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

Options

A. `d^(2) sp^(3) ` , inner orbital complex , paramagnetic , 2.87 B.M.
B. `d^(2) sp^(3)` , inner orbital complex , diamagnetic , zero magnetic moment .
C. `d^(2) sp^(3)` , outer orbital complex , paramagnetic , 3.87 B.M.
D. `dsp^(2)` , outer orbital complex , diamagnetic , zero magnetic moment

Explanation

To solve the question regarding the complex ion \([Mn(CN)_6]^{3-}\), we need to analyze its properties step by step.

### Step 1: Determine the oxidation state of manganese (Mn).
1. Let the oxidation state of manganese be \(X\).
2. Each cyanide ion (CN) has a charge of \(-1\). Since there are 6 cyanide ions, their total contribution to the charge is \(-6\).
3. The overall charge of the complex is \(-3\).
4. Set up the equation based on the charges:
\[ X + (-6) = -3 \]
5. Solving for \(X\):
\[ X - 6 = -3 \implies X = -3 + 6 = +3 \]
Thus, the oxidation state of manganese in \([Mn(CN)_6]^{3-}\) is \(+3\).

### Step 2: Determine the electronic configuration of \(Mn^{3+}\).
1. The atomic number of manganese (Mn) is 25, and its ground state electronic configuration is:
\[ [Ar] 3d^5 4s^2 \]
2. For \(Mn^{3+}\), we need to remove 3 electrons. The electrons are removed first from the 4s and then from the 3d subshell:
\[ Mn^{3+} \rightarrow [Ar] 3d^4 \]

### Step 3: Analyze the coordination environment.
1. The complex \([Mn(CN)_6]^{3-}\) is octahedral because it has six ligands (CN).
2. Cyanide (CN) is a strong field ligand, which causes significant splitting of the d-orbitals.

### Step 4: Determine the splitting of d-orbitals and electron filling.
1. In an octahedral field, the 3d orbitals split into two sets: \(t_{2g}\) (lower energy) and \(e_g\) (higher energy).
2. For \(Mn^{3+}\) with \(3d^4\):
- The first three electrons will fill the \(t_{2g}\) orbitals singly (due to Hund's rule).
- The fourth electron will pair up in one of the \(t_{2g}\) orbitals because CN is a strong field ligand.
3. Thus, the electron configuration in the crystal field will be:
- \(t_{2g}^4\) and \(e_g^0\).

### Step 5: Determine hybridization.
1. The hybridization for \([Mn(CN)_6]^{3-}\) can be determined by the number of orbitals involved:
- Since we have 4 electrons in \(t_{2g}\) and the complex is octahedral, the hybridization is \(d^2sp^3\).

### Step 6: Determine the magnetic properties.
1. Since there are 2 unpaired electrons in the \(t_{2g}\) orbitals, the complex is paramagnetic.
2. The magnetic moment can be calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \(n\) is the number of unpaired electrons.
3. For \(n = 2\):
\[ \mu = \sqrt{2(2 + 2)} = \sqrt{8} = 2.83 \text{ (approximately)} \]

### Conclusion:
The correct statements regarding \([Mn(CN)_6]^{3-}\) are:
- The oxidation state of manganese is \(+3\).
- The hybridization is \(d^2sp^3\).
- It is an inner orbital complex.
- It is paramagnetic with a magnetic moment of approximately \(2.83 \, \mu_B\).