Pick out the correct statement with respect `[Mn(CN)_(6)]^(4-)`

Options

A. It is `sp^(3)d^(2)` hybridised and octahedral
B. it is `sp^(3)d^(2)` hybridised and tetrahedral
C. It is `d^(2)sp^(3)` hybridised and octahedral
D. It is `dsp^(2)` hybridised and square planer

Answer

To solve the question regarding the coordination compound \([Mn(CN)_{6}]^{4-}\), we will follow these steps:

### Step 1: Determine the Oxidation State of Manganese
To find the oxidation state of manganese (Mn) in the complex \([Mn(CN)_{6}]^{4-}\), we can set up the equation based on the charges of the ligands and the overall charge of the complex.

Let \(x\) be the oxidation state of Mn. The cyanide ion (CN) has a charge of -1, and since there are 6 cyanide ligands, their total contribution to the charge is -6. The overall charge of the complex is -4.

Thus, we can write the equation:
\[ x + 6(-1) = -4 \]
\[ x - 6 = -4 \]
\[ x = +2 \]

**Hint:** Remember that the charge of the ligands contributes to the overall charge of the complex.

### Step 2: Identify the Geometry of the Complex
Since there are 6 ligands surrounding the manganese ion, the geometry of the complex will be octahedral. Octahedral complexes are formed when there are 6 ligands.

**Hint:** The number of ligands can help you determine the geometry of the coordination compound.

### Step 3: Determine the Hybridization of Manganese
In the case of \([Mn(CN)_{6}]^{4-}\), we need to consider the electron configuration of manganese. Manganese has the electron configuration of \([Ar] 3d^5 4s^2\).

Since Mn is in the +2 oxidation state, we remove two electrons from the outermost shell:
- The configuration becomes \(3d^5\) (4s electrons are removed).

Cyanide (CN) is a strong field ligand, which means it will cause pairing of the electrons in the d-orbitals. Therefore, the electron configuration in the presence of CN will be:
- \(3d^6\) (with all 5 d-electrons paired and one additional electron paired).

For hybridization, we have:
- 2 electrons from the s-orbital,
- 3 electrons from the p-orbitals,
- 2 electrons from the d-orbitals.

Thus, the hybridization is \(d^2sp^3\).

**Hint:** Strong field ligands can cause pairing of electrons, which affects the hybridization.

### Step 4: Conclusion
Based on the above analysis, we can conclude that:
- The oxidation state of manganese is +2.
- The geometry of the complex is octahedral.
- The hybridization of the complex is \(d^2sp^3\).

**Final Answer:** The correct statement with respect to \([Mn(CN)_{6}]^{4-}\) is that it has an octahedral geometry and \(d^2sp^3\) hybridization.