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 $A l$ , 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):

N_{2}

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C l_{2}

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N H_{3}

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H_{2}

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Solution

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.

$2 N a_{2} O_{2} + 2 H_{2} O ⟶ 4 N a O H + O_{2}$

Al is attacked by $N a O H$ and thus the gas (D) is $H_{2}$ :

$2 A l + 2 N a O H + 2 H_{2} O ⟶ 2 N a A l O_{2} + 3 H_{2}$

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

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