# One of the following combinations which illustrates the law of reciprocal proportion is(A)N2O3,N2O4,N2O5 (B)NaCl,NaBr,NaI(C)CS2,CO2,SO2 (D)PH3,P2O3,P2O5

Answer is (C) CS2, CO2, SO2

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 CS2:               Ratio of weight of C:S = 12:32
In CO2:
​Ratio of weight of C:O = 12:16

Thus here           S:O = 32:16 = 2:1

Now in SO2 ​ 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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