Base hydrolysis of $\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}$ is an overall second order reaction, whereas that of $[\mathrm{Co}(\mathrm{CN})]^{3-}$ is of first order. The rates depend in both cases solely on the concentrations of the cobalt complex. This may be due to(A) Presence of ionizable proton in $\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}$ but not in $\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}$(B) $\mathrm{S_{N}}1_{\mathrm{CB}}$ mechanism in the case of $\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}$ only(C) $\mathrm{S_{N}}1_{\mathrm{CB}}$ mechanism in the case of $\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}$, only(D) $\mathrm{S_{N}}1_{\mathrm{CB}}$ mechanism in both the complexesCorrect explanation(s) is/are
(a)A and B
(b)A and C
(c)B only
(d)A and D
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
[CoCl(NH₃)₅]²⁺ has acidic N–H protons, so OH⁻ forms the amido conjugate base, which then loses Cl⁻ (SN1CB, second order). [Co(CN)₆]³⁻ has no such proton. A and B (a).