Question

Nonmetal A, a major gaseous component of air, combines with ${\mathrm{H}}_2$ in the mole ratio 1:3 in the presence of iron catalyst to give a gas B which is highly soluble in water. Gas A forms a reddish brown oxide C. When C is dissolved in water in the presence of air an acid D is formed which is a strong oxidizing agent.

(i) Identify A, B, C and D

(ii) What is the nature of the aqueous solution of gas B?

(iii) To which group and period of the period table does the nonmetal A belong?

Answer 1

1. Non-metals

Substances that accept electrons to obtain stable electronic configuration are called non-metals. Therefore, non-metals have an affinity to electrons. They generally have more than 4 electrons in the valence shell. Non-metals can form ionic or covalent bonds. They form ionic bonds with metals and covalent bonds with non-metals.

2. Identification of A, B, C, and D

(i) A major constituent of air that reacts with Hydrogen $\left({\mathrm{H}}_2\right)$ in a 1:3 ratio is Nitrogen $\left({\mathrm{N}}_2\right)$ gas.

Therefore, A is Nitrogen.

Nitrogen reacts with Hydrogen in the presence of a catalyst to form Ammonia gas$\left({\mathrm{NH}}_3\right)$. Ammonia is soluble in water and is a covalent compound. The reaction can be represented as:

$$\begin{gathered} {\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\stackrel{\mathrm{Fe}}{\to }2{\mathrm{NH}}_3\left(\mathrm{g}\right) \\ \left(\mathrm{Nitrogen}\right)\left(\mathrm{Hydrogen}\right)\left(\mathrm{Ammonia}\right) \end{gathered}$$

Therefore, B is Ammonia.

The reddish-brown oxide formed by Nitrogen is Nitrogen dioxide$\left({\mathrm{NO}}_2\right)$. The reaction can be represented as:

$$\begin{gathered} {\mathrm{N}}_2\left(\mathrm{g}\right)+2{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{NO}}_2\left(\mathrm{g}\right) \\ \left(\mathrm{Nitrogen}\right)\left(\mathrm{Oxygen}\right)(\mathrm{Nitrogen}\mathrm{dioxide}) \end{gathered}$$

Therefore, C is Nitrogen dioxide.

Nitrogen dioxide dissolves in water $\left({\mathrm{H}}_2\mathrm{O}\right)$ to form Nitric acid$\left({\mathrm{HNO}}_3\right)$. The reaction can be represented as:

$$\begin{gathered} 4{\mathrm{NO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 4{\mathrm{HNO}}_3\left(\mathrm{aq}\right) \\ (\mathrm{Nitrogen}(\mathrm{Water})(\mathrm{Oxygen})(\mathrm{Nitric}\mathrm{acid}) \\ \mathrm{dioxide}) \end{gathered}$$

Therefore, D is Nitric acid.

3. Nature of aqueous solution of B

(ii) Gas B is Ammonia. It dissolves in water to form Ammonium hydroxide$\left({\mathrm{NH}}_4\mathrm{OH}\right)$. Ammonium hydroxide is a base and therefore, the aqueous solution of Ammonia will have a basic nature. The dissolution of Ammonia can be represented as:

$$\begin{gathered} {\mathrm{NH}}_3\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{NH}}_4\mathrm{OH}\left(\mathrm{aq}\right) \\ \left(\mathrm{Ammonia}\right)\left(\mathrm{Water}\right)(\mathrm{Ammonium}\mathrm{hydroxide}) \end{gathered}$$

Therefore, the aqueous solution of gas B will be basic.

4. Group and period of non-metal A

(iii) Non-metal A is Nitrogen which has the electronic configuration,$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^3$. It has 2 electronic shells and therefore, it belongs to the 2nd period. The group is determined by the number of valence electrons. Nitrogen contains 5 electrons in the valence shell and therefore, it belongs to the 15th group.

Therefore, non-metal A belongs to the 2nd period and 15th group.