The number of ions given by $[P t (N H_{3})_{6}] C l_{4}$ in aqueous solution will be:

two

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three

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five

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eleven.

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Solution

The correct option is C five
$[P t (N H_{3})_{6}] C l_{4} ⇌ [P t (N H_{3})_{6}]^{4 +} + 4 C l^{-}$
Total number of ions given ( by dissociation of one molecule of $[P t (N H_{3})_{6}] C l_{4}$ in aqueous solution) $=$ 5. This includes one $[P t (N H_{3})_{6}]^{4 +}$ ion and four $C l^{-}$ ions.

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