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

Options

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

Answer

To determine the number of unpaired electrons in the complex ion \([Ni(H_2O)_6]^{2+}\), we can follow these steps:

### Step 1: Determine the oxidation state of Nickel (Ni)
The complex ion is \([Ni(H_2O)_6]^{2+}\). Since water \((H_2O)\) is a neutral ligand, the oxidation state of Ni can be calculated as follows:

Let the oxidation state of Ni be \(x\). The overall charge of the complex is \(+2\).

\[ x + 6(0) = +2 \implies x = +2 \]

### Step 2: Write the electronic configuration of Ni
Nickel (Ni) has an atomic number of 28. The electronic configuration of Ni is:

\[ Ni: [Ar] 3d^8 4s^2 \]

### Step 3: Determine the electronic configuration of \(Ni^{2+}\)
When Ni loses two electrons to form \(Ni^{2+}\), the electrons are removed first from the 4s orbital. Therefore, the electronic configuration of \(Ni^{2+}\) is:

\[ Ni^{2+}: [Ar] 3d^8 \]

### Step 4: Analyze the 3d orbitals
In the case of \(Ni^{2+}\), we have 8 electrons in the 3d subshell. The distribution of these electrons in the 3d orbitals can be shown as follows:

- The 3d subshell can hold a maximum of 10 electrons (5 orbitals).
- According to Hund's rule, the electrons will fill each orbital singly before pairing begins.

The filling of the 3d orbitals for \(Ni^{2+}\) will look like this:

- 3d orbital filling: ↑ ↑ ↑ ↑ ↑ (5 orbitals)
- After filling 5 orbitals with 5 electrons, the next 3 electrons will pair up in the first 3 orbitals.

Thus, the arrangement will be:

\[ \text{3d: } ↑↓ ↑↓ ↑ ↑ \]

### Step 5: Count the unpaired electrons
From the arrangement above, we can see that there are 2 unpaired electrons in the 3d orbitals.

### Conclusion
The number of unpaired electrons in \([Ni(H_2O)_6]^{2+}\) is **2**.

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