Question

Consider the following compounds : \(\underline{\text{K}}\text{O}_{2}\), \(\text{H}_{2}\underline{\text{O}}_{2}\) and \(\text{H}_{2}\underline{\text{S}}\text{O}_{4}\). The oxidation state of the underlined elements in them are, respectively,

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

Hint:

KO2 is a superoxide with K at +1, H2O2 is a peroxide with O at minus 1, and S in H2SO4 is +6.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The sum of oxidation states in a neutral compound is zero.

Step 2: Calculation
In \(\text{KO}_2\), potassium is an alkali metal, so its state is \(+1\).
In \(\text{H}_2\text{O}_2\): \(2(+1)+2x=0\), so \(x=-1\) for oxygen.
In \(\text{H}_2\text{SO}_4\): \(2(+1)+x+4(-2)=0\), so \(x=+6\) for sulfur.

Step 3: Compare the options
The set \(+1,-1,+6\) matches option (C).

Final Answer:
The oxidation states are +1 for K, minus 1 for O and +6 for S. \[ \boxed{\text{(C) }\text{+1, -1 and +6}} \]