Cr in `[Cr(NH_3)_6] Br_3` has number of unpaired electron :

Options

A. 4
B. 3
C. 1
D. 2

Answer

To determine the number of unpaired electrons in the complex ion \([Cr(NH_3)_6]Br_3\), we will follow these steps:

### Step 1: Determine the oxidation state of Chromium (Cr)
In the complex \([Cr(NH_3)_6]Br_3\):
- Ammonia (NH₃) is a neutral ligand, contributing 0 to the oxidation state.
- Bromide (Br⁻) has a charge of -1. Since there are three bromide ions, they contribute a total of -3.

Let the oxidation state of chromium be \(x\). The equation for the oxidation state is:
\[ x + 0 - 3 = 0 \]
\[ x = +3 \]

### Step 2: Write the electronic configuration of Cr in its +3 oxidation state
The atomic number of chromium (Cr) is 24, and its electronic configuration is:
\[ \text{Cr: } [Ar] 4s^1 3d^5 \]

When chromium is in the +3 oxidation state, it loses three electrons. The electrons are removed first from the 4s orbital and then from the 3d orbital:
- Remove 1 electron from 4s: \(4s^0\)
- Remove 2 electrons from 3d: \(3d^3\)

Thus, the electronic configuration of \(Cr^{3+}\) is:
\[ \text{Cr}^{3+}: [Ar] 3d^3 \]

### Step 3: Determine the number of unpaired electrons
In the \(3d\) subshell, the distribution of electrons is as follows:
- The \(3d\) subshell can hold a maximum of 10 electrons, but in \(Cr^{3+}\), there are 3 electrons.
- According to Hund's rule, the electrons will occupy the orbitals singly before pairing up.

The configuration for \(3d^3\) will be:
- \(3d_{xy}^1\)
- \(3d_{xz}^1\)
- \(3d_{yz}^1\)

This means all three electrons are unpaired.

### Conclusion
The number of unpaired electrons in \([Cr(NH_3)_6]Br_3\) is **3**.

### Final Answer
**3 unpaired electrons.**

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