The only cations present in a slightly acidic solution are $F e^{3 +}$ , $Z n^{2 +}$ and $C u^{2 +}$ . The reagent which when added in excess to this solution would identify the $F e^{3 +}$ ion in one step is :

2

H C l

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6

N H_{3}

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6

N a O H

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H_{2} S

gas

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Solution

The correct option is B $6$ M $N H_{3}$
Cupric ion reacts with excess of ammonia solution to give a deep blue color of tetra ammine cupric ion.
$C u^{2 +} + 4 N H_{4} O H \to [C u (N H_{3})_{4}]^{2 +} Deep blue color + 4 H_{2} O$
Zn(II) ions react with ammonia to form white gelatinous ppt of $Z n (O H)_{2}$ which is soluble in excess ammonia.
$Z n^{2 +} + 2 N H_{3} + 2 H_{2} O \to Z n (O H)_{2} ↓ + 2 N H_{4}^{+}$
$Z n (O H) ↓ 4 N H_{3} \to [Z n (N H_{3})_{4}]^{2 +} + 2 O H^{-}$
When the above reactions are not observed, $F e^{3 +}$ ion is confirmed.

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