The number of chloride ions which would be precipitated, when $CrC{l}_3.4N{H}_3$ is treated with silver nitrate solution:

The correct option is C 1

Chromium has coordination number six.

Therefore to write its formula, we will first add the maximum number of neutral elements inside the coordination sphere.

Hence, the formula is $\left[Cr\right(N{H}_3{)}_{4}C{l}_2]Cl$.

As 1 $C{l}^{-}$ ion is available for dissociation only one $C{l}^{-}$ ion will precipitate when it reacts with $AgN{O}_3$.

Hence, the correct option is (C).