The number of unpaired electrons in the complex ion `[CoF_(6)]^(3-)` is

Options

A. 3
B. 2
C. 4
D. 0

Explanation

To determine the number of unpaired electrons in the complex ion \([CoF_6]^{3-}\), we can follow these steps:

### Step 1: Determine the oxidation state of cobalt in the complex
The complex ion \([CoF_6]^{3-}\) has a total charge of -3. Fluoride (\(F^-\)) is a monodentate ligand with a charge of -1. Since there are 6 fluoride ions, the total contribution from fluoride is \(6 \times (-1) = -6\).

Let the oxidation state of cobalt be \(x\). The equation for the charge balance is:
\[ x + (-6) = -3 \]
Solving for \(x\):
\[ x - 6 = -3 \implies x = +3 \]
Thus, cobalt is in the +3 oxidation state.

### Step 2: Write the electron configuration of cobalt in its +3 oxidation state
Cobalt has an atomic number of 27, and its ground state electron configuration is:
\[ [Ar] 4s^2 3d^7 \]
When cobalt loses 3 electrons to form \(Co^{3+}\), it loses the 2 electrons from the 4s orbital and 1 electron from the 3d orbital:
\[ Co^{3+}: [Ar] 3d^6 \]

### Step 3: Determine the nature of the ligand and its effect on electron pairing
Fluoride (\(F^-\)) is a weak field ligand. Weak field ligands do not cause pairing of electrons in the d-orbitals. Therefore, the electrons in the \(3d\) orbitals will remain unpaired.

### Step 4: Fill the d-orbitals and count the unpaired electrons
The \(3d\) subshell can hold a maximum of 10 electrons. For \(Co^{3+}\) with \(3d^6\), the filling of the \(3d\) orbitals will be as follows:

- 1st orbital: 2 electrons (paired)
- 2nd orbital: 2 electrons (paired)
- 3rd orbital: 2 electrons (unpaired)

The electron configuration in the \(3d\) orbitals will look like this:
\[ \text{3d: } \uparrow\downarrow \, \uparrow\downarrow \, \uparrow \, \uparrow \, \uparrow \]
This means we have 4 unpaired electrons (the last three electrons in the \(3d\) orbitals are unpaired).

### Conclusion
The number of unpaired electrons in the complex ion \([CoF_6]^{3-}\) is **4**.