Which of the electronic configuration according to crystal field theory of the compound `Rh^(+2)` with `CN=6` is correct when `DeltagtP` ? .

Options

A. `a.(t_(2g)^(6) e_(g)^(1))`
B. `b. (t_(2g)^(5) e_(g)^(2))`
C. `c. (t_(2g)^(5) e_(g)^(1))`
D. `d. (t_(2g)^(3) e_(g)^(3))`

Explanation

To determine the electronic configuration of the compound Rhodium (Rh) in the +2 oxidation state with a coordination number of 6 (CN=6) according to crystal field theory, we will follow these steps:

### Step 1: Determine the electronic configuration of Rhodium (Rh)
Rhodium (Rh) has the atomic number 45. Its ground state electronic configuration is:
\[ \text{Rh: } [Kr] 4d^8 5s^1 \]

### Step 2: Find the electronic configuration for Rhodium in the +2 oxidation state
When Rhodium loses 2 electrons to form Rh\(^{+2}\), the electrons are removed first from the 5s orbital and then from the 4d orbital. Therefore, the electronic configuration for Rh\(^{+2}\) will be:
\[ \text{Rh}^{+2}: [Kr] 4d^6 \]

### Step 3: Understand the crystal field splitting in octahedral complexes
In an octahedral field, the d-orbitals split into two sets: the lower energy \(t_{2g}\) (which consists of three orbitals) and the higher energy \(e_g\) (which consists of two orbitals). The splitting occurs due to the presence of ligands around the central metal ion.

### Step 4: Determine the type of ligands and their effect on electron pairing
Given that the problem states that the crystal field splitting energy (\(\Delta\)) is greater than the pairing energy (P), we are dealing with strong field ligands. Strong field ligands favor low spin complexes, which means that electrons will pair up in the lower energy \(t_{2g}\) orbitals before occupying the higher energy \(e_g\) orbitals.

### Step 5: Fill the d-orbitals according to the crystal field theory
Since we have 6 electrons to place in the d-orbitals, we will fill the \(t_{2g}\) orbitals first:
1. Fill the three \(t_{2g}\) orbitals with 6 electrons:
- \(t_{2g}\) will have 6 electrons, all paired: \(t_{2g}^6\)
2. The \(e_g\) orbitals will have 0 electrons: \(e_g^0\)

### Step 6: Write the final electronic configuration
The final electronic configuration of Rh\(^{+2}\) in an octahedral field with strong field ligands is:
\[ t_{2g}^6 e_g^0 \]

### Conclusion
Thus, the correct electronic configuration of Rh\(^{+2}\) with CN=6 when \(\Delta > P\) is:
\[ t_{2g}^6 e_g^0 \]