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With excess of $Cl_2$ ammonia forms

$\begin{array}{1 1}(a)\;NH_4Cl&(b)\;N_2\\(c)\;NCl_3&(d)\;NClNH_3\end{array}$

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1 Answer

With excess of $Cl_2$ ammonia forms $NCl_3$
Hence (c) is the correct answer.
answered Nov 28, 2013 by sreemathi.v
 
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