NCERT Class 11 Solutions Hydrogen

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NCERT Class 11 Solutions Hydrogen - NCERT Solutions prepared for CBSE students by the best teachers in Delhi.

Class XI Chapter 9 – Hydrogen Chemistry

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 [1s1]. 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 : 1s1

Li : [He] 2s1

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 : 1s1

F : 1s2 2s2 2p5

Cl : 1s2 2s2 2p6 3s2 3p5

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 (H2) 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 (H2SO4) 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 9.7: 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 (H2), 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) electronrich 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 (B2H6). In B2H6, 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.


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