Electronic configuration of `[Cu(NH_(3))_(6)]^(2+)` on the basis of crystal field splitting theory is

Options

A. `t_(2g)^(4) e_(g)^(5)`
B. `t_(2g)^(6) e_(g)^(3)`
C. `t_(2g)^(9) e_(g)^(0)`
D. `t_(2)^(5) e_(g)^(4)`

Explanation

To determine the electronic configuration of the complex ion \([Cu(NH_3)_6]^{2+}\) based on crystal field splitting theory, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Oxidation State of Copper:**
- The complex ion is \([Cu(NH_3)_6]^{2+}\).
- Ammonia (\(NH_3\)) is a neutral ligand, contributing 0 to the overall charge.
- Therefore, the oxidation state of copper (Cu) can be calculated as:
\[ x + 0 = +2 \implies x = +2 \]
- Thus, the oxidation state of copper in this complex is +2.

2. **Find the Electronic Configuration of Copper:**
- The atomic number of copper (Cu) is 29. The ground state electronic configuration of copper is:
\[ [Ar] 3d^{10} 4s^1 \]
- For \(Cu^{2+}\), we need to remove two electrons. The electrons are removed first from the 4s orbital and then from the 3d orbital:
\[ [Ar] 3d^{10} 4s^1 \rightarrow [Ar] 3d^9 \]
- Therefore, the electronic configuration of \(Cu^{2+}\) is \(3d^9\).

3. **Identify the Geometry of the Complex:**
- The complex \([Cu(NH_3)_6]^{2+}\) is octahedral because it has six ligands surrounding the central metal ion.

4. **Apply Crystal Field Splitting Theory:**
- 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 can be represented as:
\[ t_{2g} \text{ (lower energy)} \quad e_g \text{ (higher energy)} \]

5. **Fill the Electrons According to the Splitting:**
- The \(Cu^{2+}\) ion has a total of 9 electrons in the \(d\) orbitals.
- According to the crystal field theory, we fill the \(t_{2g}\) orbitals first:
- Fill \(t_{2g}\) with 6 electrons:
\[ t_{2g}: \uparrow \uparrow \uparrow \uparrow \uparrow \uparrow \]
- Then fill the \(e_g\) with the remaining 3 electrons:
\[ e_g: \uparrow \uparrow \uparrow \]
- Thus, the distribution of electrons is:
- \(t_{2g} = 6\)
- \(e_g = 3\)

6. **Write the Final Electronic Configuration:**
- The final electronic configuration based on crystal field splitting theory for \([Cu(NH_3)_6]^{2+}\) is:
\[ t_{2g}^6 \, e_g^3 \]

### Final Answer:
The electronic configuration of \([Cu(NH_3)_6]^{2+}\) is \(t_{2g}^6 \, e_g^3\).