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NCERT Solutions Class 11 Chemistry Chapter 9 Hydrogen– Here are all the NCERT solutions for Class 11 Chemistry Chapter 9. This solution contains questions, answers, images, explanations of the complete chapter 1 titled Hydrogen taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Chemistry, then you must come across chapter 9 Hydrogen After you have studied the lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrogen in one place.

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NCERT Solutions Class 11 Chemistry Chapter 9 Hydrogen

Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Chemistry for Class 11 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 9 Hydrogen , Chemistry, Class 11.

Class 11
Subject Chemistry
Book Chemistry Part I
Chapter Number 9
Chapter Name

Hydrogen

NCERT Solutions Class 11 Chemistry chapter 9 Hydrogen

Class 11, Chemistry chapter 9, Hydrogen solutions are given below in PDF format. You can view them online or download PDF file for future use.

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Question & Answer

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

Ans : Hydrogen is the first element of the periodic table. Its electronic configuration is [\( 1 \mathrm{s}^{1}\)]. Due to the presence of only one electron in its 1s shell, hydrogen exhibits a dual behaviour, i.e., it resembles both alkali metals and halogens.
Resemblance with alkali metals:
1. Like alkali metals, hydrogen contains one valence electron in its valence shell.
\( \begin{array}{l}{\mathrm{H} : 1 \mathrm{s}^{1}} \\ {\mathrm{Li} :[\mathrm{He}] 2 \mathrm{s}^{1}} \\ {\mathrm{Na} :[\mathrm{Ne}] 3 s^{1}}\end{array}\)
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.
\( \begin{array}{l}{H : 1 s^{1}} \\ {F : 1 s^{2} 2 s^{2} 2 p^{5}} \\ {C l : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}}\end{array}\)
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. 
Q.2: Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes?

Ans : Hydrogen has three isotopes. They are: \( \begin{array}{l}{\text { 1. Protium, }^{1} \mathrm{H}} \\ {\text { 2. Deuterium, }^{2} \mathrm{H}_{\text { or } \mathrm{D}, \text { and }}} \\ {\text { 3. Tritium, }^{3} \mathrm{H}_{\text { or } \mathrm{T}}}\end{array}\)
The mass ratio of protium, deuterium and tritium  is 1:2:3. 
Q.3: Why does hydrogen occur in a diatomic form rather than in a mono-atomic form under normal conditions?

Ans : The ionization enthalpy of hydrogen atom is very high (\( 1312 \mathrm{kJ} \mathrm{mol}^{-1}\)). Hence, it is very hard to remove its only electron. As a result, its tendency to exist in the mono-atomic form is rather low. Instead, hydrogen forms a covalent bond with another hydrogen atom and exists as a diatomic \( H_{2}\) molecule. 
Q.4: How can the production of dihydrogen, obtained from ‘coal gasification’, be increased?

Ans : Dihydrogen is produced by coal gasification method as:
\( \begin{array}{l}{\mathrm{C}_{(s)}+\mathrm{H}_{2} \mathrm{O}_{(g)} \stackrel{1270 \mathrm{k}}{\longrightarrow} \mathrm{CO}_{(g)}+\mathrm{H}_{2(\mathrm{g})}} \\ {\text { (coal) }}\end{array}\)
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.
\( \mathrm{CO}_{(g)}+\mathrm{H}_{2} \mathrm{O}_{(g)} \frac{673 \mathrm{K}}{\text { Cualyst }}\)\( \mathrm{CO}_{2(\mathrm{g})}+\mathrm{H}_{2(\mathrm{g})}\)
This reaction is called the water-gas shift reaction. Carbon dioxide is removed by scrubbing it with a solution of sodium arsenite. 
Q.5: Describe the bulk preparation of dihydrogen by electrolytic method. What is the role of an electrolyte in this process ?

Ans : 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:
\( 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2}+2 \mathrm{OH}^{-}\)
At the anode, oxidation of OH– ions takes place as:
\( 2 \mathrm{OH}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\frac{1}{2} \mathrm{O}_{2}+2 \mathrm{e}^{-}\)
Net reaction can be represented as:
\( \mathrm{H}_{2} \mathrm{O}_{(l)} \longrightarrow \mathrm{H}_{2(\mathrm{x})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{g})}\)
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.

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