Number of unpaired electrons present in`[Ni(H_(2)O)_(6)]^(3+)`

Options

A. Two
B. One
C. Four
D. Three

Answer

To determine the number of unpaired electrons in the complex ion \([Ni(H_2O)_6]^{3+}\), we can follow these steps:

### Step 1: Identify the oxidation state of Nickel
The complex ion is \([Ni(H_2O)_6]^{3+}\). Since water (\(H_2O\)) is a neutral ligand, the oxidation state of Nickel (Ni) in this complex is +3.

### Step 2: Write the electron configuration of Nickel
Nickel has an atomic number of 28. The electron configuration of neutral Nickel (Ni) is:
\[ \text{Ni: } [Ar] 3d^8 4s^2 \]

### Step 3: Adjust the electron configuration for the +3 oxidation state
When Nickel loses 3 electrons to form \([Ni]^{3+}\), the electrons are removed first from the 4s orbital and then from the 3d orbital. Therefore, the electron configuration for \([Ni]^{3+}\) becomes:
\[ \text{Ni}^{3+}: [Ar] 3d^7 \]

### Step 4: Determine the arrangement of electrons in the 3d subshell
The 3d subshell can hold a maximum of 10 electrons. For \(3d^7\), we can visualize the filling of the 3d orbitals (considering Hund's rule and the Pauli exclusion principle):
- The 3d orbitals are filled as follows:
- 1st orbital: ↑
- 2nd orbital: ↑
- 3rd orbital: ↑
- 4th orbital: ↑
- 5th orbital: ↑
- 6th orbital: ↑ (pairing starts here)
- 7th orbital: ↑

This results in the following filling:
- \(3d_{xy}\): ↑
- \(3d_{yz}\): ↑
- \(3d_{zx}\): ↑
- \(3d_{x^2-y^2}\): ↑
- \(3d_{z^2}\): ↑

### Step 5: Count the number of unpaired electrons
From the arrangement above, we can see that there are 3 unpaired electrons in the 3d subshell.

### Final Answer
The number of unpaired electrons present in \([Ni(H_2O)_6]^{3+}\) is **3**.

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