Question

Which of the following statements is correct with respect to \([\text{Mn(CN)}_6]^{3-}\) ?

• A. It is \(sp^3d^2\) hybridised and tetrahedral.
• B. It is \(d^2sp^3\) hybridised and octahedral.
• C. It is \(dsp^2\) hybridised and square planar
• D. It is \(sp^3d^2\) hybridised and octahedral.

Hint:

CN- is a strong field ligand, so d electrons pair up and the complex is inner orbital, d2sp3.

Answer 1

The Correct Option is B

Step 1: Plan:
Get the oxidation number, decide spin state, then count the orbitals used.

Step 2: Steps:
Mn is $+3$, so it has $4$ d electrons. Cyanide forces pairing, so only one d orbital set is occupied by a pair while two d orbitals are left free (low spin, $\mu\approx 2.83$ BM, 2 unpaired electrons).
Those two vacant inner $d$ orbitals bond with the $s$ and three $p$ orbitals, giving $d^2sp^3$. Six hybrid orbitals point to the corners of an octahedron.

Final Answer:
The complex is $d^2sp^3$ hybridised and octahedral, option (B). \[ \boxed{d^2sp^3\ \text{and octahedral}} \]