Among the given options, the ion that forms a complex with a coordination number of 2 with an excess of $\text{CN}^{-}$ ions is $(3)$ $\text{Ag}^{+}$ (silver ion).
$\text{Ag}^{+}$ (silver ion) forms a complex with cyanide ion, $\text{CN}^{-}$, with a coordination number of 2. The resulting complex is called $\text{Ag(CN)}_{2}^{-}$, and it has a linear structure. The cyanide ion acts as a strong ligand and can donate two electrons to form a stable complex with $\text{Ag}^{+}$.
On the other hand, $\text{Ni}^{2+}$ (nickel ion) typically forms complexes with a coordination number of 4 or 6 when cyanide ions are present. $\text{Cu}^{+}$ (copper ion) also forms complexes with a coordination number of 4 when cyanide ions are present. $\text{Fe}^{2+}$ (iron ion) does not form a stable complex with cyanide ion at all.
Therefore, the correct answer is $(3)$ $\text{Ag}^{+}$, as it forms a complex with a coordination number of 2 with an excess of $\text{CN}^{-}$ ions. |
