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Hydrides of Nitrogen

A colorless s...

Question

A colorless sait $(X)$ is soluble in water and also in alcohol and amines. On strong heating, $(X)$ gives a brown gas $(Y)$ and a grey residue. $(X)$ dissolves in ammonia to give a solution $(Z)$ which gives silver mirror with aldehydes. A solution of $(X)$ is easily reduced by iron(II) sulphate. A solution of $(X)$ also gives a brick red precipitate with potassium dichromate solution. Hence,choose the correct alternative :

A

X = P b {(N O_{3})}_{2}

,

Y = N O_{2}

,

Z = A g_{2} O

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B

X = A g N O_{3}

,

Y = N O

,

Z = {[A g {(N H_{3})}_{2}]}^{+}

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C

X = A g N O_{3}

,

Y = N O_{2}

,

Z = A g_{2} O

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D

X = A g N O_{3}

,

Y = N O_{2}

,

Z = {[A g {(N H_{3})}_{2}]}^{+}

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Solution

The correct option is B $X = A g N O_{3}$ , $Y = N O_{2}$ , $Z = {[A g {(N H_{3})}_{2}]}^{+}$
Option (D) is correct.
$\begin{matrix} A g N O_{3} & Δ - \to & 2 A g ↓ & + & 2 N O_{2} ↑ & + & O_{2} ↑ (X) & (g r e y r e s i d u e) & (Y) \end{matrix}$
$A g N O_{3} + N H_{4} O H ⟶ {[A g {(N H_{3})}_{2}]}^{+}$
$2 {[A g {(N H_{3})}_{2}]}^{+} + C H_{3} C H O + H_{2} O ⟶ 2 A g ↓ + C H_{3} C O O^{-} + 3 N H_{4}^{+} + N H_{3}$
$A g^{+} + F e^{2 +} ⟶ A g ↓ + F e^{3 +}$
$\begin{matrix} 2 A g N O_{3} & + & K_{2} C r O_{4} & ⟶ & A g_{2} C r O_{4} & ↓ & + & 2 K N O_{3} (B r i c k r e d) \end{matrix}$

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