The number of unpaired electrons presenet in `[NiF_(6)]^(2-)` is ……

Explanation

To determine the number of unpaired electrons in the complex ion \([NiF_6]^{2-}\), we will follow these steps:

### Step 1: Determine the oxidation state of Nickel (Ni)
In the complex \([NiF_6]^{2-}\), we need to find the oxidation state of Nickel. The fluoride ion (F) has a charge of -1. Since there are six fluoride ions, their total contribution to the charge is -6. The overall charge of the complex is -2.

Let \( x \) be the oxidation state of Nickel. We can set up the equation:
\[ x + 6(-1) = -2 \]
\[ x - 6 = -2 \]
\[ x = +4 \]

### Step 2: Determine the electronic configuration of Nickel
The atomic number of Nickel (Ni) is 28. The ground state electronic configuration of Nickel is:
\[ [Ar] 3d^8 4s^2 \]
When Nickel is in the +4 oxidation state, it loses 4 electrons (2 from 4s and 2 from 3d), resulting in:
\[ Ni^{4+}: [Ar] 3d^8 \]

### Step 3: Determine the geometry and hybridization of \([NiF_6]^{2-}\)
The complex \([NiF_6]^{2-}\) has an octahedral geometry due to the coordination number of 6. In an octahedral field, the \(d\) orbitals split into two sets: \(t_{2g}\) (lower energy) and \(e_g\) (higher energy).

### Step 4: Fill the \(d\) orbitals according to the crystal field splitting
For \(Ni^{4+}\) with \(3d^8\), the electrons will fill the \(t_{2g}\) and \(e_g\) orbitals. The filling will occur as follows:
- The \(t_{2g}\) orbitals will be filled first. Since \(t_{2g}\) can hold a maximum of 6 electrons, all 8 electrons will fill the \(t_{2g}\) orbitals first, leading to pairing of electrons.
- The \(e_g\) orbitals will remain empty.

### Step 5: Count the number of unpaired electrons
Since all 8 electrons are paired in the \(t_{2g}\) orbitals, there are no unpaired electrons in the complex \([NiF_6]^{2-}\).

### Conclusion
The number of unpaired electrons in \([NiF_6]^{2-}\) is **0**.

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