Question

A hydride (X) of Group 15 element is distinctly basic and has unexpeclably high boiling point. It reacts with NaOCl to give another hydride (Y), which is a strong reducing agent and is used in organic analysis. X and Y are

• A. $PH_3, P_2H_4$
• B. $NH_3, N_2H_4$
• C. $AsH_3, As_2H_4$
• D. $NH_3, NH_4Cl$

Answer 1

Answer: (B)

$NH_3, N_2H_4$

Answer 2

Out of all the hydrides of group 15 elements only $NH_3$ and $PH_3$ are basic. Further boiling pt. of $NH_3$ is higher than $PH_3$ and it is distinctly basic whereas $PH_3$ is weakly basic. Hence X is $NH_3$. {\underset{\mathrm{ (X)}}{ {2NH_3}}$+ NaOCl ->$\underset{\mathrm{ (Y) }}{ { N_2H_4}} + NaCl + H_2O}