Which of the following complexes will give a white precipitate on treatment with a solution of barium nitrate?

[C r (N H_{3})_{4} S O_{4}] C l

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[C o (N H_{3})_{4} C l_{2}] N O_{2}

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[C r (N H_{3})_{4} C l_{2}] S O_{4}

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[C r C l_{2} (H_{2} O)_{4}] C l

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Solution

The correct option is A $[C r (N H_{3})_{4} C l_{2}] S O_{4}$
When $[C r (N H_{3})_{4} C l_{2}] S O_{4}$ dissolves it yields $[C r (N H_{3})_{4} C l_{2}]^{+ 2}$ and $S O_{4}^{2 -}$ .

The $S O_{4}^{2 -}$ ion thus formed reacts with the $B a^{2 +}$ ion to form a white precipitate of $B a S O_{4}$ .

$B a^{2 +} + S O_{4}^{2 -} ⇌ B a S O_{4} ↓$ Barium sulphate (white precipitate)

Hence, option $C$ is correct.

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