Question

Consider the following compounds: K$_2$O$_2$, H$_2$O$_2$, and H$_2$SO$_4$
The oxidation states of the underlined elements in them are, respectively:

• A. +2, -2, and +6
• B. +1, -2, and +4
• C. +4, -4, and +6
• D. +1, -1, and +6

Hint:

In peroxides, the oxidation state of oxygen is -1. In sulfuric acid, sulfur has an oxidation state of +6, and hydrogen generally has an oxidation state of +1.

Answer 1

The Correct Option is A

Calculate the oxidation states of the underlined elements in each compound: 1. K\(_2\)O\(_2\):
- Oxygen in peroxides has an oxidation state of -1.
- Potassium (K) has an oxidation state of +1 to balance the charges in K\(_2\)O\(_2\).
- Thus, the oxidation states are K = +1 and O = -1. The underlined element is K, with an oxidation state of +1.
2. H\(_2\)O\(_2\): - Hydrogen (H) has an oxidation state of +1.
- Oxygen (O) in peroxides has an oxidation state of -1.
- Thus, the oxidation states in H\(_2\)O\(_2\) are H = +1 and O = -1. The underlined element is oxygen, with an oxidation state of -1.
3. H\(_2\)SO\(_4\):
- Hydrogen (H) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- In H\(_2\)SO\(_4\), sulfur (S) has an oxidation state of +6. This is calculated by balancing the compound: (2 × +1) + S + (4 × -2) = 0, so S = +6.
Thus, the oxidation states for the underlined elements are: - K = +1 - O = -1 - S = +6. The correct answer is (1) +1, -1, and +6.