A complex compound of chromium chromium contains five `NH_(3)` molecules, one nitro group and two chloride ions for one `Cr^(3+)` cation. One molecule of this compound produces three ions in aq. Solution, on reacting with excess of `AgNO_(3)` solution, two moles of `AgCl` get precipitated
Q. Magnetic moment off complex compound is:

Options

A. 0 BM
B. `sqrt(24)` BM
C. `sqrt(15)` BM
D. `sqrt(3)` BM

Answer

To determine the magnetic moment of the complex compound of chromium, we will follow these steps:

### Step 1: Identify the Composition of the Complex
The complex contains:
- 1 Chromium ion (Cr³⁺)
- 5 Ammonia (NH₃) ligands
- 1 Nitro (NO₂) group
- 2 Chloride (Cl⁻) ions

### Step 2: Write the Formula of the Complex
From the information given, the complex can be represented as:
\[ \text{[Cr(NH}_3\text{)}_5\text{NO}_2\text{]Cl}_2 \]
This shows that the chromium is coordinated with 5 ammonia molecules and 1 nitro group, while the 2 chloride ions are outside the coordination sphere.

### Step 3: Determine the Number of Ions in Solution
When this complex dissolves in water, it produces 3 ions:
- 1 complex ion \([Cr(NH_3)_5NO_2]^{+}\)
- 2 chloride ions \((2Cl^{-})\)

This confirms that the complex behaves as expected in solution.

### Step 4: Determine the Oxidation State of Chromium
To find the oxidation state of chromium in the complex:
- Ammonia (NH₃) is neutral, contributing 0.
- Nitro (NO₂) has a charge of -1.
- Each chloride ion (Cl⁻) has a charge of -1.

Let the oxidation state of chromium be \(x\):
\[ x + 5(0) + (-1) + 2(-1) = 0 \]
\[ x - 1 - 2 = 0 \]
\[ x = +3 \]

Thus, chromium is in the +3 oxidation state.

### Step 5: Determine the Electron Configuration of Cr³⁺
The electron configuration of neutral chromium (Cr) is:
\[ [Ar] 4s^1 3d^5 \]

For Cr³⁺, we remove 3 electrons:
- Remove 1 from 4s and 2 from 3d:
\[ Cr^{3+} : [Ar] 3d^3 \]

### Step 6: Determine the Number of Unpaired Electrons
In the 3d subshell, the configuration is \(3d^3\). In an octahedral field, the 3 electrons will occupy the lower energy \(d_{2g}\) orbitals:
- The arrangement will be:
\[ d_{2g}^3 \]
This means there are 3 unpaired electrons.

### Step 7: Calculate the Magnetic Moment
The magnetic moment (\(\mu\)) can be calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \(n\) is the number of unpaired electrons.

Substituting \(n = 3\):
\[ \mu = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \]

### Conclusion
The magnetic moment of the complex compound is:
\[ \mu = \sqrt{15} \text{ Bohr magneton} \]