1. Interpret the composition

Given composition of complex $A$ is: $H_{12}O_6Cl_3Cr$

This corresponds to:

• $6$ oxygen atoms and $12$ hydrogens $\Rightarrow 6H_2O$
• $3$ chlorine atoms
• $1$ chromium atom

So the overall composition is: $CrCl_3\cdot 6H_2O$

All four options have this same overall composition; they differ only in how many water molecules are inside/outside the coordination sphere.

2. Effect of treatment with conc. $H_2SO_4$

Concentrated $H_2SO_4$ removes only the water of crystallization (outer-sphere water), not the coordinated water.

So if the complex loses $13.5\%$ of its original mass, that percentage must correspond to the mass of outer-sphere water molecules.

3. Calculate molar mass of the complex

Using given atomic masses:

• $Cr = 52$
• $Cl = 35$
• $H_2O = 18$

Total molar mass of $CrCl_3\cdot 6H_2O$ is: $52 + 3(35) + 6(18) = 52 + 105 + 108 = 265$

4. Mass lost on dehydration

Given loss is $13.5\%$ of original mass: $\text{mass lost} = \frac{13.5}{100}\times 265 = 35.775 \approx 36$

Since one water molecule has mass $18$, mass loss of about $36$ means: $\frac{36}{18} = 2$

So the complex contains 2 water molecules as water of crystallization.

5. Match with options

Now check which option has exactly 2 outer-sphere water molecules:

• A: $[Cr(H_2O)_5Cl]Cl_2\cdot H_2O$
  Outer-sphere water $=1$
• B: $[Cr(H_2O)_4Cl_2]Cl\cdot 2H_2O$
  Outer-sphere water $=2$
• C: $[Cr(H_2O)_3Cl_3]\cdot 3H_2O$
  Outer-sphere water $=3$
• D: $[Cr(H_2O)_6]Cl_3$
  Outer-sphere water $=0$

Therefore, the correct formula is: $[Cr(H_2O)_4Cl_2]Cl\cdot 2H_2O$

6. Check percentage exactly for option B

For option B, outer water mass $=2\times 18=36$

Percentage loss: $\frac{36}{265}\times 100 = 13.58\% \approx 13.5\%$

This matches the given data.

Final Answer

Option B: $[Cr(H_2O)_4Cl_2]Cl\cdot 2H_2O$