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:
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:
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- +2, -2, and +6
- +1, -2, and +4
- +4, -4, and +6
- +1, -1, and +6
The Correct Option is A
Solution and Explanation
- 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.
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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.
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