A Complex P of compositon `Cr(H_2O)_(6)Br_(n)` has a spin only magnetic moment of `3.83 BM`. It reacts with `AgNO_3` and shows geometrical isomerism. The IUPAC nomenclature of P is

Options

A. Hexaaqua chromium (III) bromide
B. Dibromidotetraaqua chromium (IV) bromide dihydrate
C. Tetraaquadibromido chromium (IV) bromide dihydrate
D. Tetraaquadibromido chromium (III) bromide dihydrate

Answer

To solve the problem, we need to determine the IUPAC nomenclature of the complex \( P \) with the composition \( \text{Cr(H}_2\text{O)}_6\text{Br}_n \) based on the given information.

### Step 1: Determine the oxidation state of chromium
The spin-only magnetic moment is given as \( 3.83 \, \text{BM} \). The formula for the magnetic moment is:

\[ \mu = \sqrt{n(n + 2)} \]

where \( n \) is the number of unpaired electrons. Setting \( \mu = 3.83 \):

\[ 3.83 = \sqrt{n(n + 2)} \]

Squaring both sides:

\[ 14.6689 = n(n + 2) \]

This simplifies to:

\[ n^2 + 2n - 14.6689 = 0 \]

Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ n = \frac{-2 \pm \sqrt{4 + 58.6756}}{2} = \frac{-2 \pm \sqrt{62.6756}}{2} \]

Calculating the discriminant:

\[ \sqrt{62.6756} \approx 7.92 \]

Thus,

\[ n \approx \frac{-2 + 7.92}{2} \approx 2.96 \approx 3 \]

So, \( n = 3 \). This means there are 3 unpaired electrons.

### Step 2: Determine the oxidation state of chromium
The electronic configuration of chromium is \( [Ar] 3d^5 4s^1 \). In the \( +3 \) oxidation state, chromium loses 3 electrons, resulting in \( 3d^5 \) configuration with 3 unpaired electrons.

### Step 3: Determine the composition of the complex
The complex is \( \text{Cr(H}_2\text{O)}_6\text{Br}_n \). Since the oxidation state of chromium is \( +3 \) and water is neutral, the total charge contributed by bromine must be \( -3 \) to balance the \( +3 \) charge of chromium. Therefore, \( n = 3 \).

### Step 4: Analyze the geometrical isomerism
The complex can show geometrical isomerism, which is possible if it has the form \( MA_4B_2 \) where \( A \) and \( B \) are different ligands. In this case, we have:

- 6 water molecules (neutral ligands)
- 3 bromide ions (anionic ligands)

This means we can have the following arrangement:

- \( \text{Cr(H}_2\text{O)}_4\text{Br}_2 \) (inside the coordination sphere)
- 2 bromides outside the coordination sphere.

This allows for cis and trans isomers.

### Step 5: Write the IUPAC name
The IUPAC name of the complex can be constructed as follows:

1. **Identify the ligands**:
- 4 water molecules are called "aqua".
- 2 bromide ions are called "bromo".

2. **Order of naming**:
- Ligands are named in alphabetical order, ignoring any prefixes (di-, tri-, etc.).
- The central metal is named after the ligands.

3. **Oxidation state**:
- The oxidation state of chromium is \( +3 \).

Thus, the IUPAC name is:

**Tetraquabromochromium(III)**.

### Final Answer
The IUPAC nomenclature of the complex \( P \) is **Tetraquabromochromium(III)**.

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