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

Let's calculate the oxidation states of the underlined elements in each compound: 1. K\(_2\)O\(_2\):
- The oxidation state of oxygen in peroxides is always -1. - To balance the charges, the oxidation state of potassium (K) in K\(_2\)O\(_2\) is +1.
-
Thus, the oxidation states of the elements are K = +1 and O = -1. The underlined element here is K, and its oxidation state is +1.
2. H\(_2\)O\(_2\): - The oxidation state of hydrogen (H) is +1.
- The oxidation state of oxygen (O) in peroxides is -1.
-
Thus, the oxidation states of the elements in H\(_2\)O\(_2\) are H = +1 and O = -1. The underlined element is oxygen, and its oxidation state is -2 (as per typical peroxide rule).
3. H\(_2\)SO\(_4\):
- The oxidation state of hydrogen (H) is +1.
- The oxidation state of oxygen (O) is -2.
- In H\(_2\)SO\(_4\), sulfur (S) has an oxidation state of +6, as the sum of oxidation states of hydrogen (2 × +1) and oxygen (4 × -2) must balance the compound.
Therefore, S = +6.

Thus, the correct oxidation states for the underlined elements in these compounds are: - K = +1, O = -1, and S = +6. The correct answer is (1) +2, -2, and +6.