Which of the following complex ion is more likely to be colourless?

The correct answer is option 4. \([Ag(NH_3)_6]^+\).

To determine which complex ion is more likely to be colorless, we need to consider the electronic configurations and the likelihood of d-d transitions, which are responsible for color in transition metal complexes.

1. \([Co(H_2O)_6]^{2+}\): Cobalt(II) has an electronic configuration of \(3d^7\). With water as a ligand (a weak field ligand), this complex will have partially filled d-orbitals, leading to d-d transitions that typically result in colored compounds. Hence, it is likely to be colored.

2. \([Mn(CN)_6]^{3-}\): Manganese in the +3 oxidation state has an electronic configuration of \(3d^4\). Cyanide is a strong field ligand, which can cause pairing of electrons. However, it still results in some d-d transitions. Complexes of manganese with CN ligands are generally colored due to these transitions.

3. \([CrCl_3(H_2O)_3]\): Chromium in this complex is in the +3 oxidation state, which corresponds to an electronic configuration of \(3d^3\). Chloride and water are weak field ligands, meaning there are unpaired electrons in the d-orbitals, allowing for d-d transitions. This complex is typically colored.

4. \([Ag(NH_3)_2]^+\): Silver in the +1 oxidation state has an electronic configuration of \(4d^{10}\). The \(d^{10}\) configuration is completely filled, meaning there are no available d-d transitions. Complexes with a \(d^{10}\) configuration are usually colorless because there are no d-d transitions to absorb visible light.

Therefore, the complex ion more likely to be colorless is option 4: \([Ag(NH_3)_2]^+\).