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 \( \text{Cr(H}_2\text{O)}_6\text{Br}_n \) based on the given information.

### Step-by-Step Solution:

1. **Determine the Magnetic Moment**:
The magnetic moment is given as \( 3.83 \, \text{BM} \). The formula for the spin-only magnetic moment is:
\[ \mu = \sqrt{n(n + 2)} \, \text{BM} \]
where \( n \) is the number of unpaired electrons.

2. **Calculate the Number of Unpaired Electrons**:
Rearranging the formula, we can set \( \mu = 3.83 \):
\[ 3.83 = \sqrt{n(n + 2)} \]
Squaring both sides:
\[ 14.6689 = n(n + 2) \]
Solving the quadratic equation \( n^2 + 2n - 14.6689 = 0 \) gives us \( n \approx 3 \) (since \( n \) must be a whole number).

3. **Determine the Oxidation State of Chromium**:
The electronic configuration of chromium is \( [Ar] 3d^5 4s^1 \). In the complex, to have 3 unpaired electrons, chromium must be in the +3 oxidation state:
\[ \text{Cr}^{3+} \rightarrow [Ar] 3d^3 \]

4. **Determine the Composition of the Complex**:
The complex is \( \text{Cr(H}_2\text{O)}_6\text{Br}_3 \). Since bromide ions are negatively charged, we need 3 bromide ions to balance the +3 charge from chromium.

5. **Geometrical Isomerism**:
The complex \( \text{Cr(H}_2\text{O)}_6\text{Br}_3 \) can exhibit geometrical isomerism. This is because it can have different spatial arrangements of the ligands around the chromium ion. Specifically, it can show cis and trans isomerism.

6. **IUPAC Nomenclature**:
According to IUPAC rules, the naming of the complex involves:
- Naming the ligands first in alphabetical order.
- Using prefixes to indicate the number of each type of ligand.
- Indicating the oxidation state of the metal in Roman numerals.

The ligands are:
- Aqua (for \( \text{H}_2\text{O} \)) - 6 water molecules = Hexa
- Bromido (for \( \text{Br}^- \)) - 3 bromide ions = Tri

Therefore, the IUPAC name for the complex is:
\[ \text{[Cr(H}_2\text{O)}_6\text{Br}_3] \text{ is called Tetraamminechromium(III) bromide.} \]

### Final Answer:
The IUPAC nomenclature of the complex \( \text{Cr(H}_2\text{O)}_6\text{Br}_3 \) is **Hexaqua-bromochromium(III)**.