Hydrogen

Chemistry Class 11

NCERT

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

Solution :

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 behavior, i.e., it resembles both alkali metals and halogens. $\textbf{Resemblance with alkali metals:}$$\\ 1. Like alkali metals, hydrogen contains one valence electron in its valence shell.\\ H:1s^1$$\\$ $Li:[He]2S^1$$\\ Na:[Ne]3s^{-1}$$\\$ $\vdots$$\\ \vdots$$\\$ 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.$\\$ $\textbf{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.

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

Solution :

Hydrogen has three isotopes. They are:$\\$ 1. Protium, ; $^1_1 H$$\\ 2. Deuterium, _1^ 2 H or D, and \\ 3. Tritium, _1^ 3 H or T$$\\$ The mass ratio of protium, deuterium and tritium is $1:2:3.$

3   Why does hydrogen occur in a diatomic form rather than in a monoatomic form under nomal conditions?

Solution :

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.

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

Solution :

Dihydrogen is produced by coal gasification method as:$\\$ $C(s)+H_2O(g) \overset{1270K} {\to} CO(g)+H_2(g)$$\\ (coal)\\ The yield of dihydrogen (obtained from coal gasification) can be increased by reacting carbon monoxide (formed during the reaction) with steam in the presence of iron chromate as a catalyst.\\ CO(s)+H_2O(g) \overset{673K}{\underset{\text{Catalyst}} {\to}}CO_2(g)+H_2(g)$$\\$ This reaction is called the water-gas shift reaction. Carbon dioxide is removed by scrubbing it with a solution of sodium arsenite.

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

Solution :

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:$\\$ $2 H_ 2 O + 2 e^- \to 2 H _2 + 2 OH ^-$$\\ At the anode, oxidation of OH^- ions takes place as:\\ 2OH^- \to H_2O+\dfrac{1}{2}O_2 +2e^-$$\\$ $\therefore$ Net reaction can be represented as:$\\$ $H_2O(l)\to H_2(g)+\dfrac{1}{2}O_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.

6   Complete the following reactions:$\\$ (i) $H_ 2( g ) + M _m O_ o ( s )\overset{\Delta}\longrightarrow$ $\\$ (ii) $CO ( g ) + H _2( g ) \overset{\Delta}{\underset{\text{catalyst}}{\longrightarrow }}$ $\\$ (iii) $C _3 H _8( g ) + 3 H _2 O ( g ) \overset{\Delta}{\underset{\text{catalyst}}{\longrightarrow }}$ $\\$ (iv) $Zn ( s ) + NaOH ( aq ) \overset{heat}\longrightarrow$

Solution :

(i) $H _2( g ) + M_ m O _o ( s ) \longrightarrow mM ( s ) + H _2 O ( l )$ $\\$ (ii) $CO ( g ) + 2 H _2( g )\overset{\Delta}{\underset{\text{catalyst}}{\longrightarrow }} CH _3 OH ( l )$ $\\$ (iii) $C _3 H _8( g ) + 3 H _2 O ( g ) \overset{\Delta}{\underset{\text{catalyst}}{\longrightarrow }}3 CO ( g ) +7 H _2( g )$ $\\$ (iv) $Zn ( s ) + NaOH ( aq )\overset{heat}\longrightarrow \underset{\text{Sodium zincate}}{NaZnO_ 2( aq ) }+ H _2( g )$

7   Discuss the consequences of high enthalpy of $H-H$ bond in terms of chemical reactivity of dihydrogen.

Solution :

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.

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

Solution :

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.

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

Solution :

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.$\\$ $B _2 H_ 6 + 2 NMe \longrightarrow 2 BH _3 . NMe _3$ $\\$ $B _2 H_ 6 + 2 CO \longrightarrow 2 BH_ 3 . CO$

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.

Solution :

For carbon hydrides of type $C _n H _{2n+2 }$, the following hydrides are possible for $\\$ $n = 1 \Rightarrow CH_ 4$ $\\$ $n = 2 \Rightarrow C_ 2 H_ 6$ $\\$ $n = 3 \Rightarrow C_ 3 H_ 8$ $\\$ 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.