The compound in which cation is isoelectronic with anion is

 $$\begin{gathered} \left(1\right) \mathrm{NaCl}\to {\mathrm{Na}}^{+} + {\mathrm{Cl}}^{-} \\ (11-1) (17+1) \mathrm{electrons} \\ \mathrm{Both} \mathrm{Cation} \& \mathrm{anion} \mathrm{have} \mathrm{different} \mathrm{no}.\mathrm{of} \mathrm{electrons} \\ \left(2\right) \mathrm{CsF}\to {\mathrm{Cs}}^{+} + {\mathrm{F}}^{-} \\ (55-1) (9+1)\mathrm{electrons} \\ \mathrm{Both} \mathrm{Cation} \& \mathrm{anion} \mathrm{have} \mathrm{different} \mathrm{no}.\mathrm{of} \mathrm{electrons} \\ \left(3\right) \mathrm{NaI}\to {\mathrm{Na}}^{+} + {\mathrm{I}}^{-} \\ (11-1) (53+1)\mathrm{electrons} \\ \mathrm{Both} \mathrm{Cation} \& \mathrm{anion} \mathrm{have} \mathrm{different} \mathrm{no}.\mathrm{of} \mathrm{electrons} \end{gathered}$$

$$\begin{gathered} \left(4\right) {\mathrm{K}}_2\mathrm{S}\to {\mathrm{K}}^{+}+ {\mathrm{S}}^{-2} \\ \mathrm{Number} \mathrm{of} \mathrm{electrons} \mathrm{in} {\mathrm{K}}^{+} =\mathrm{Z}-1=19-1 = 18 \\ \mathrm{Number} \mathrm{of} \mathrm{electrons} \mathrm{in} {\mathrm{S}}^{-2} = \mathrm{Z}+2 =16+2 =18 \\ \mathrm{Hence} \mathrm{both} \mathrm{cation} \& \mathrm{anion} \mathrm{have} \mathrm{same} \mathrm{no}.\mathrm{of} \mathrm{electrons}\left(\mathrm{Isoelectronic}\right) \end{gathered}$$