The crystal field configuration of complexes `[Ru(en)_3]Cl_2` and `[Fe(H_2O)_6]^(2+)` respectively is:

Options

A. `t_(2g)^4,e_g^2`and `t_(2g)^6,e_g^0`
B. `t_(2g)^6,e_g^0`and `t_(2g)^4,e_g^2`
C. `t_(2g)^4,e_g^2`and `t_(2g)^4,e_g^2`
D. `t_(2g)^6,e_g^0`and `t_(2g)^6,e_g^0`

Answer

To determine the crystal field configuration of the complexes \([Ru(en)_3]Cl_2\) and \([Fe(H_2O)_6]^{2+}\), we will follow these steps:

### Step 1: Determine the oxidation states of the central metal ions.

1. **For \([Ru(en)_3]Cl_2\)**:
- Ethylene diamine (en) is a neutral ligand, contributing 0 charge.
- The overall charge of the complex is -2 from the two chloride ions (Cl\(^-\)).
- Let the oxidation state of Ru be \(x\). The equation becomes:
\[ x + 0 - 2 = -2 \implies x = +2 \]

2. **For \([Fe(H_2O)_6]^{2+}\)**:
- Water (H\(_2\)O) is also a neutral ligand, contributing 0 charge.
- The overall charge of the complex is +2.
- Let the oxidation state of Fe be \(y\). The equation becomes:
\[ y + 0 = +2 \implies y = +2 \]

### Step 2: Write the electron configurations of the metal ions.

1. **Ruthenium (Ru)**:
- Atomic number of Ru = 44.
- The electron configuration of neutral Ru is \([Kr] 5s^2 4d^6\).
- For Ru in +2 oxidation state, we remove 2 electrons from the 5s orbital:
\[ \text{Configuration: } [Kr] 4d^6 \]

2. **Iron (Fe)**:
- Atomic number of Fe = 26.
- The electron configuration of neutral Fe is \([Ar] 4s^2 3d^6\).
- For Fe in +2 oxidation state, we remove 2 electrons from the 4s orbital:
\[ \text{Configuration: } [Ar] 3d^6 \]

### Step 3: Determine the crystal field splitting and filling of electrons.

1. **For \([Ru(en)_3]Cl_2\)**:
- Ethylene diamine is a strong field ligand, causing pairing of electrons.
- The 4d orbitals split into \(t_{2g}\) and \(e_g\) levels.
- Fill the 6 electrons in the \(t_{2g}\) and \(e_g\) orbitals:
- All 6 electrons will pair in the \(t_{2g}\) level:
\[ t_{2g}^6 \quad e_g^0 \]

2. **For \([Fe(H_2O)_6]^{2+}\)**:
- Water is a weak field ligand, leading to less pairing.
- The 3d orbitals also split into \(t_{2g}\) and \(e_g\) levels.
- Fill the 6 electrons:
- Electrons will occupy each orbital singly before pairing:
\[ t_{2g}^4 \quad e_g^2 \]

### Final Answer:
- The crystal field configurations are:
- For \([Ru(en)_3]Cl_2\): \(t_{2g}^6 \quad e_g^0\)
- For \([Fe(H_2O)_6]^{2+}\): \(t_{2g}^4 \quad e_g^2\)