When chlorine is passed through molten tin, the product obtained is :

S n C l_{4}

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[S n C l_{6}]^{2 -}

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[S n C l_{4}]^{2 -}

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S n C l_{2}

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Solution

The correct option is A $S n C l_{4}$
Chlorine is passed through molten tin, the product obtained is:

$S n + 2 C l_{2} \to S n C l_{4}$ .

Hence,option A is correct.

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Similar questions

S n C l_{4} + 2 C l^{- 1} \to {[S n C l_{6}]}_{6}^{2 -}

S n C l_{4}

When Sn reacts with calculated amount of $H g C l_{2}$ and excess $H g C l_{2}$ , then the product will be respectively.

Valency of tin in $S n C l_{2}$ and $S n C l_{4}$ is _______ and _______ respectively.

S n C l_{2} + C l_{2} \to S n C l_{4} + 2 e^{-}

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