One of the following combinations which illustrates the law of reciprocal proportion is

(A)N_{2}O_{3},N_{2}O_{4},N_{2}O_{5}

(B)NaCl,NaBr,NaI

(C)CS_{2},CO_{2},SO_{2}

(D)PH_{3},P_{2}O_{3},P_{2}O_{5}

_{2}, CO

_{2}, SO

_{2}

According to law of reciprocal proportions, if there are three atoms A, B and C, and if A forms molecule by combining with B and separately with C also, then if B and C also combine together, the ratio of the masses in which they do so are either the same or a simple multiple of the mass ratio in which they combine.

Thus in part (C), there are three atoms in all which are C, S and O.

In CS

_{2}: Ratio of weight of C:S = 12:32

In CO

_{2}: Ratio of weight of C:O = 12:16

Thus here S:O = 32:16 = 2:1

Now in SO

_{2} Ratio of weight of S:O = 32:32 = 1:1

The ratio between the two ratios of S and O is

$\frac{2}{1}:\frac{1}{1}$

which is 2:1 and it is a multiple ratio.

Now in option (A), there are only two atoms involved i.e. N and O.

In option (B), Third molecule must have been formed by any two halogens.

In option (D), P is forming one molecule with O and other with H, so third molecule must have been formed by H and O.

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