Question

A soft silver-white metal (A) burns with golden yellow flame to give a yellow powder B which on treatment with water liberates oxygen giving a clear solution C. The solution placed on Al attacks this metal, liberating a gas D and forming a water soluble compound E. The metal (A) dissolves in liquid $N{H}_3$ to form 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

Answer: (D)

$H_2$

Sodium always burns with yellow flame. according to passage a soft silver white metal (A) burns with golden yellow flame to give a yellow poweder B which on treatment with water liberates oxygen giving a clear solution .

$2N{a}_2+O_2\to 2N{a}_{2}O$ (yellow powder)

Which on treatment with water liberates oxygen giving a clear solution.

$2N{a}_2{O}_2+2H_{2}O\to 4NaOH+O_2$

The solution placed on AI attacks this metal,liberating a gas D and forming a water soluble compound E.

$2Al+2NaOH+2H_{2}O\to 2NaAl{O}_2+3H_2$

So, D is $H_2$.