Q33 / 150 · MHT CET 2026 (13 April Shift 2) · Chemistry
Which of the following statements is correct with respect to $[\text{Mn(CN)}_6]^{3-}$ ?
- AIt is $sp^3 d^2$ hybridised and tetrahedral.
- Correct answerBIt is $d^2 sp^3$ hybridised and octahedral.
- CIt is $dsp^2$ hybridised and square planar
- DIt is $sp^3 d^2$ hybridised and octahedral.
Solution
The given complex is $[\text{Mn(CN)}_6]^{3-}$.
Let the oxidation state of Mn be $x$.
$x + 6(-1) = -3 \Rightarrow x = +3$
The electronic configuration of Mn is $[\text{Ar}] 3d^5 4s^2$.
The electronic configuration of $\text{Mn}^{3+}$ is $[\text{Ar}] 3d^4$.
Since $\text{CN}^-$ is a strong field ligand, it causes pairing of electrons in the $3d$ orbitals against Hund's rule. The four electrons occupy the $t_{2g}$ orbitals, leaving two $e_g$ orbitals empty.
These two empty $3d$ orbitals, along with one $4s$ and three $4p$ orbitals, undergo $d^2sp^3$ hybridization.
The geometry corresponding to $d^2sp^3$ hybridization is octahedral.
Answer: It is $d^2 sp^3$ hybridised and octahedral.
Let the oxidation state of Mn be $x$.
$x + 6(-1) = -3 \Rightarrow x = +3$
The electronic configuration of Mn is $[\text{Ar}] 3d^5 4s^2$.
The electronic configuration of $\text{Mn}^{3+}$ is $[\text{Ar}] 3d^4$.
Since $\text{CN}^-$ is a strong field ligand, it causes pairing of electrons in the $3d$ orbitals against Hund's rule. The four electrons occupy the $t_{2g}$ orbitals, leaving two $e_g$ orbitals empty.
These two empty $3d$ orbitals, along with one $4s$ and three $4p$ orbitals, undergo $d^2sp^3$ hybridization.
The geometry corresponding to $d^2sp^3$ hybridization is octahedral.
Answer: It is $d^2 sp^3$ hybridised and octahedral.
Hence, the correct answer is option: 'B'.