• # question_answer Which one of the following ions will give a coloured solution? A) $C{{u}^{+}}$                               B)  $F{{e}^{2}}^{+}$C) $Z{{n}^{2+}}$                              D) $A{{g}^{+}}$

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].