According to crystal field therory the electronic configuration of the `[Cr(H_(2)O)_(6)]^(2+)` ion when `(Delta ltP)`
`(Delta = CFSE, P= "Pairing energy")` .

Options

A. `3 (t_(2g)^(3) e_(g)^(1))`
B. `4 (t_(2g)^(3) e_(g)^(1))`
C. `3(e_(g)^(1) t_(2g)^(1))`
D. `3(e_(g)^(3) t_(2g)^(1))`

Explanation

To determine the electronic configuration of the \([Cr(H_2O)_6]^{2+}\) ion according to crystal field theory when the crystal field splitting energy (\(\Delta\)) is less than the pairing energy (P), we can follow these steps:

### Step 1: Determine the oxidation state of chromium in the complex.
- The complex is \([Cr(H_2O)_6]^{2+}\).
- Water (\(H_2O\)) is a neutral ligand, so it does not contribute to the charge.
- The overall charge of the complex is \(+2\).
- Therefore, the oxidation state of chromium (Cr) is \(+2\).

### Step 2: Write the electronic configuration of chromium in its elemental form.
- The atomic number of chromium (Cr) is 24.
- The ground state electronic configuration of Cr is:
\[ [Ar] 3d^5 4s^1 \]

### Step 3: Determine the electronic configuration of \(Cr^{2+}\).
- When chromium loses two electrons to form \(Cr^{2+}\), it loses the \(4s\) electron first and then one \(3d\) electron.
- Thus, the electronic configuration of \(Cr^{2+}\) is:
\[ [Ar] 3d^4 \]

### Step 4: Understand the crystal field splitting in octahedral complexes.
- In an octahedral field, the \(d\) orbitals split into two sets: \(t_{2g}\) (lower energy) and \(e_g\) (higher energy).
- The \(t_{2g}\) orbitals can hold a maximum of 6 electrons, while the \(e_g\) orbitals can hold a maximum of 4 electrons.

### Step 5: Apply the condition \((\Delta < P)\).
- Since \(\Delta < P\), this indicates that the ligands (in this case, water) are weak field ligands, leading to a high-spin configuration.
- In a high-spin configuration, electrons will occupy the higher energy \(e_g\) orbitals before pairing occurs in the lower energy \(t_{2g}\) orbitals.

### Step 6: Fill the \(d\) orbitals according to the high-spin configuration.
- We have 4 electrons to place in the \(3d\) orbitals.
- Following Hund's rule, we fill the \(t_{2g}\) orbitals first:
- Place one electron in each of the three \(t_{2g}\) orbitals: \(t_{2g}^3\).
- Then place the fourth electron in one of the \(e_g\) orbitals: \(e_g^1\).

### Final Electronic Configuration:
- The final electronic configuration of the \([Cr(H_2O)_6]^{2+}\) ion is:
\[ t_{2g}^3 e_g^1 \]

### Conclusion:
- The electronic configuration of the \([Cr(H_2O)_6]^{2+}\) ion is \(t_{2g}^3 e_g^1\).

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