Chemistry MCQs for NEET — Practice Questions with Answers

Ammonia generally prepared by Haber's process

Among the given compounds, $N_3H$ (hydrazoic acid) is the acidic one. It is a weak acid but more acidic than the other nitrogen hydrides listed. $NH_3$ (ammonia) is basic, $N_2H_4$ (hydrazine) and $N_2H_2$ (diimide) are not acidic.

Laughing gas, or nitrous oxide ($N_2O$), is prepared by heating ammonium nitrate ($NH_4NO_3$). The reaction is: $$ NH_4NO_3 ightarrow N_2O + 2H_2O $$

$ HO - SO_2 - OH + PCl_5 \rightarrow Cl - SO_2 - OH + POCl_3 + HCl $ $ HO - SO_2 - OH + 2PCl_5 \rightarrow Cl - SO_2 - Cl + 2POCl_3 + 2HCl $

$ H_2O $ containing hydrogen bond,

$ SO_2 + 2H_2S \rightarrow 3S + 2H_2O $

The boiling point of hydrides generally increases as we move down the group in the periodic table due to an increase in molecular weight. However, $H_2O$ has an exceptionally high boiling point due to hydrogen bonding. Among $H_2S$, $H_2Se$, and $H_2Te$, $H_2S$ has the lowest molecular weight, and hence, the lowest boiling point.

The bond angle decreases as we move down the group in the periodic table due to increasing atomic size and decreasing bond pair-bond pair repulsion. Thus, the bond angle is minimum for $ H_2 Te $.

In the reaction $ SO_2 + 2H_2S ightarrow 3S + 2H_2O $, sulphur dioxide ( $ SO_2 $ ) acts as an oxidizing agent. It oxidizes $ H_2S $ to sulfur (S), and in the process, $ SO_2 $ itself gets reduced to $ H_2O $. Hence, $ SO_2 $ is the oxidizing agent in this reaction.

When $ SO_2 $ is passed through cupric chloride ( $ CuCl_2 $ ) solution, a white precipitate of $ Cu_2Cl_2 $ is formed, and the solution becomes colorless. The reaction is: $$ SO_2 + 2CuCl_2 + 2H_2O ightarrow Cu_2Cl_2 + 2HCl + H_2SO_4 $$. Here, $ SO_2 $ reduces $ CuCl_2 $ to $ Cu_2Cl_2 $, resulting in the formation of a white precipitate and a colorless solution.