Mercury(II) chloride on reaction with gaseous ammonia forms:

H g ({N H}_{2}) C l

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H g {({N H}_{3})}_{2} {C l}_{2}

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[H g {({N H}_{3})}_{4}] {C l}_{2}

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[H g {({N H}_{3})}_{2}] C l

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Solution

The correct option is A $H g ({N H}_{2}) C l$
Mercury(II) chloride on reaction with gaseous ammonia forms a white precipitate of $[H g ({N H}_{2}) C l]$ .
$H g C l_{2} + 2 N H_{3} \to H g ({N H}_{2}) C l + N H_{4} C l$

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