1. Use magnetic moment to find oxidation state / electron count

Given spin-only magnetic moment: $\mu = 3.83\ \text{BM}$

For spin-only moment, $\mu = \sqrt{n(n+2)}$ where $n$ is the number of unpaired electrons.

Now, $\sqrt{3(3+2)} = \sqrt{15} \approx 3.87\ \text{BM}$ which is very close to $3.83$ BM.

So, the complex has: $n=3\ \text{unpaired electrons}$

For chromium:

• $Cr^{3+}$ is $3d^3$  has 3 unpaired electrons
• $Cr^{4+}$ is $3d^2$  has 2 unpaired electrons

Hence chromium is in the $+3$ oxidation state.

So options with chromium(IV) are eliminated:

• B wrong
• D wrong

2. Use reaction with $AgNO_3$ to determine ionisable chloride

The complex has composition: $Cr(H_2O)_6Cl_n$

It reacts with $AgNO_3$, so at least one $Cl^-$ must be outside the coordination sphere as counter ion, because only free chloride gives $AgCl$ precipitate readily.

Also, the complex shows geometrical isomerism.

Let us test possibilities:

• $[Cr(H_2O)_6]Cl_3$ : octahedral with all six ligands same inside sphere  no geometrical isomerism.
• $[Cr(H_2O)_5Cl]Cl_2\cdot H_2O$ : type $MA_5B$  no geometrical isomerism.
• $[Cr(H_2O)_4Cl_2]Cl\cdot 2H_2O$ : type $MA_4B_2$  shows cis/trans geometrical isomerism.

Therefore the complex must be: $[Cr(H_2O)_4Cl_2]Cl\cdot 2H_2O$

This matches chromium in +3 oxidation state: $x + 2(-1) = +1 \Rightarrow x=+3$ inside the complex ion $[Cr(H_2O)_4Cl_2]^+$.

3. Write the correct IUPAC name

For $[Cr(H_2O)_4Cl_2]Cl\cdot 2H_2O$

Name of cationic complex:

• $4$ aqua ligands  tetraaqua
• $2$ chlorido ligands  dichlorido
• metal with oxidation state $+3$  chromium(III)

Then outside-sphere anion:

• chloride

Then water of crystallization:

• dihydrate

So the name is: $\boxed{\text{Tetraaquadichloridochromium(III) chloride dihydrate}}$

This corresponds to Option C.

4. Check options

• A Hexaaqua chromium(III) chloride  would be $[Cr(H_2O)_6]Cl_3$, no geometrical isomerism.
• B chromium(IV)  inconsistent with magnetic moment.
• C correct.
• D chromium(IV)  inconsistent with magnetic moment.

Therefore, the correct answer is: $\boxed{\text{C}}$