Question

One of the following combinations illustrate law of reciprocal proportions :

• A. $N_2{O}_3,N_2{O}_4,N_2{O}_5$
• B. $NaCl,NaBr,NaI$
• C. $C{S}_2,C{O}_2,S{O}_2$
• D. $P{H}_3,P_2{O}_3,P_2{O}_5$

Answer 1

Answer: (A)

Explanation:

• The law of reciprocal proportion states that if two different elements combine separately with a fixed mass of a $3^{rd}$ element, the ratio of masses in which they do so are either the same as or multiple of the ratio of the masses in which they combine with each other.
• Consider option C $⟶C{S}_2,C{O}_2,S{O}_2$
• The weight ratios in the following compounds-

$C{S}_2,C:S=12:32$

$C{O}_2,C:O=12:16$

$\therefore$ Applying the above ratios, we find

$S:O=C:OorC:S=\frac{12}{16}\times \frac{32}{12}=\frac{32}{16}=2:1$

Also in $S{O}_2$, the ratio of weight of $SO=32:16=2:1$

$\therefore$ This implies that these molecules obey the law of reciprocal proportion.

Hence, option (D) is correct.