Consider the following compounds : KO_2, H_2O_2 and H_2SO_4.The oxidation states of the underlined elements in them are, respectively,

1. +1, -1, and +6

2. +2, -2, and +6

3. +1, -2, and +4

4. +4, -4, and +6

Doubtly Asked 30 Oct 2026 NEET UG 2025 Practice test neet Save WhatsApp Report

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Answer: 1. +1, -1, and +6

Key idea: assign oxidation states from known charges.

\(\mathrm{KO_2}\) is potassium superoxide, \(\mathrm{K^+}\) and \(\mathrm{O_2^-}\): K = +1.
\(\mathrm{H_2O_2}\) is a peroxide: \(2(+1)+2x=0\), so O = -1.
\(\mathrm{H_2SO_4}\): \(2+x-8=0\), so S = +6.

+1, -1, +6: option (1).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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