Crystal field stabalisation energy for complex `[Co(CN)_(6)]^(-3)` will be :-

Options

A. `+2.4 Delta_(0)
B. `+2.4 Delta_(0)
C. `-3.6 Delta_(0)
D. `-1.8 Delta_(0)

Answer

To find the Crystal Field Stabilization Energy (CFSE) for the complex \([Co(CN)_6]^{-3}\), we will follow these steps:

### Step 1: Determine the oxidation state of cobalt in the complex.
The complex is \([Co(CN)_6]^{-3}\). Since cyanide (CN) is a ligand with a charge of -1 and there are 6 cyanide ligands, the total negative charge contributed by the ligands is -6. To balance this with the overall charge of -3, cobalt must have an oxidation state of +3.

**Hint:** Remember that the overall charge of the complex is equal to the sum of the oxidation states of the metal and the charges of the ligands.

### Step 2: Write the electron configuration of cobalt in the +3 oxidation state.
Cobalt (Co) has an atomic number of 27. The ground state electron configuration of cobalt is:
\[ \text{Co: } [Ar] 3d^7 4s^2 \]

In the +3 oxidation state, cobalt loses 3 electrons (2 from 4s and 1 from 3d), resulting in:
\[ \text{Co}^{3+}: [Ar] 3d^6 \]

**Hint:** When determining the electron configuration for an ion, remember to remove electrons from the outermost shell first.

### Step 3: Determine the number of electrons in the d-orbitals after splitting.
In the octahedral field created by the six cyanide ligands, the 3d orbitals split into two sets: the lower energy \(T_{2g}\) (dxy, dyz, dzx) and the higher energy \(E_g\) (dx^2-y^2, dz^2).

Since cyanide is a strong field ligand, it causes pairing of electrons. Therefore, the 6 electrons in \(3d\) will fill the \(T_{2g}\) orbitals completely:
- \(T_{2g}\): 6 electrons (fully filled)
- \(E_g\): 0 electrons (empty)

**Hint:** Strong field ligands lead to pairing of electrons in the lower energy orbitals before occupying the higher energy orbitals.

### Step 4: Calculate the Crystal Field Stabilization Energy (CFSE).
The formula for CFSE is given by:
\[ \text{CFSE} = (n_{T_{2g}} \times -0.4) + (n_{E_g} \times 0.6) \]
Where:
- \(n_{T_{2g}}\) = number of electrons in \(T_{2g}\)
- \(n_{E_g}\) = number of electrons in \(E_g\)

Substituting the values:
- \(n_{T_{2g}} = 6\)
- \(n_{E_g} = 0\)

Thus,
\[ \text{CFSE} = (6 \times -0.4) + (0 \times 0.6) = -2.4 \]

**Hint:** Make sure to multiply the number of electrons by their respective stabilization energies correctly.

### Conclusion
The Crystal Field Stabilization Energy (CFSE) for the complex \([Co(CN)_6]^{-3}\) is \(-2.4 \Delta\).

**Final Answer:** \(-2.4 \Delta\)