Why is there only one ionization isomer of [Cr(NH3)5(NO2)]Cl2?
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 $\ce{[Cr(NH3)5(NO2)]Cl2}$.