Answer
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Hint: If a compound forms a white precipitate with $AgN{{O}_{3}}$ then the product formed is silver chloride having formula $AgCl$. If a compound forms a white precipitate with $BaC{{l}_{2}}$ then the product formed is barium sulfate having formula $BaS{{O}_{4}}$.
Complete step by step solution:
The given coordination complex has molecular composition as $M.5N{{H}_{3}}.Cl.S{{O}_{4}}$. In the coordination compound, we know that not all the ions or elements take place in the reaction, only the elements or ions that are outside the coordination bracket.
So the question says that $M.5N{{H}_{3}}.Cl.S{{O}_{4}}$ has two isomers i.e, A and B.
So, the A isomer combined with $AgN{{O}_{3}}$ gives white precipitate, and white is formed by the formation of silver chloride, having formula $AgCl$. This means that in the isomer A chloride ion was outside the coordination bracket and its exchanges with nitrate ion.
So, the B isomer combines with $BaC{{l}_{2}}$ gives white precipitate, and white is formed by the formation of barium sulfate having formula $BaS{{O}_{4}}$. This means that in the isomer B, sulfate ion was outside the coordination bracket and its exchanges with chloride ion.
So, the isomer A can be $[M{{(N{{H}_{3}})}_{5}}S{{O}_{4}}]Cl$ and the isomer B can be $[M{{(N{{H}_{3}})}_{5}}Cl]S{{O}_{4}}$. So, in these two isomers the ions are changing so they are ionization isomers.
Therefore, the correct answer is an option (b)- Ionisation isomerism.
Note: Linkage isomers are those in which the same ligand can join from two different atoms. Coordinate isomers are those in which the positive and negative ligands of the complex interchange. Geometrical isomers are those in which the cis and trans-form can be written.
Complete step by step solution:
The given coordination complex has molecular composition as $M.5N{{H}_{3}}.Cl.S{{O}_{4}}$. In the coordination compound, we know that not all the ions or elements take place in the reaction, only the elements or ions that are outside the coordination bracket.
So the question says that $M.5N{{H}_{3}}.Cl.S{{O}_{4}}$ has two isomers i.e, A and B.
So, the A isomer combined with $AgN{{O}_{3}}$ gives white precipitate, and white is formed by the formation of silver chloride, having formula $AgCl$. This means that in the isomer A chloride ion was outside the coordination bracket and its exchanges with nitrate ion.
So, the B isomer combines with $BaC{{l}_{2}}$ gives white precipitate, and white is formed by the formation of barium sulfate having formula $BaS{{O}_{4}}$. This means that in the isomer B, sulfate ion was outside the coordination bracket and its exchanges with chloride ion.
So, the isomer A can be $[M{{(N{{H}_{3}})}_{5}}S{{O}_{4}}]Cl$ and the isomer B can be $[M{{(N{{H}_{3}})}_{5}}Cl]S{{O}_{4}}$. So, in these two isomers the ions are changing so they are ionization isomers.
Therefore, the correct answer is an option (b)- Ionisation isomerism.
Note: Linkage isomers are those in which the same ligand can join from two different atoms. Coordinate isomers are those in which the positive and negative ligands of the complex interchange. Geometrical isomers are those in which the cis and trans-form can be written.
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