Question:medium

Given below are two statements.
Statement I : In H\(_2\)O\(_2\), each oxygen atom is assigned an oxidation number of -1, In RbO\(_2\), each oxygen atom is assigned an oxidation number of - \(\frac{1}{2}\).
Statement II : Representation of HAuCl\(_4\) and MnO\(_2\) in stock notation is HAu(III)Cl\(_4\) and Mn(II)O\(_2\), respectively.
Examine the above statements and choose the correct answer.

Show Hint

Memorize the special cases for oxygen's oxidation number:
  • \textbf{-2} in most oxides (e.g., H\(_2\)O).
  • \textbf{-1} in peroxides (e.g., H\(_2\)O\(_2\)).
  • \textbf{-1/2} in superoxides (e.g., KO\(_2\)).
  • \textbf{+2} in OF\(_2\) (with a more electronegative element).
Updated On: Apr 28, 2026
  • Both Statement I and Statement II are correct
  • Both Statement I and Statement II are incorrect
  • Statement I is correct but Statement II is incorrect
  • Statement I is incorrect but Statement II is correct
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
Oxidation numbers depend on the electronegativity and bonding environment. Stock notation is a method of representing the oxidation state of a metal in a compound using Roman numerals in parentheses.
Step 2: Key Formula or Approach:
Calculate oxidation numbers using standard rules (peroxide = -1, superoxide = -1/2) and apply Roman numerals for the Stock notation based on the calculated oxidation state of the central metal.
Step 3: Detailed Explanation:
1. Statement I: $H_{2}O_{2}$ is hydrogen peroxide. In peroxides, the oxidation state of oxygen is strictly $-1$. $RbO_{2}$ is rubidium superoxide, containing the $Rb^{+}$ cation and the $O_{2}^{-}$ superoxide anion. The charge of $-1$ is shared by 2 oxygen atoms, so each $O$ has an average oxidation state of $-1/2$. (Statement I is True)

2. Statement II: Let's calculate the oxidation states for the metals:
In $HAuCl_{4}$: $H$ is $+1$ and each $Cl$ is $-1$. Let Au be $x$.
$1 + x + 4(-1) = 0 \Rightarrow x - 3 = 0 \Rightarrow x = +3$. So $HAu(III)Cl_{4}$ is correct.
In $MnO_{2}$: oxygen is usually $-2$. Let Mn be $x$.
$x + 2(-2) = 0 \Rightarrow x - 4 = 0 \Rightarrow x = +4$. The stock notation should be $Mn(IV)O_{2}$, not $Mn(II)O_{2}$. (Statement II is False)
Step 4: Final Answer:
Statement I is correct but Statement II is incorrect.
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