NCERT Solution - Hydrogen Class 11 Notes | EduRev

 Page 1


  CHAPTER-9 NCERT SOLUTIONS 
 
Question 9.1: 
Justify the position of hydrogen in the periodic table on the basis of its electronic configuration. 
 
Answer: 
Hydrogen is the first element of the periodic table. Its electronic configuration 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 valency shell. 
H : 1s
1
 
Li : [He] 2s
1
 
 
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. 
H : 1s
1
 
F : 1s
2
 2s
2
 2p
5
 
Cl : 1s
2
 2s
2
 2p
6
 3s
2
 3p
5
 
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. 
 
Question 9.2: 
Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes? 
 
Answer: 
Hydrogen has three isotopes. They are: 
1. Protium, , 
2. Deuterium, or D, and 
3. Tritium, or T 
The mass ratio of protium, deuterium and tritium  is 1:2:3. 
  
Question 9.3: 
Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal 
conditions? 
 
Answer: 
The ionization enthalpy of hydrogen atom is very high (1312 kJ mol
–1
). Hence, it is very hard to 
remove its only electron. As a result, its tendency to exist in the monoatomic form is rather low. 
Page 2


  CHAPTER-9 NCERT SOLUTIONS 
 
Question 9.1: 
Justify the position of hydrogen in the periodic table on the basis of its electronic configuration. 
 
Answer: 
Hydrogen is the first element of the periodic table. Its electronic configuration 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 valency shell. 
H : 1s
1
 
Li : [He] 2s
1
 
 
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. 
H : 1s
1
 
F : 1s
2
 2s
2
 2p
5
 
Cl : 1s
2
 2s
2
 2p
6
 3s
2
 3p
5
 
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. 
 
Question 9.2: 
Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes? 
 
Answer: 
Hydrogen has three isotopes. They are: 
1. Protium, , 
2. Deuterium, or D, and 
3. Tritium, or T 
The mass ratio of protium, deuterium and tritium  is 1:2:3. 
  
Question 9.3: 
Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal 
conditions? 
 
Answer: 
The ionization enthalpy of hydrogen atom is very high (1312 kJ mol
–1
). Hence, it is very hard to 
remove its only electron. As a result, its tendency to exist in the monoatomic form is rather low. 
Instead, hydrogen forms a covalent bond with another hydrogen atom and exists as a diatomic 
(H
2
) molecule. 
 Question 9.4: 
How can the production of dihydrogen, obtained from ‘coal gasification’, be increased? 
 
Answer: 
Dihydrogen is produced by coal gasification method as: 
 
The yield of dihydrogen (obtained from coal gasification) can be increased by reacting carbon 
monoxide (formed during the reaction) with steam in the presence of iron chromate as a catalyst. 
 
This reaction is called the water-gas shift reaction. Carbon dioxide is removed by scrubbing it 
with a solution of sodium arsenite. 
  
Question 9.5: 
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 acidified or alkaline water using platinum 
electrodes. Generally, 15 – 20% of an acid (H
2
SO
4
) or a base (NaOH) is used. 
Reduction of water occurs at the cathode as: 
 
At the anode, oxidation of OH
–
 ions takes place as: 
 
Net reaction can be represented as: 
 
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. 
  
Question 9.6: 
Complete the following reactions: 
(i)  
(ii)  
(iii)  
(iv)  
 
Answer: 
(i)  
Page 3


  CHAPTER-9 NCERT SOLUTIONS 
 
Question 9.1: 
Justify the position of hydrogen in the periodic table on the basis of its electronic configuration. 
 
Answer: 
Hydrogen is the first element of the periodic table. Its electronic configuration 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 valency shell. 
H : 1s
1
 
Li : [He] 2s
1
 
 
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. 
H : 1s
1
 
F : 1s
2
 2s
2
 2p
5
 
Cl : 1s
2
 2s
2
 2p
6
 3s
2
 3p
5
 
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. 
 
Question 9.2: 
Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes? 
 
Answer: 
Hydrogen has three isotopes. They are: 
1. Protium, , 
2. Deuterium, or D, and 
3. Tritium, or T 
The mass ratio of protium, deuterium and tritium  is 1:2:3. 
  
Question 9.3: 
Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal 
conditions? 
 
Answer: 
The ionization enthalpy of hydrogen atom is very high (1312 kJ mol
–1
). Hence, it is very hard to 
remove its only electron. As a result, its tendency to exist in the monoatomic form is rather low. 
Instead, hydrogen forms a covalent bond with another hydrogen atom and exists as a diatomic 
(H
2
) molecule. 
 Question 9.4: 
How can the production of dihydrogen, obtained from ‘coal gasification’, be increased? 
 
Answer: 
Dihydrogen is produced by coal gasification method as: 
 
The yield of dihydrogen (obtained from coal gasification) can be increased by reacting carbon 
monoxide (formed during the reaction) with steam in the presence of iron chromate as a catalyst. 
 
This reaction is called the water-gas shift reaction. Carbon dioxide is removed by scrubbing it 
with a solution of sodium arsenite. 
  
Question 9.5: 
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 acidified or alkaline water using platinum 
electrodes. Generally, 15 – 20% of an acid (H
2
SO
4
) or a base (NaOH) is used. 
Reduction of water occurs at the cathode as: 
 
At the anode, oxidation of OH
–
 ions takes place as: 
 
Net reaction can be represented as: 
 
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. 
  
Question 9.6: 
Complete the following reactions: 
(i)  
(ii)  
(iii)  
(iv)  
 
Answer: 
(i)  
 
(iii)  
 
  
Question 9.7: 
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
–1
). This indicates that 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
2
), hydrides with elements, and a large number 
of covalent bonds. 
Since ionization enthalpy is very high, hydrogen does not possess metallic characteristics (lustre, 
ductility, etc.) like metals. 
 
Question 9.8: 
What do you understand by (i) electron-deficient, (ii) electron-precise, and (iii) electron-rich 
compounds of hydrogen? Provide justification with suitable examples. 
 
Answer: 
Molecular hydrides are classified on the basis of the presence of the total number of electrons 
and bonds in their Lewis structures as: 
1. Electron-deficient hydrides 
2. Electron-precise hydrides 
3. Electron-rich hydrides 
An electron-deficient hydride has very few electrons, less than that required for representing its 
conventional Lewis structure e.g. diborane (B
2
H
6
). In B
2
H
6
, there are six bonds in all, out 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 sufficient number of electrons to be represented by its 
conventional Lewis structure e.g. CH
4
. 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. NH
3.
 
 
There are three regular bonds in all with a lone pair of electrons on the nitrogen atom. 
 
Question 9.9: 
Page 4


  CHAPTER-9 NCERT SOLUTIONS 
 
Question 9.1: 
Justify the position of hydrogen in the periodic table on the basis of its electronic configuration. 
 
Answer: 
Hydrogen is the first element of the periodic table. Its electronic configuration 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 valency shell. 
H : 1s
1
 
Li : [He] 2s
1
 
 
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. 
H : 1s
1
 
F : 1s
2
 2s
2
 2p
5
 
Cl : 1s
2
 2s
2
 2p
6
 3s
2
 3p
5
 
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. 
 
Question 9.2: 
Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes? 
 
Answer: 
Hydrogen has three isotopes. They are: 
1. Protium, , 
2. Deuterium, or D, and 
3. Tritium, or T 
The mass ratio of protium, deuterium and tritium  is 1:2:3. 
  
Question 9.3: 
Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal 
conditions? 
 
Answer: 
The ionization enthalpy of hydrogen atom is very high (1312 kJ mol
–1
). Hence, it is very hard to 
remove its only electron. As a result, its tendency to exist in the monoatomic form is rather low. 
Instead, hydrogen forms a covalent bond with another hydrogen atom and exists as a diatomic 
(H
2
) molecule. 
 Question 9.4: 
How can the production of dihydrogen, obtained from ‘coal gasification’, be increased? 
 
Answer: 
Dihydrogen is produced by coal gasification method as: 
 
The yield of dihydrogen (obtained from coal gasification) can be increased by reacting carbon 
monoxide (formed during the reaction) with steam in the presence of iron chromate as a catalyst. 
 
This reaction is called the water-gas shift reaction. Carbon dioxide is removed by scrubbing it 
with a solution of sodium arsenite. 
  
Question 9.5: 
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 acidified or alkaline water using platinum 
electrodes. Generally, 15 – 20% of an acid (H
2
SO
4
) or a base (NaOH) is used. 
Reduction of water occurs at the cathode as: 
 
At the anode, oxidation of OH
–
 ions takes place as: 
 
Net reaction can be represented as: 
 
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. 
  
Question 9.6: 
Complete the following reactions: 
(i)  
(ii)  
(iii)  
(iv)  
 
Answer: 
(i)  
 
(iii)  
 
  
Question 9.7: 
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
–1
). This indicates that 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
2
), hydrides with elements, and a large number 
of covalent bonds. 
Since ionization enthalpy is very high, hydrogen does not possess metallic characteristics (lustre, 
ductility, etc.) like metals. 
 
Question 9.8: 
What do you understand by (i) electron-deficient, (ii) electron-precise, and (iii) electron-rich 
compounds of hydrogen? Provide justification with suitable examples. 
 
Answer: 
Molecular hydrides are classified on the basis of the presence of the total number of electrons 
and bonds in their Lewis structures as: 
1. Electron-deficient hydrides 
2. Electron-precise hydrides 
3. Electron-rich hydrides 
An electron-deficient hydride has very few electrons, less than that required for representing its 
conventional Lewis structure e.g. diborane (B
2
H
6
). In B
2
H
6
, there are six bonds in all, out 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 sufficient number of electrons to be represented by its 
conventional Lewis structure e.g. CH
4
. 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. NH
3.
 
 
There are three regular bonds in all with a lone pair of electrons on the nitrogen atom. 
 
Question 9.9: 
What characteristics do you expect from an electron-deficient hydride with respect to its structure 
and chemical reactions? 
 
Answer: 
An electron-deficient hydride does not have sufficient electrons to form a regular bond in which 
two electrons are shared by two atoms e.g., B
2
H
6
, Al
2
H
6
 etc. 
These hydrides cannot be represented by conventional Lewis structures. B
2
H
6
, for example, 
contains four regular bonds and two three centered-two electron bond. Its structure can be 
represented as: 
 
Since these hydrides are electron-deficient, they have a tendency to accept electrons. Hence, 
they act as Lewis acids. 
 
  
Question 9.10: 
Do you expect the carbon hydrides of the type (C
n
H
2n + 2
) to act as ‘Lewis’ acid or base? Justify 
your answer. 
 
Answer: 
For carbon hydrides of type C
n
H
2n + 2
, the following hydrides are possible for 
 
For a hydride to act as a Lewis acid i.e., electron accepting, it should be electron-deficient. Also, 
for it to act as a Lewis base i.e., electron donating, it should be electron-rich. 
Taking C
2
H
6
 as an example, the total number of electrons are 14 and the total covalent bonds 
are seven. Hence, the bonds are regular 2e
–
-2 centered bonds. 
 
Hence, hydride C
2
H
6
 has sufficient electrons to be represented by a conventional Lewis 
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. 
  
Question 9.11: 
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: 
Page 5


  CHAPTER-9 NCERT SOLUTIONS 
 
Question 9.1: 
Justify the position of hydrogen in the periodic table on the basis of its electronic configuration. 
 
Answer: 
Hydrogen is the first element of the periodic table. Its electronic configuration 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 valency shell. 
H : 1s
1
 
Li : [He] 2s
1
 
 
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. 
H : 1s
1
 
F : 1s
2
 2s
2
 2p
5
 
Cl : 1s
2
 2s
2
 2p
6
 3s
2
 3p
5
 
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. 
 
Question 9.2: 
Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes? 
 
Answer: 
Hydrogen has three isotopes. They are: 
1. Protium, , 
2. Deuterium, or D, and 
3. Tritium, or T 
The mass ratio of protium, deuterium and tritium  is 1:2:3. 
  
Question 9.3: 
Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal 
conditions? 
 
Answer: 
The ionization enthalpy of hydrogen atom is very high (1312 kJ mol
–1
). Hence, it is very hard to 
remove its only electron. As a result, its tendency to exist in the monoatomic form is rather low. 
Instead, hydrogen forms a covalent bond with another hydrogen atom and exists as a diatomic 
(H
2
) molecule. 
 Question 9.4: 
How can the production of dihydrogen, obtained from ‘coal gasification’, be increased? 
 
Answer: 
Dihydrogen is produced by coal gasification method as: 
 
The yield of dihydrogen (obtained from coal gasification) can be increased by reacting carbon 
monoxide (formed during the reaction) with steam in the presence of iron chromate as a catalyst. 
 
This reaction is called the water-gas shift reaction. Carbon dioxide is removed by scrubbing it 
with a solution of sodium arsenite. 
  
Question 9.5: 
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 acidified or alkaline water using platinum 
electrodes. Generally, 15 – 20% of an acid (H
2
SO
4
) or a base (NaOH) is used. 
Reduction of water occurs at the cathode as: 
 
At the anode, oxidation of OH
–
 ions takes place as: 
 
Net reaction can be represented as: 
 
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. 
  
Question 9.6: 
Complete the following reactions: 
(i)  
(ii)  
(iii)  
(iv)  
 
Answer: 
(i)  
 
(iii)  
 
  
Question 9.7: 
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
–1
). This indicates that 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
2
), hydrides with elements, and a large number 
of covalent bonds. 
Since ionization enthalpy is very high, hydrogen does not possess metallic characteristics (lustre, 
ductility, etc.) like metals. 
 
Question 9.8: 
What do you understand by (i) electron-deficient, (ii) electron-precise, and (iii) electron-rich 
compounds of hydrogen? Provide justification with suitable examples. 
 
Answer: 
Molecular hydrides are classified on the basis of the presence of the total number of electrons 
and bonds in their Lewis structures as: 
1. Electron-deficient hydrides 
2. Electron-precise hydrides 
3. Electron-rich hydrides 
An electron-deficient hydride has very few electrons, less than that required for representing its 
conventional Lewis structure e.g. diborane (B
2
H
6
). In B
2
H
6
, there are six bonds in all, out 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 sufficient number of electrons to be represented by its 
conventional Lewis structure e.g. CH
4
. 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. NH
3.
 
 
There are three regular bonds in all with a lone pair of electrons on the nitrogen atom. 
 
Question 9.9: 
What characteristics do you expect from an electron-deficient hydride with respect to its structure 
and chemical reactions? 
 
Answer: 
An electron-deficient hydride does not have sufficient electrons to form a regular bond in which 
two electrons are shared by two atoms e.g., B
2
H
6
, Al
2
H
6
 etc. 
These hydrides cannot be represented by conventional Lewis structures. B
2
H
6
, for example, 
contains four regular bonds and two three centered-two electron bond. Its structure can be 
represented as: 
 
Since these hydrides are electron-deficient, they have a tendency to accept electrons. Hence, 
they act as Lewis acids. 
 
  
Question 9.10: 
Do you expect the carbon hydrides of the type (C
n
H
2n + 2
) to act as ‘Lewis’ acid or base? Justify 
your answer. 
 
Answer: 
For carbon hydrides of type C
n
H
2n + 2
, the following hydrides are possible for 
 
For a hydride to act as a Lewis acid i.e., electron accepting, it should be electron-deficient. Also, 
for it to act as a Lewis base i.e., electron donating, it should be electron-rich. 
Taking C
2
H
6
 as an example, the total number of electrons are 14 and the total covalent bonds 
are seven. Hence, the bonds are regular 2e
–
-2 centered bonds. 
 
Hence, hydride C
2
H
6
 has sufficient electrons to be represented by a conventional Lewis 
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. 
  
Question 9.11: 
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-deficient 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
2.87
, YbH
2.55
, TiH
1.5 – 1.8
 etc. 
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. 
  
Question 9.12: 
How do you expect the metallic hydrides to be useful for hydrogen storage? Explain. 
 
Answer: 
Metallic hydrides are hydrogen deficient, 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. 
  
Question 9.13: 
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 (435.88 kJ mol
–1
). 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. 
  
Question 9.14: 
Among NH
3
, H
2
O and HF, which would you expect to have highest magnitude of hydrogen 
bonding and why? 
 
Answer: 
The extent of hydrogen bonding depends upon electronegativity and the number of hydrogen 
atoms available for bonding. Among nitrogen, fluorine, 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
2
O > NH
3
. 
But, the actual order is H
2
O > HF > NH
3
. 
Although fluorine 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.