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Chlorides of Phosphorus

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Question

Read the following scheme:
Colourless (A) $N a O H ⟶$ (B) gas + (C) aq (alkaline due to excess NaOH added)
(B) + $N a ⟶ (D) + H_{2}$
C (aq) + $Z n △ ⟶$ (B)
(A) $△ ⟶ (E) + H_{2} O$
(E) + (D) $⟶$ (F)
(F) + $H_{2} S O_{4} ⟶$ (G)
(G) contains 97.67% N, rest hydrogen.
Compound (A) to (G) are identified as:
(A) : $N H_{4} N O_{2}$ , (B) : $N H_{3}$ , (C) : $N a N O_{3}$ , (D) : $N a N H_{2}$ , (E) : $N_{2} O$ , (F) : $N_{3} N a$ (sodium azide) and (G) : $N_{3} H$ (hydroazoic acid)
If true enter 1, else enter 0.

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Solution

The colourless compound (A) is $N H_{4} N O_{2}$
The gas (B) is $N H_{3}$
The compound (C) is $N a N O_{3}$
The compounnd (D) is $N a N H_{2}$
The gaseous compound (E) is $N_{2} O$
The compound (F) is $N_{3} N a$ (sodium azide)
The compound (G) is $N_{3} H$ (hydroazoic acid)
Ammonium nitrite reacts with NaOH to form ammonia and sodium nitrate.
$N H_{4} N O_{2} N a O H ⟶ N H_{3} + N a N O_{3} a q$ (alkaline due to excess NaOH added)
Sodium metal reacts with ammonia to form sodaamide and hydrogen gas.
$2 N H_{3} + 2 N a ⟶ 2 N a N H_{2} + H_{2}$
When sodium nitrite is heated with zinc, ammonia is obtained.
$N a N O_{3} (a q) + Z n △ ⟶ N H_{3}$
When ammonium nitrite is heated, dintorgen monoxide gas is obtained along withwater.
$N H_{4} N O_{2} △ ⟶ N_{2} O + H_{2} O$
Dinitrogen monoxide reacts with sodaamide to form sodium azide.
$N_{2} O + N a N H_{2} ⟶ N_{3} N a + H_{2} O$
Sodium azide reacts with sulphuic acid to form hydrazic acid.
$2 N_{3} N a + H_{2} S O_{4} ⟶ 2 N_{3} H + N a_{2} S O_{4}$
$N_{3} H$ contains 97.67% N, rest hydrogen.

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