The conductivity measurement of a coordination compound of cobalt (III) shows that it dissociates into 3 ions in solution. The compound is

Co-ordinate compound of Co (III) dissociates into 3 ions in the solution means it has two ionisable-ions out side the coordination sphere with oxidation state of cobalt $(\mathrm{Co})=+3$ In option Name : Hexaammine cobalt (III) chloride Formula : $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ $\because$ Oxidation state of $\mathrm{Co}=(+) 3$ In option (b) Name : Pentaammine sulphatocobalt (III) chloride Formula : $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{SO}_{4}\right)\right] \mathrm{Cl}$ $\because$ Oxidation state of $\mathrm{Co}=+3$ In option (c) Name : Pentaamminechloridocobalt (III) sulphate Formula : $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{SO}_{4}$ $\because$ Oxidation state of $\mathrm{Co}=+3$ In option (d) Name : Pentaamminechloridocobalt (III) chloride Formula: $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$ $\because$ Oxidation state of $\mathrm{Co}=+3$ Dissociation of this compound is shown below: $\left[\mathrm{CO}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \rightarrow \mathrm{CO} \underbrace{\left.\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right]^{2+}+2 \mathrm{Cl}^{-}}_{3 \text { ions }}$ This compound can give total 3 -ions, as follows. one $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right]^{2+}$ and two $\mathrm{Cl}^{-}$ ions. Hence, option (d) is the correct answer.