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.