In `[NiCl_(4)]^(2-)` ,the number of unparied electron is

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Answer

To determine the number of unpaired electrons in the complex ion \([NiCl_4]^{2-}\), we can follow these steps:

### Step 1: Determine the oxidation state of Nickel
The first step is to find the oxidation state of nickel in the complex. The formula for the complex is \([NiCl_4]^{2-}\).

- Let the oxidation state of nickel be \(X\).
- The oxidation state of each chloride ion (Cl) is \(-1\).
- Since there are 4 chloride ions, the total contribution from chloride is \(4 \times (-1) = -4\).
- The overall charge of the complex is \(-2\).

Setting up the equation:
\[ X + (-4) = -2 \]
Solving for \(X\):
\[ X = -2 + 4 = +2 \]
Thus, the oxidation state of nickel in \([NiCl_4]^{2-}\) is \(+2\).

### Step 2: Write the electron configuration of Nickel
The atomic number of nickel (Ni) is 28. The electron configuration of neutral nickel is:
\[ [Ar] 3d^8 4s^2 \]
For the nickel ion \(Ni^{2+}\), we remove two electrons (the 4s electrons are removed first):
\[ Ni^{2+} : [Ar] 3d^8 \]

### Step 3: Analyze the ligand field
Chloride ion (Cl\(^-\)) is a weak field ligand. Weak field ligands do not cause significant splitting of the d-orbitals, which means that the energy difference between the split d-orbitals is small.

### Step 4: Splitting of d-orbitals
In an octahedral field, the \(3d\) orbitals split into two sets:
- \(t_{2g}\) (lower energy)
- \(e_g\) (higher energy)

However, since we have a tetrahedral complex here, the splitting is different:
- The \(3d\) orbitals split into \(e\) (higher energy) and \(t_2\) (lower energy).

### Step 5: Filling the d-orbitals
For \(Ni^{2+}\) with \(3d^8\):
- The electrons will fill the orbitals according to Hund's rule and the Pauli exclusion principle.
- In a weak field, the electrons will occupy the higher energy \(e\) orbitals before pairing occurs in the lower energy \(t_2\) orbitals.

Filling the \(3d\) orbitals:
- The first 6 electrons will fill the \(t_2\) orbitals (3 electrons in \(t_2\) and 5 in \(e\)).
- The last 2 electrons will pair up in the \(t_2\) orbitals.

### Step 6: Count the unpaired electrons
After filling the orbitals:
- The \(t_2\) orbitals will have 2 paired electrons.
- The \(e\) orbitals will have 2 unpaired electrons.

Thus, the total number of unpaired electrons in \([NiCl_4]^{2-}\) is **2**.

### Final Answer
The number of unpaired electrons in \([NiCl_4]^{2-}\) is **2**.

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