Which of these statements about `[Co(CN)_6]^(3-)` is true?

Options

A. `[Co(CN)_(6)]^(3-)` has no unpaired electron and will be in a low-spin configuration.
B. `[Co(CN)_(6)]^(3-)` has four unpaired electrons and will be in a low-spin configuration.
C. `[Co(CN)_(6)]^(3-)` has four unpaired electrons and will be in a high-spin configuration.
D. `[Co(CN)_(6)]^(3-)` has no unpaired electrons and will be in a high-spin configuration.

Explanation

To determine which statement about the complex ion \([Co(CN)_6]^{3-}\) is true, we can follow these steps:

### Step 1: Calculate the Oxidation State of Cobalt
In the complex \([Co(CN)_6]^{3-}\), we need to find the oxidation state of cobalt (Co). The cyanide ion (CN) has a charge of -1. Since there are six cyanide ligands, the total contribution from the ligands is:
\[ 6 \times (-1) = -6 \]
Let the oxidation state of cobalt be \(X\). The overall charge of the complex is -3. Therefore, we can set up the equation:
\[ X - 6 = -3 \]
Solving for \(X\):
\[ X = -3 + 6 = +3 \]
Thus, the oxidation state of cobalt in \([Co(CN)_6]^{3-}\) is +3.

### Step 2: Determine the Electronic Configuration of Cobalt
The atomic number of cobalt (Co) is 27. The electronic configuration of neutral cobalt is:
\[ [Ar] 3d^7 4s^2 \]
When cobalt is in the +3 oxidation state, it loses three electrons. The electrons are removed first from the 4s orbital, followed by the 3d orbital:
\[ [Ar] 3d^6 4s^0 \]
So, the electronic configuration of \(Co^{3+}\) is \([Ar] 3d^6\).

### Step 3: Analyze the Ligand Field Strength
The ligand in this complex is cyanide (CN), which is a strong field ligand. Strong field ligands cause pairing of electrons in the lower energy d-orbitals.

### Step 4: Electron Pairing in d-Orbitals
In the case of \(Co^{3+}\) with a \(3d^6\) configuration, the strong field ligand CN will cause the six electrons to pair up in the lower energy orbitals. The arrangement in the d-orbitals will be:
- Three orbitals (dxy, dyz, dzx) will each have paired electrons.
- The higher energy orbitals (dx²-y² and dz²) will remain empty.

This results in:
- 0 unpaired electrons.

### Step 5: Determine the Spin State
Since all the d-electrons are paired, the complex is classified as low spin.

### Conclusion
The correct statement about \([Co(CN)_6]^{3-}\) is that it has no unpaired electrons and is in a low spin configuration.