Write the following ionic equation in the molecular form if the reactants are chlorides.
$2 F e^{3 +} + S n^{2 +} \to 2 F e^{2 +} + S n^{4 +}$ .

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Solution

For writing the reactants in molecular forms, the requisite number of chloride ions are added.
$2 F e^{3 +} + 6 C l^{-}      + S n^{2 +} + 2 C l^{-}     $
or $2 F e C l_{3}$ $+$ $S n C l_{2}$
Similarly $8 C l^{-}$ ions are added on RHS to neutralise the charges.
$2 F e^{2 +} + 4 C l^{-}      + S n^{4 +} + 4 C l^{-}     $
or $2 F e C l_{2}$ $+$ $S n C l_{4}$
Thus, the balanced molecular equation is
$2 F e C l_{3} + S n C l_{2} = 2 F e C l_{2} + S n C l_{4}$

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