\[ \begin{array}{l} \left[\mathrm{CoF}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+} \text{ and } \left[\mathrm{Cr}\left(\mathrm{OH}_{2}\right)_{6}\right]^{2+} ; \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{OH}_{2}\right)\right]^{3+} \text{ and } \left[\mathrm{Cr}\left(\mathrm{OH}_{2}\right)_{6}\right]^{2+} \\ \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+} \text{ and } \left[\mathrm{Cr}\left(\mathrm{OH}_{2}\right)_{6}\right]^{2+} ; \left[\mathrm{CoI}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+} \text{ and } \left[\mathrm{Cr}\left(\mathrm{OH}_{2}\right)_{6}\right]^{2+} \end{array} \]
The electron transfer rate will be fastest in the pair.
(a)$\left[\mathrm{CoF}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}$ and $\left[\mathrm{Cr}\left(\mathrm{OH}_{2}\right)_{6}\right]^{2+}$
(b)$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{OH}_{2}\right)^{3+}\right]$ and $\left[\mathrm{Cr}\left(\mathrm{OH}_{2}\right)_{6}\right]^{2+}$
(c)$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ and $\left[\mathrm{Cr}\left(\mathrm{OH}_{2}\right)_{6}\right]^{2+}$
(d)$\left[\mathrm{CoI}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}$ and $\left[\mathrm{Cr}\left(\mathrm{OH}_{2}\right)_{6}\right]^{2+}$
Answer
Answer: D ✓ checked by 4AB · confidence high
Explanation
Cr(II) reduces these Co(III) complexes by the inner-sphere route through a bridging ligand. Halide bridges work best for the heavier, more polarisable halogen: I⁻ > Br⁻ > Cl⁻ > F⁻; NH₃ and H₂O cannot bridge. So [CoI(NH₃)₅]²⁺ reacts fastest.