Balance the following and choose the quantity which is the sum of the coefficients of reactants and products:
$P t C l_{4} + X e F_{2} \to P t F_{6} + C l F + X e$

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Solution

The correct option is A 16
Oxidation number of each element:
$+ 4 P t {- 1 C l}_{4} + + 2 X e {- 1 F}_{2} \to + 6 P t {- 1 F}_{6} + + 1 C l - 1 F + 0 X e$
Increase in oxidation number:
$+ 4 P t \to + 6 P t$
$1 (+ 6 - (+ 4)) = 2$
$- 1 C l \to + 1 C l$
$4 (+ 1 - (- 1)) = 8$
total increase in oxidation number $= 10 (i)$
Decrease in oxidation number:
$+ 2 X e \to 0 X e$
total decrease in oxidation number $= 2 (i i)$
To make equal changes in oxidation:
$5 \times (i i) = (i)$
$P t C l_{4} + 5 X e F_{2} \to P t F_{6} + C l F + X e$

after balancing other elements we get,
$P t C l_{4} + 5 X e F_{2} \to P t F_{6} + 4 C l F + 5 X e$

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