A) \[Zn{{(N{{O}_{3}})}_{2}}\]
B) \[LiN{{O}_{3}}\]
C) \[CrC{{l}_{3}}\]
D) Potash alum
Correct Answer: C
Solution :
\[CrC{{l}_{3}},\ \text{as}\ C{{r}^{+3}}\ \Rightarrow \ 3{{d}^{3}},\] has unpaired electron in d-orbital as a result it will show paramagnetism and thus forms coloured complexes.
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