What is magnetic moment of complex `[Co(CN)_6]^(4-)`:

Answer

To determine the magnetic moment of the complex \([Co(CN)_6]^{4-}\), we can follow these steps:

### Step 1: Determine the oxidation state of cobalt in the complex
The overall charge of the complex is \(-4\). The cyanide ion (CN\(^-\)) has a charge of \(-1\). Since there are six CN\(^-\) ligands, their total contribution to the charge is \(-6\).

Let \(x\) be the oxidation state of cobalt. The equation can be set up as follows:
\[ x + 6(-1) = -4 \]
This simplifies to:
\[ x - 6 = -4 \]
Solving for \(x\):
\[ x = +2 \]
Thus, the oxidation state of cobalt in \([Co(CN)_6]^{4-}\) is +2.

### Step 2: Write the electron configuration of cobalt
The atomic number of cobalt (Co) is 27. The electron configuration of neutral cobalt is:
\[ [Ar] \, 3d^7 \, 4s^2 \]
For cobalt in the +2 oxidation state, we remove two electrons from the 4s orbital:
\[ Co^{2+}: [Ar] \, 3d^7 \]

### Step 3: Analyze the electron configuration in the presence of ligands
Cyanide (CN\(^-\)) is a strong field ligand, which means it will cause pairing of electrons in the d-orbitals. The 3d orbitals split into two sets: \(T_{2g}\) and \(E_g\).

- The \(T_{2g}\) orbitals can hold 6 electrons (3 pairs).
- The \(E_g\) orbitals can hold 4 electrons (2 pairs).

For \(Co^{2+}\) with 7 electrons in the 3d subshell, the filling will be as follows:
1. Fill the \(T_{2g}\) orbitals first:
- 6 electrons fill the \(T_{2g}\) (3 pairs).
2. The remaining 1 electron will go into one of the \(E_g\) orbitals.

Thus, the electron distribution will look like this:
- \(T_{2g}\): 6 electrons (3 pairs)
- \(E_g\): 1 electron (1 unpaired)

### Step 4: Count the number of unpaired electrons
From the above distribution, we see that there is **1 unpaired electron** in the \(E_g\) orbitals.

### Step 5: Calculate the magnetic moment
The formula for calculating the magnetic moment (\(\mu\)) is given by:
\[ \mu = \sqrt{n(n + 2)} \, \text{Bohr magneton} \]
where \(n\) is the number of unpaired electrons. Here, \(n = 1\).

Substituting the value of \(n\):
\[ \mu = \sqrt{1(1 + 2)} = \sqrt{3} \, \text{Bohr magneton} \]
Calculating this gives:
\[ \mu \approx 1.732 \, \text{Bohr magneton} \]

### Final Answer
The magnetic moment of the complex \([Co(CN)_6]^{4-}\) is approximately **1.732 Bohr magneton**.

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