How many isomers are possible for $\ce{[Cr(NH3)5(NO2)]Cl2}$? Consider all possible types of isomerism and draw each isomer.

My texbooks presents only three possible isomers: $\ce{[Cr(NH3)5(NO2)]Cl2}$ (the original compound), $\ce{[Cr(NH3)5(ONO)]Cl2}$ (linkage isomerism through $\ce{NO2}$), $\ce{[Cr(NH3)5Cl]Cl(NO2)}$ (ionization isomerism).

However, why is $\ce{[Cr(NH3)5Cl]Cl(NO2)}$ the only possible ionization isomer? Why can't $\ce{[Cr(NH3)4Cl(NO2)]Cl(NH3)}$, $\ce{[Cr(NH3)3Cl2(NO2)](NH3)2}$, or $\ce{[Cr(NH3)4Cl2](NH3)(NO2)}$ form as well.

The proposed ionization isomers were constructed by substituting one, or both, of the $\ce{Cl-}$ ions in the complex ion structure with either $\ce{NH3}$ or $\ce{NO2-}$; the answer seems to indicate that $\ce{NH3}$ cannnot under ionization isomerism with $\ce{Cl-}$.


1 Answer 1


Because $\ce{NH3}$ is not a counter ion and thus would not be bound to the complex if not bonded to the center ion. In $\ce{[Cr(NH3)5(NO2)]Cl2}$ the $\ce{Cl-}$ ions do not form dative bonds with the $\ce{Cr^{3+}}$ ion but are attracted by the positive charge of the $\ce{[Cr(NH3)5(NO2)]^2+}$ complex. If you were to form$\ce{[Cr(NH3)4C(NO2)]Cl(NH3)}$, the $\ce{NH3}$ would not be datively bonded to the $\ce{Cr^3+}$ center it would have no bond or attraction to the complex and thus would be a free ion, not part of the complex anymore. Thus the compound would now be $\ce{[Cr(NH3)4(NO2)Cl]Cl}$ as opposed to $\ce{[Cr(NH3)5(NO2)]Cl2}$.


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