Complex `Cr(H_2 O)_6 Cl_n` shows geometrical isomerism and also reacts with `AgNO_3 solution. Given : Spin only magnetic moment = 3.8 B.M. What is the IUPAC name of the complex.

Options

A. Hexaaquachromium(III) chloride
B. Tetraaquadichloridochromium(III) chloride dihydrate
C. Hexaaquachromium(IV) chloride
D. Tetraaquadichloridochromium(IV) chloride dihydrate

Explanation

To solve the problem, we need to determine the IUPAC name of the complex `Cr(H2O)6Cl_n` based on the information provided.

### Step-by-Step Solution:

1. **Determine the Spin-Only Magnetic Moment**:
The given spin-only magnetic moment is 3.8 B.M. We can use the formula for the magnetic moment:
\[ \mu = \sqrt{n(n+2)} \]
where \( n \) is the number of unpaired electrons.

2. **Calculate the Number of Unpaired Electrons**:
Setting \(\mu = 3.8\):
\[ 3.8 = \sqrt{n(n+2)} \]
Squaring both sides:
\[ 14.44 = n(n+2) \]
Rearranging gives us:
\[ n^2 + 2n - 14.44 = 0 \]
Solving this quadratic equation using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ n = \frac{-2 \pm \sqrt{4 + 57.76}}{2} = \frac{-2 \pm \sqrt{61.76}}{2} \]
\[ n \approx \frac{-2 + 7.86}{2} \approx 2.93 \quad \text{(taking the positive root)} \]
Rounding gives \( n = 3 \).

3. **Determine the Value of n**:
Since \( n \) represents the number of unpaired electrons, we have \( n = 3 \). Therefore, the complex can be represented as:
\[ Cr(H_2O)_6Cl_3 \]

4. **Check for Geometrical Isomerism**:
The complex has six coordination sites (octahedral geometry) and contains two different types of ligands (water and chloride). This allows for geometrical isomerism (cis and trans forms).

5. **Determine the Oxidation State of Chromium**:
The oxidation state of chromium can be calculated as follows:
- The charge of \( H_2O \) is 0 (neutral).
- The charge of \( Cl \) is -1.
Let the oxidation state of \( Cr \) be \( x \):
\[ x + 6(0) + 3(-1) = 0 \]
\[ x - 3 = 0 \implies x = +3 \]

6. **Construct the IUPAC Name**:
The ligands are named first in alphabetical order:
- \( H_2O \) is called "aqua".
- \( Cl \) is called "chlorido".

The IUPAC name is constructed as follows:
- The number of each ligand is indicated by prefixes: "hexa" for six \( H_2O \) and "tri" for three \( Cl \).
- The central metal is named next, followed by its oxidation state in Roman numerals.

Therefore, the IUPAC name is:
\[ \text{hexaquachloridocromium(III)} \]

### Final Answer:
The IUPAC name of the complex \( Cr(H_2O)_6Cl_3 \) is **hexaquachloridocromium(III)**.