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, so K is +1; H2O2 is a peroxide, so O is -1.

Answer 1

The Correct Option is C

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

Step 2: KO2 and H2O2
\(\text{KO}_2\) is potassium superoxide with \(\text{O}_2^{-}\), so K is \(+1\). In \(\text{H}_2\text{O}_2\), with H \(=+1\): \(2(+1)+2x=0\), so O is \(-1\).

Step 3: H2SO4
With H \(=+1\) and O \(=-2\): \(2+x+4(-2)=0\), so S is \(+6\).

Step 4: Result
The oxidation states are \(+1\), \(-1\) and \(+6\).

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