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: Understanding the Concept:
Hybridisation in a complex depends on the metal oxidation state and whether the ligand is strong or weak field. Strong field ligands pair up the d electrons and free inner d orbitals.

Step 2: Find the metal configuration:
Charge: \(x + 6(-1) = -3\), so \(x = +3\). Mn(III) is \(3d^4\).

Step 3: Fill the orbitals:
\(\text{CN}^-\) is a strong field ligand, so the four d electrons pair up in two of the three \(t_{2g}\) orbitals. This leaves two \(3d\) orbitals empty.
Two \(3d\), one \(4s\) and three \(4p\) orbitals mix to give six \(d^2sp^3\) hybrid orbitals, which accept six ligand pairs.

Step 4: Conclusion:
The complex is \(d^2sp^3\) and octahedral (low spin, inner orbital). Options A and D say \(sp^3d^2\), which is used with weak ligands such as \(\text{F}^-\). Option A also says tetrahedral, and option C says square planar, which fits \(d^8\) ions.

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