MHTCET 15 May 2024 Shift 2 Paper

Concept:For octahedral complexes, the type of hybridization depends on the oxidation state of the metal and the nature of the ligand.Explanation:Let the oxidation state of cobalt be $x$. For $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$: $x + 6(0) = +3$, so $x = +3$. Thus, cobalt is in the $\mathrm{Co}^{3+}$ state. Electronic configuration of Co ($Z=27$): $[\mathrm{Ar}]\,3d^{7}4s^{2}$. For $\mathrm{Co}^{3+}$: $[\mathrm{Ar}]\,3d^{6}$. $\mathrm{NH}_3$ is a strong field ligand, so it causes pairing of electrons. In the octahedral field, the configuration becomes $t_{2g}^{6}e_{g}^{0}$. This leaves two inner $3d$ orbitals vacant. Therefore, the hybridization involves two $3d$, one $4s$, and three $4p$ orbitals. So, the hybridization is $d^{2}sp^{3}$, which is an inner orbital complex.Answer:Option A: $d^{2}sp^{3}$