A) \[C{{u}^{+}}\]
B) \[F{{e}^{2}}^{+}\]
C) \[Z{{n}^{2+}}\]
D) \[A{{g}^{+}}\]
Correct Answer: B
Solution :
Since\[\text{F}{{\text{e}}^{\text{2+}}}\]has unpaired electrons so, d-d transitions would take place and hence, its solution would be coloured. \[C{{u}^{+}}\to 3{{d}^{10}}\] \[F{{e}^{2+}}\to 3{{d}^{6}}\] \[Z{{n}^{2+}}\to 3{{d}^{10}}\] \[A{{g}^{+}}\to 4{{d}^{10}}\] Hence, the correct option is [b].
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