A) \[CuCl\]
B) \[Cu{{F}_{2}}\]
C) \[A{{g}_{2}}S{{O}_{4}}\]
D) \[Mg{{F}_{2}}\]
Correct Answer: B
Solution :
In\[Cu{{F}_{2}}\],\[Cu\]is present as\[C{{u}^{2+}}\]. \[C{{u}^{2+}}=[Ar]\,\,3{{d}^{9}},\,\,4{{s}^{0}}\] Since, in it one unpaired electron is present, it is coloured.
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