Pick out the correct statement with respect to `[Mn(CN)_(6)]^(3-)`:

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 planar

Explanation

To solve the question regarding the complex ion \([Mn(CN)_6]^{3-}\), we will follow these steps:

### Step 1: Determine the oxidation state of manganese (Mn)
Let the oxidation state of Mn be \(X\). The cyanide ion (CN) has a charge of \(-1\), and since there are 6 cyanide ions, their total contribution is \(-6\). The overall charge of the complex is \(-3\).

The equation can be set up as follows:
\[ X + 6(-1) = -3 \]
Solving for \(X\):
\[ X - 6 = -3 \implies X = +3 \]

### Step 2: Write the electronic configuration of \(Mn^{3+}\)
The atomic number of manganese (Mn) is 25. The electronic configuration of neutral Mn is:
\[ [Ar] 3d^5 4s^2 \]
When Mn loses 3 electrons to form \(Mn^{3+}\), it loses the two 4s electrons and one 3d electron:
\[ Mn^{3+} \text{ configuration: } [Ar] 3d^4 \]

### Step 3: Identify the hybridization
Cyanide (CN) is a strong field ligand, which means it will cause pairing of electrons in the d-orbitals. In the case of \(Mn^{3+}\), the 3d orbitals will be filled as follows:
- The 3d orbitals will have 4 electrons, and since CN is a strong field ligand, the pairing will occur, resulting in:
\[ \text{3d: } \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow \uparrow \]
This means we have 2 paired electrons and 2 unpaired electrons in the 3d orbitals.

### Step 4: Determine the hybridization type
Since there are 6 ligands (CN) surrounding the Mn ion, and the hybridization involves 2 d-orbitals, 1 s-orbital, and 3 p-orbitals, the hybridization can be described as:
\[ d^2sp^3 \]

### Step 5: Determine the geometry
With \(d^2sp^3\) hybridization and 6 ligands, the geometry of the complex is octahedral.

### Conclusion
The correct statement regarding \([Mn(CN)_6]^{3-}\) is:
- **Option C:** It is \(d^2sp^3\) hybridized and octahedral.