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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

Vedclass · Accepted Answer

pentaamminechloridocobalt$(III)$ chloride

Vedclass · Answer

(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

Similar Questions

$CuSO_4$ reacts with $KCN$ solution and forms

Consider the following complexes:
$(a) K_2PtCl_6$
$(b) PtCl_4 \cdot 2NH_3$
$(c) PtCl_4 \cdot 3NH_3$
$(d) PtCl_4 \cdot 5NH_3$
Their electrical conductances in aqueous solutions are:

Assertion : $NF_3$ is a weaker ligand than $N(CH_3)_3$.
Reason : $NF_3$ ionizes to give $F^{-}$ ions in aqueous solution.

Which is the incorrect observation?

$A$ solution contains $2.675 \ g$ of $CoCl_3 \cdot 6NH_3$ $(M = 267.5 \ g/mol)$. When passed through a cation exchanger,the chloride ions liberated are treated with excess $AgNO_3$ to yield $4.78 \ g$ of $AgCl$ $(M = 143.5 \ g/mol)$. What is the formula of the coordination compound?

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