Identify the ionization isomer of [Cr(H_2O)_4 | vedclass

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(B) Step $1$: Ionization isomers are compounds that produce different ions in solution. This occurs when a counter ion in the coordination compound ex

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$[Cr(H_2O)_4 Cl_2](NO_2)$

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(B) Step $1$: Ionization isomers are compounds that produce different ions in solution. This occurs when a counter ion in the coordination compound exchanges its position with a ligand present in the coordination sphere.Step $2$: The given complex is $[Cr(H_2O)_4 Cl(NO_2)] Cl$. Here, $Cl^-$ is the counter ion.Step $3$: To form an ionization isomer, the $Cl^-$ ion outside the coordination sphere must exchange with a ligand inside the sphere (e.g., $NO_2^-$).Step $4$: Swapping $Cl^-$ with $NO_2^-$ gives the complex $[Cr(H_2O)_4 Cl_2](NO_2)$.Step $5$: Thus, option $B$ is the correct ionization isomer.

Complex	Structure and magnetic property
$i$. $NiCl_4^{2-}$	$p$. tetrahedral and diamagnetic
$ii$. $Ni(CO)_4$	$q$. tetrahedral and paramagnetic
$iii$. $PtCl_4^{2-}$	$r$. square planar and diamagnetic
$iv$. $Ni(CN)_4^{2-}$	$s$. square planar and paramagnetic

Identify the ionization isomer of $[Cr(H_2O)_4 Cl(NO_2)] Cl$ from the following.

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