Give the total number of `t_(2g)` and `e_(g)` electrons in `[NiF_(6)]^(2-)` .

Explanation

To solve the problem of determining the total number of \( t_{2g} \) and \( e_g \) electrons in the complex \([NiF_6]^{2-}\), we can follow these steps:

### Step 1: Determine the oxidation state of Nickel in \([NiF_6]^{2-}\)

The overall charge of the complex is \(-2\). Fluorine (F) has a charge of \(-1\), and there are six fluorine atoms. Therefore, the total charge contributed by fluorine is:

\[ 6 \times (-1) = -6 \]

Let the oxidation state of Nickel (Ni) be \( x \). The equation for the oxidation state is:

\[ x + (-6) = -2 \]

Solving for \( x \):

\[ x = -2 + 6 = +4 \]

Thus, the oxidation state of Nickel in \([NiF_6]^{2-}\) is \( +4 \).

### Step 2: Determine the electronic configuration of Nickel in the +4 oxidation state

Nickel has an atomic number of 28, and its ground state electronic configuration is:

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

In the +4 oxidation state, Nickel loses four electrons. The electrons are removed first from the 4s orbital and then from the 3d orbital. Therefore, the electronic configuration for \( Ni^{4+} \) becomes:

\[ 3d^6 \]

### Step 3: Identify the ligand field and determine the splitting of d-orbitals

The ligand in this complex is fluoride (F\(^-\)), which is a weak field ligand. However, since Nickel is in a +4 oxidation state, it has a high effective nuclear charge, which can lead to a stronger field effect.

The geometry of the complex is octahedral 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 Hund's rule and the Pauli exclusion principle

For \( Ni^{4+} \) with a \( 3d^6 \) configuration, the electrons will fill the \( t_{2g} \) and \( e_g \) orbitals as follows:

1. The first three electrons will fill the three \( t_{2g} \) orbitals (each getting one electron).
2. The next three electrons will pair up in the \( t_{2g} \) orbitals due to the strong field effect, as the energy gap is large.
3. The \( e_g \) orbitals will remain empty.

Thus, the filling will be:

- \( t_{2g} \): 6 electrons (all paired)
- \( e_g \): 0 electrons

### Step 5: Conclusion

The total number of \( t_{2g} \) and \( e_g \) electrons in \([NiF_6]^{2-}\) is:

- \( t_{2g} = 6 \)
- \( e_g = 0 \)

Therefore, the total is:

\[ \text{Total} = t_{2g} + e_g = 6 + 0 = 6 \]

### Final Answer

The total number of \( t_{2g} \) and \( e_g \) electrons in \([NiF_6]^{2-}\) is **6**.

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