Question

When burned sulfur forms a gaseous product X which can be oxidized to produce a gas Y. Gas Y reacts with water to produce a product Z. Which row correctly shows the oxidation state of sulfur in X, Y and Z?

| X| Y| Z
---|---|---|---
A| -2| +4| +4
B| -2| +4| +6
C| +4| +6| +4
D| +4| +6| +6

A.A
B.B
C.C
D.D

Answer 1

Reaction of sulfur with oxygen from sulfur dioxide. When oxidized further it forms the higher oxide of sulfur. The reaction of that oxide with water forms sulfuric acid.

Complete step by step solution:
Whenever oxidation of sulfur takes place firstly the formation of sulfur dioxide occurs. The molecular formula for sulfur dioxide is ${\text{S}}{{\text{O}}_2}$. Let us calculate the oxidation state of sulfur in this. Since oxygen has $- 2$ charge and there are 2 oxygen atoms. The overall charge on oxygen is $- 4$. Hence the oxidation state of sulfur will be $+ 4$. Now when this sulfur dioxide is further oxidized it forms a higher oxide of sulfur named as sulfur trioxide. The molecular formula for sulfur trioxide is ${\text{S}}{{\text{O}}_3}$. Similarly the oxygen has $- 2$ charge and there are 3 oxygen atoms. The overall charge on oxygen is $- 6$ . Hence the oxidation state of sulfur will be $+ 6$. Now when we hydrolyse this gas or react it with water. It forms a well known sulfur acid known as sulfuric acid. The molecular formula for sulfuric acid is. ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$. Now in the same way overall charge on oxygen will be $- 2 \times 4 = - 8$. Hydrogen has a $+ 1$ charge and there are two hydrogen atoms. Net charge on hydrogen will be $+ 2$. Hence total charge becomes -6 and the oxidation state of sulfur will be 6.

So, the correct option is D that is 4, 6, 6.

Note: Sulfur is a 16 group element and has 6 valence shell electrons. The maximum oxidation state of sulfur can be 6 because after losing 6 electrons it will acquire the Noble gas electronic configuration similar to argon so it will not lose any more electrons because it attained stability.