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(D) coordination compound of $Co(III)$ that dissociates into $3$ ions in solution must have two ionizable ions outside the coordination sphere, given

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pentaamminechloridocobalt$(III)$ chloride

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(D) coordination compound of $Co(III)$ that dissociates into $3$ ions in solution must have two ionizable ions outside the coordination sphere, given the oxidation state of cobalt is $+3$.$(A)$ $[Co(NH_{3})_{6}]Cl_{3} \rightarrow [Co(NH_{3})_{6}]^{3+} + 3Cl^{-}$ ($4$ ions)$(B)$ $[Co(NH_{3})_{5}(SO_{4})]Cl \rightarrow [Co(NH_{3})_{5}(SO_{4})]^{+} + Cl^{-}$ ($2$ ions)$(C)$ $[Co(NH_{3})_{5}Cl]SO_{4} \rightarrow [Co(NH_{3})_{5}Cl]^{2+} + SO_{4}^{2-}$ ($2$ ions)$(D)$ $[Co(NH_{3})_{5}Cl]Cl_{2} \rightarrow [Co(NH_{3})_{5}Cl]^{2+} + 2Cl^{-}$ ($3$ ions)Thus, the compound that dissociates into $3$ ions is $[Co(NH_{3})_{5}Cl]Cl_{2}$.Hence, option $(D)$ is the correct answer.

The conductivity measurement of a coordination compound of cobalt$(III)$ shows that it dissociates into $3$ ions in solution. The compound is

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