
Which of the following isomeric pairs shows ionization isomerism
A \[\left[ Co{{\left( N{{H}_{3}} \right)}_{6}} \right]\left[ Cr{{\left( CN \right)}_{6}} \right]\]and
\[\left[ Cr{{\left( N{{H}_{3}} \right)}_{6}} \right]\left[ Co{{\left( CN \right)}_{6}} \right]\]
B \[\left[ Cr{{\left( {{H}_{2}}O \right)}_{6}} \right]C{{l}_{3}}\]and
\[\left[ Cr{{\left( {{H}_{2}}O \right)}_{5}}Cl \right]C{{l}_{2}}\text{ }{{H}_{2}}O\]
C \[\left[ Pt{{\left( N{{H}_{3}} \right)}_{2}}C{{l}_{2}} \right]\]and
\[\left[ Pt{{\left( N{{H}_{3}} \right)}_{4}} \right]\left[ PtC{{l}_{4}} \right]\]
D \[\left[ Co{{\left( N{{H}_{3}} \right)}_{5}}Br \right]S{{O}_{4}}\] and
\[\left[ Co{{\left( N{{H}_{3}} \right)}_{5}}S{{O}_{4}} \right]Br\]
Answer
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Hint:The coordination compounds are those complex compounds in which central metal has two types of valence one is primary and the other is secondary. And when secondary valence is changed to the primary valence or vice versa then the resulting compound is said to be an ionization isomer (same molecular formula but different bonding of ligands with central metal atom) which is a structural isomer.
Complete answer:In the first option, both the anionic and cationic part of the coordinate compound is complex (the first part is always cation and the second one is anionic). Now the bonding of ligands of the cationic part changed with the bonding of ligands of the anionic part so, this type of isomerism is known as coordination isomerism (structural isomerism). Also, the third option is not showing ionization isomerism as it does not contain primary valence and it will show geometrical isomerism.
The primary valence of any complex compound is equal to its oxidation number whereas the secondary valence of any compound is equal to the coordination number (number of coordination bonds). In the third option secondary valence (Br) is exchanged with the primary valence (SO4). So, as per the hint, it is an ionization isomer.
Thus the correct option is D.
Note: In option second secondary (Cl) and primary valence (\[{{H}_{2}}O\]) are exchanged so, it should be an ionization isomer but as in this water molecule get exchange so, the isomerism is said to be hydrate isomer but not ionization isomers.
Complete answer:In the first option, both the anionic and cationic part of the coordinate compound is complex (the first part is always cation and the second one is anionic). Now the bonding of ligands of the cationic part changed with the bonding of ligands of the anionic part so, this type of isomerism is known as coordination isomerism (structural isomerism). Also, the third option is not showing ionization isomerism as it does not contain primary valence and it will show geometrical isomerism.
The primary valence of any complex compound is equal to its oxidation number whereas the secondary valence of any compound is equal to the coordination number (number of coordination bonds). In the third option secondary valence (Br) is exchanged with the primary valence (SO4). So, as per the hint, it is an ionization isomer.
Thus the correct option is D.
Note: In option second secondary (Cl) and primary valence (\[{{H}_{2}}O\]) are exchanged so, it should be an ionization isomer but as in this water molecule get exchange so, the isomerism is said to be hydrate isomer but not ionization isomers.
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