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Some Basic Concepts in Chemistry: Chemistry | JEE Main

A sample of a metal oxide has formula

\mathrm{M}_{0.83}\mathrm{O}_{1.00}

. The metal

\mathrm{M}

can exist in two oxidation states

+2

and

+3

. In the sample of

\mathrm{M}_{0.83}\mathrm{O}_{1.00}

, the percentage of metal ions existing in

+2

oxidation state is %. (nearest integer)

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Correct answer

In the non-stoichiometric oxide $\mathrm{M}_{0.83}\mathrm{O}_{1.00}$ , approximately $59\%$ of the metal ions exist in the $+2$ oxidation state.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Formula of non-stoichiometric oxide is $\mathrm{M}_{0.83}\mathrm{O}_{1.00}$ . Metal ions are in $+2$ and $+3$ oxidation states. Oxide ion is $\mathrm{O}^{2-}$ .

02find

Calculate the percentage of metal ions existing in $+2$ oxidation state: $\frac{n(\mathrm{M}^{2+})}{n(\mathrm{M}_{\text{total}})} \times 100$ .

03strategise

Apply the principle of electrical neutrality: total positive charge from $\mathrm{M}^{2+}$ and $\mathrm{M}^{3+}$ must equal total negative charge from $\mathrm{O}^{2-}$ . Let the moles of $\mathrm{M}^{2+}$ be $x$ ; then $\mathrm{M}^{3+}$ is $(0.83 - x)$ .

04execute

Balance charges: $2x + 3(0.83 - x) = 2(1.00) \implies 2.49 - x = 2 \implies x = 0.49$ . Then the percentage is $\frac{0.49}{0.83} \times 100 \approx 59.036\% \approx 59\%$ .

✓verify

If $\mathrm{M}^{2+} = 0.49$ , then $\mathrm{M}^{3+} = 0.83 - 0.49 = 0.34$ . Total charge = $2(0.49) + 3(0.34) = 0.98 + 1.02 = 2.00$ , matching the charge of one $\mathrm{O}^{2-}$ ion perfectly.

Hints that build this answer step by step

If there are $x$ moles of $\mathrm{M}^{2+}$ and $y$ moles of $\mathrm{M}^{3+}$ per mole of oxide $\mathrm{M}_{0.83}\mathrm{O}_{1.00}$ , which pair of equations correctly represents total metal ions and electrical neutrality?
$x + y = 0.83$ and $2x + 3y = 2.00$
Solving $x + y = 0.83$ and $2x + 3y = 2.00$ , what is the value of $x$ (the amount of $\mathrm{M}^{2+}$ )?
$0.49$
What is the percentage of metal ions present as $\mathrm{M}^{2+}$ , calculated as $\frac{x}{x+y} \times 100$ ?
$59\%$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is the total negative charge set to 2?

Because each formula unit contains 1.00 oxide ion, and each oxide ion carries a -2 charge: $1.00 \times 2 = 2$ .

Why do we divide by 0.83 instead of 1?

The question asks for the percentage of *metal ions* in the +2 state, so the denominator must be the total number of metal ions ( $0.83$ ), not the formula unit total or oxygen atoms.

Answer

In the non-stoichiometric oxide $\mathrm{M}_{0.83}\mathrm{O}_{1.00}$ , approximately $59\%$ of the metal ions exist in the $+2$ oxidation state.

Why each option works or fails

Step-by-step solution

given: Formula of non-stoichiometric oxide is $\mathrm{M}_{0.83}\mathrm{O}_{1.00}$ . Metal ions are in $+2$ and $+3$ oxidation states. Oxide ion is $\mathrm{O}^{2-}$ .
find: Calculate the percentage of metal ions existing in $+2$ oxidation state: $\frac{n(\mathrm{M}^{2+})}{n(\mathrm{M}_{\text{total}})} \times 100$ .
strategise: Apply the principle of electrical neutrality: total positive charge from $\mathrm{M}^{2+}$ and $\mathrm{M}^{3+}$ must equal total negative charge from $\mathrm{O}^{2-}$ . Let the moles of $\mathrm{M}^{2+}$ be $x$ ; then $\mathrm{M}^{3+}$ is $(0.83 - x)$ .
execute: Balance charges: $2x + 3(0.83 - x) = 2(1.00) \implies 2.49 - x = 2 \implies x = 0.49$ . Then the percentage is $\frac{0.49}{0.83} \times 100 \approx 59.036\% \approx 59\%$ .
verify: If $\mathrm{M}^{2+} = 0.49$ , then $\mathrm{M}^{3+} = 0.83 - 0.49 = 0.34$ . Total charge = $2(0.49) + 3(0.34) = 0.98 + 1.02 = 2.00$ , matching the charge of one $\mathrm{O}^{2-}$ ion perfectly.

Some Basic Concepts in Chemistry: Chemistry | JEE Main

What feels right?

In the non-stoichiometric oxide $\mathrm{M}_{0.83}\mathrm{O}_{1.00}$ , approximately $59\%$ of the metal ions exist in the $+2$ oxidation state.

Why each option works or fails

Solution

01given

02find

03strategise

04execute

✓verify

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At $600\text{K}$ , the root mean square (rms) speed of gas $\text{X}$ …

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