Question

The ionization isomer of $\left[Cr\right(H_{2}O{)}_{4}Cl\left(N{O}_2\right)]Cl$ is :

• A. $\left[Cr\right(H_{2}O{)}_4\left(O_{2}N\right)]C{l}_2$
• B. $\left[Cr\right(H_{2}O{)}_{4}C{l}_2\left]\right(N{O}_2)$
• C. $\left[Cr\right(H_{2}O{)}_{4}Cl\left(ONO\right)]Cl$
• D. $\left[Cr\right(H_{2}O{)}_{4}C{l}_2\left(N{O}_2\right)]H_{2}O$

Answer 1

The correct option is B $\left[Cr\right(H_{2}O{)}_{4}C{l}_2\left]\right(N{O}_2)$

Ionization isomers are identical except for a ligand has exchanged places with anion or neutral molecule that was originally outside the coordination complex.

Thus, in Ionization isomerism the central ion and the other ligands are identical.

So, according to the definition, $C{l}^{-}$ is replaced by $N{O}_2^{-}$ in ionization sphere.

Thus the correct answer is option B.