${C l}_{2}$ reacts with excess of ammonia gives:

N H_{4} C l

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N {C l}_{3}

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N_{2} C l

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none of these

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Solution

The correct option is A $N H_{4} C l$
$C l_{2}$ reacts with excess of ammonia to give $N H_{4} C l + N_{2}$
$3 C l_{2} + 8 N H_{3} (e x c e s s) \to 6 N H_{4} C l + N_{2}$

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