck’s Method
Na2O2 + H2SO4 → Na2SO4 + H2O2
Auto oxidation of 2 ethylanthraquinol
OH
O
C2H5
O2
H2/Pd
C2H5
+ H2O2
OH
Preparation
of H2O2
Acidifying Barium peroxide
BaO2 ⋅ 8H2O + H2SO4 → BaSO4 + H2O2 + 8H2O
Electrolysis of 50% H2SO4 using Pt as anode and graphite cathode
2H2SO4 → 2H+ + 2HSO −4
At cathode : 2H+ + 2e– → H2↑
At anode: 2HSO −4 → H2S2O8 + 2e–
H2S2O8 + H2O → H2SO5 + H2SO4
By bubbling CO2 through a paste of BaO2
BaO2 + H2O + CO2 → BaCO3↓ + H2O2
Action of phosphoric acid on BaO2
3BaO2 + 2H3PO4 → Ba3(PO4)2↓ + 3H2O2
Hydrogen
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Std. XI Sci. Success Chemistry - II
5.
HORIZON Publications
Chemical properties of H2O2 :
Decomposes rapidly or heating in presence of metals like Ca, Fe, Cu, Au,
Ag, Pt light etc.
∆
H2O2 + H2O2 
→ 2H2O + O2 ∆H = – 196 kJ
Eg. Of auto oxidation and auto reduction of one molecule of H2O2.
neutralizes alkalies
NaOH + H2O2 → NaHO2 + H2O
As oxidizing agent
2Fe2+ + 2H+ + H2O2 → 2Fe3+ + 2H2O
Chemical properties of
H2O2
(acidic medium)
Oxidizing action in basic medium
2Fe2+ + H2O2 → 2Fe3+ + 2OH–
Mn2+ + H2O2 → Mn4+ + 2OH–
I2 + H2O2 + 2OH– → 2I– + 2H2O + O2
2MnO −4 + 3H2O2 → 2MnO2 + 3O2 + 2H2O + 2OH–
Reducing action in acidic medium
2MnO −4 + 6H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2
HOCl + H2O2 → H3O+ + Cl– + O2
Called as antichlor as reduces Cl2 to HCl.
H2O2 + Cl2 → 2HCl + O2↑
Acts as bleaching agent
H2O2 → H2O + O
(O↓ + coloured matter → colourless matter)
Forms glycol with alkenes
CH2
CH2OH
||
|
CH2 + H2O2 → CH2OH
Ethylene
Ethylene glycol
6.
Uses of H2O2 :
In pollution control
treatment of domestic
and industrial effluent
Oxidation of cyanide
Restoration of
aerobic conditions to
sewage waste
Hydrogen
Bleaching agent for
textiles, paper pulp, oils,
fats, leather etc.
H2O2
Antiseptic (perhydrol)
Manufacturing
chemicals like sodium
perborate percarbonate
In synthesis of hydroquinone
tartaric acid and in
pharmaceuticals
Hair bleach, mild
disinfectant
9.27