Write the given complex clearly

The complex is $[\mathrm{Cr(H_2O)_4Cl(NO_2)}]Cl$

Here:

• Coordination sphere: $[\mathrm{Cr(H_2O)_4Cl(NO_2)}]^+$
• Counter ion outside the bracket: $\mathrm{Cl^-}$

Understand ionisation isomerism

Ionisation isomers are formed when:

• one ligand inside the coordination sphere, and
• one counter ion outside the coordination sphere exchange places.

So, the outside $\mathrm{Cl^-}$ must come inside the coordination sphere, and one inner anionic ligand must go outside.

Identify the inner anionic ligands

Inside the complex, the ligands are:

• $4\,\mathrm{H_2O}$ : neutral
• $\mathrm{Cl^-}$ : anionic
• $\mathrm{NO_2^-}$ : anionic

Since the outer ion is $\mathrm{Cl^-}$, ionisation isomerism will occur by exchanging outer $\mathrm{Cl^-}$ with inner $\mathrm{NO_2^-}$.

That gives: $[\mathrm{Cr(H_2O)_4Cl_2}]NO_2$

Check charge balance

Let oxidation state of Cr be $x$.

In the original complex: $x+0-1-1=+1$ $x=+3$

For $[\mathrm{Cr(H_2O)_4Cl_2}]NO_2$: $+3+0-1-1=+1$ so the complex cation is $+1$, balanced by $\mathrm{NO_2^-}$ outside.

Hence it is valid.

Check the options

• A: $[\mathrm{Cr(H_2O)_4(O_2N)}]Cl_2$

  This changes the composition/charge pattern and does not represent simple exchange of inner and outer ions. Not correct.

• B: $[\mathrm{Cr(H_2O)_4Cl_2}](NO_2)$

  This is exactly the ionisation isomer obtained by exchanging inner $\mathrm{NO_2^-}$ with outer $\mathrm{Cl^-}$. Correct.

• C: $[\mathrm{Cr(H_2O)_4Cl(ONO)}]Cl$

  This is linkage isomerism ($\mathrm{NO_2^-}$ bound through O instead of N), not ionisation isomerism. Not correct.

• D: $[\mathrm{Cr(H_2O)_4Cl_2}(NO_2)]H_2O$

  This is not an ionisation isomer; it changes the formulation incorrectly. Not correct.

Final answer

The ionisation isomer is $[\mathrm{Cr(H_2O)_4Cl_2}](NO_2)$

So, Option B is correct.