If $F e^{3 +}$ and $C r^{3 +}$ both are present in group III of qualitative analysis, then distinction can be made by :

addition of

N H_{4} O H

in presence of

N H_{4} C l

, when only

F e {(O H)}_{3}

is precipitated

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addition of

N H_{4} O H

in presence of

N H_{4} C l

, when

C r {(O H)}_{3}

and

F e {(O H)}_{3}

both are precipitated; and on adding

B r_{2}

water and

N a O H

C r {(O H)}_{3}

dissolves

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precipitates of

C r {(O H)}_{3}

and

F e {(O H)}_{3}

as obtained in (B) are treated with

c o n c . H C l

when only

F e {(O H)}_{3}

dissolves

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both (B) and (C) are correct

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Solution

The correct option is B addition of $N H_{4} O H$ in presence of $N H_{4} C l$ , when $C r {(O H)}_{3}$ and $F e {(O H)}_{3}$ both are precipitated; and on adding $B r_{2}$ water and $N a O H$ , $C r {(O H)}_{3}$ dissolves
If $F e^{3 +}$ and $C r^{3 +}$ both are present, then very first solid ammonium chloride and ammonium hydroxide is added slowly till the solution smells of ammonia. $F e^{3 +}$ and $C r^{3 +}$ preicipitates in the hydroxide form.

For identification precipitate is treated with $N a O H$ and $B r_{2}$ water, yellow colouration confirms $C r^{3 +}$ ion.

$2 N a O H + B r_{2} \to N a B r O + N a B r + H_{2} O$

$n a B r O \to N a B r + [O]$

$2 C r (O H)_{3} + 4 N a O H + 3 [O] \to 2 N a_{2} C r O_{4} + 5 H_{2} O$

Solution is acidified and treated with lead accetate solution.

$N a_{2} C r O_{4} + P b (C H_{3} C O O)_{2} \to P b C r O_{4} + 2 C H_{3} C O O N a$
(Yellow ppt)

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