Question

X + Y \( \rightarrow \) oleum The sum of oxidation states of central atom in X and Y is

• A. 12
• B. 10
• C. 06
• D. 08

Hint:

Recognize the chemical formula of oleum as \( H_2S_2O_7 \) or \( H_2SO_4 \cdot SO_3 \), which implies the reactants X and Y are sulfuric acid (\( H_2SO_4 \)) and sulfur trioxide (\( SO_3 \)). Then, apply the rules for assigning oxidation states to determine the oxidation state of the central atom (sulfur) in each reactant and sum these values. Remember that the sum of oxidation states in a neutral compound is zero.

Answer 1

The Correct Option is A

Oleum is also known as fuming sulfuric acid.
Its chemical formula is \( H_2S_2O_7 \) or can be represented as \( H_2SO_4 \cdot SO_3 \).
This indicates that oleum is formed by the reaction of sulfuric acid (\( H_2SO_4 \)) with sulfur trioxide (\( SO_3 \)).
Therefore, X and Y are \( H_2SO_4 \) and \( SO_3 \) (or vice versa).
The central atom in both compounds is sulfur (S).
Let's determine the oxidation state of sulfur in each compound: **In \( H_2SO_4 \):** The oxidation state of hydrogen (H) is +1.
The oxidation state of oxygen (O) is -2.
Let the oxidation state of sulfur (S) be \( x \).
The sum of the oxidation states in a neutral molecule is zero: $$ 2(+1) + x + 4(-2) = 0 $$ $$ 2 + x - 8 = 0 $$ $$ x - 6 = 0 $$ $$ x = +6 $$ The oxidation state of sulfur in \( H_2SO_4 \) is +6.
**In \( SO_3 \):** The oxidation state of oxygen (O) is -2.
Let the oxidation state of sulfur (S) be \( y \).
The sum of the oxidation states in a neutral molecule is zero: $$ y + 3(-2) = 0 $$ $$ y - 6 = 0 $$ $$ y = +6 $$ The oxidation state of sulfur in \( SO_3 \) is +6.
The sum of the oxidation states of the central atom (sulfur) in X and Y is: $$ (+6) + (+6) = +12 $$ Therefore, the sum of the oxidation states of the central atom in X and Y is 12.