 Solution

The free end of N-bound $\mathrm{Co}-\mathrm{NCS}$ is sulfur. Cr(II) binds there, giving the remote bridge $\mathrm{C}{\mathrm{o}}^{\mathrm{III}}-\mathrm{N}-\mathrm{C}-\mathrm{S}-\mathrm{C}{\mathrm{r}}^{\mathrm{II}}$. Electron transfer and cleavage of the Co-N bond leave S-bound Cr(III), $[\mathrm{Cr}(\mathrm{SCN})(\mathrm{O}{\mathrm{H}}_2{)}_5{]}^{2+}$; labile Co(II) becomes hexaaquacobalt(II).

The much smaller rate constant shows that the efficiency of bridge formation/electron transmission is strongly sensitive to linkage orientation. Efficient donation and coupling in a Co-S-bound precursor need not be reproduced when cobalt is N-bound. Its reduction is still consistent with inner-sphere electron transfer; a slow rate alone is not a criterion for outer-sphere transfer.