Question

A soft silver-white metal (A) burns with a golden-yellow flame to give a yellow powder (B) which on treatment with water liberates oxygen giving a clear solution (C). The solution, when mixed with $Al$, attacks the metal, liberating a gas (D) and forming a water-soluble compound (E). The metal (A) dissolves in liquid $N{H}_3$ to form a deep blue solution which is a good conductor and an excellent reducing agent. Identify (D):

• A. $N_2$
• B. $C{l}_2$
• C. $N{H}_3$
• D. $H_2$

Answer 1

The correct option is D $H_2$

The intense yellow flame points to sodium metal (A). It burns in air to give a yellow powder.
Purely owing to the colour, (B) is $N{a}_2{O}_2$ and not $N{a}_{2}O$ which is white.

$2N{a}_2{O}_2+2H_{2}O⟶4NaOH+O_2$

Al is attacked by $NaOH$ and thus the gas (D) is $H_2$:

$2Al+2NaOH+2H_{2}O⟶2NaAl{O}_2+3H_2$

The question only asks you to identify (D) and nothing else.
You can disregard any information not helpful in identifying the gas.