Properties JEE 2024: Metal-Deficient Oxide Solution

What is the answer to the Properties JEE 2024 metal-deficient oxide question?

Option D, 0.75, is correct for the Properties JEE 2024 metal-deficient oxide question. One-third describes a fraction of all M ions, not a fixed ion count per formula unit.

The source is JEE Advanced 2024, Paper 2, Chemistry, under the question-bank chapter label Properties of Solids and Liquids. The bank classifies it as medium difficulty, with a 90-second target. These are question-bank classifications, not official exam statistics.

The oxide has composition:

M_{x} Y_{2} O_{4}

Y is trivalent. M occurs in divalent and trivalent forms, with one-third of the M ions divalent. Determine the composition coefficient.

The supplied choices are:

How much positive charge must the M ions supply?

All M ions together must supply two positive charge units per formula unit. Follow the official solution by accounting for oxygen and Y first. The oxide must be electrically neutral, and charges are counted per formula unit in units of the elementary charge.

Oxygen is present as oxide ions: $O^{2 -}$

Four oxygen ions therefore contribute:

Q_{O} = 4 \times (- 2) = - 8

The two trivalent Y ions contribute:

Q_{Y} = 2 \times (+ 3) = + 6

The remaining positive charge must come from M:

Q_{M} = + 8 - 2 \times 3 = + 2

This is the required total M charge, not the oxidation state of an individual M ion. The divalent and trivalent populations together must supply this charge.

How do you turn the one-third fraction into ion counts and find x?

Multiply each fraction by the total M count before calculating charge. There is not one M ion per formula unit: the normalised total is the composition coefficient. $N_{M} = x$

The denominator of the given fraction is the count of all M ions, not all ions in the oxide:

\frac{N_{M^{2 +}}}{N_{M}} = \frac{1}{3}

Keep ion, fraction, count and charge contribution separate:

Divalent M population

Ion: & M^{2 +} Fraction of M: & \frac{1}{3} Count per formula unit: & \frac{x}{3} Positive-charge contribution: & 2 (\frac{x}{3})

Trivalent M population

Ion: & M^{3 +} Fraction of M: & 1 - \frac{1}{3} = \frac{2}{3} Count per formula unit: & \frac{2 x}{3} Positive-charge contribution: & 3 (\frac{2 x}{3})

A fraction gives the share of the M population. Multiplying it by the total M count gives the count per formula unit. Multiplying that count by the ionic charge gives the charge contribution.

Add the two contributions and equate them to the required M charge:

2 (\frac{x}{3}) + 3 (\frac{2 x}{3}) = 2

\frac{2 x}{3} + 2 x = 2

\frac{8 x}{3} = 2

x = 2 \times \frac{3}{8} = \frac{3}{4} = 0.75

The final answer is option D. The one-third belongs in the population count before that count is multiplied by the divalent charge.

How can fractional ion counts satisfy both the given fraction and neutrality?

Fractional counts describe a normalised bulk composition, not physically divided ions. Check the answer in two ways: recover the stated divalent fraction, then verify that the total charge is zero. Charge neutrality alone does not prove that you used the supplied fraction correctly.

At the calculated composition, the counts are:

N_{M^{2 +}} = \frac{3 / 4}{3} = \frac{1}{4}

N_{M^{3 +}} = \frac{2 (3 / 4)}{3} = \frac{1}{2}

The divalent fraction is:

\frac{N_{M^{2 +}}}{N_{M}} = \frac{1 / 4}{3 / 4} = \frac{1}{3}

The charge check gives:

2 (\frac{1}{4}) + 3 (\frac{1}{2}) + 3 (2) - 2 (4) = \frac{1}{2} + \frac{3}{2} + 6 - 8 = 0

For whole-ion bookkeeping, scale to four formula units: one divalent M ion, two trivalent M ions, eight trivalent Y ions and sixteen oxide ions. These integer counts describe the same composition.

Why does treating every M ion as trivalent give option C?

Assigning trivalent charge to every M ion produces option C, but violates the given fraction. It keeps the oxygen and Y accounting correct while discarding the one-third divalent population. This is a method error, not a rounding issue.

The incorrect neutrality equation would be: $3 x + 6 = 8$

It gives: $3 x = 2$ $x = \frac{2}{3} \approx 0.67$

This matches option C. The error is assigning three positive charge units to ions that carry only two.

Write fraction, then count, then charge for each oxidation state. Count the two populations separately before adding their charges. A single trivalent contribution cannot represent the mixed-valence population specified here.

How do you solve two related mixed-valence practice questions?

Use the same charge accounting, changing only the unknown. Both exercises below are original practice variations from Properties of Solids and Liquids, not additional verified JEE PYQs. The first changes the given ion fraction; the second fixes the composition and asks for that fraction.

What is x if half the M ions are divalent?

The answer is 0.80. In this variation, oxygen has oxidation state minus two and Y has oxidation state plus three. Half the M ions are divalent and the rest are trivalent; determine the coefficient in:

M_{x} Y_{2} O_{4}

Oxygen contributes minus eight and Y contributes plus six. M must therefore supply two positive charge units:

Q_{O} = - 8, Q_{Y} = + 6, Q_{M} = + 2

Each M population contains half the total:

2 (\frac{x}{2}) + 3 (\frac{x}{2}) = 2

\frac{5 x}{2} = 2

x = 2 \times \frac{2}{5} = \frac{4}{5} = 0.80

What fraction of M is divalent when its coefficient is 0.9?

The answer is seven-ninths. In this variation, oxygen has oxidation state minus two, Y has oxidation state plus three, and M occurs only in divalent and trivalent forms:

M_{0.9} Y_{2} O_{4}

Let the unknown divalent fraction be: $f$

The total M count is 0.9. Multiply each population fraction by 0.9 before assigning charge:

2 (0.9 f) + 3 [0.9 (1 - f)] = 2

$1.8 f + 2.7 - 2.7 f = 2$ $0.9 f = 0.7$ $f = \frac{7}{9}$ Substitute this fraction into the two population counts. Verify that their combined positive charge is two.

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Frequently asked questions

What is the answer to the Properties JEE 2024 metal-deficient oxide question?

Option D, x = 0.75, is correct for MₓY₂O₄. Four oxide ions contribute −8 charge and two trivalent Y ions contribute +6, so M must supply +2. With one-third of M divalent and two-thirds trivalent, charge neutrality gives 2(x/3) + 3(2x/3) = 2, hence x = 0.75.

Why is the number of divalent M ions x/3 and not 1/3?

One-third is the fraction of all M ions that are divalent, not their count per formula unit. Since the total M count in MₓY₂O₄ is x, the divalent count is x/3 and the trivalent count is 2x/3.

How can an oxide formula have fractional ion counts?

Fractional ion counts express a normalised bulk composition; they do not mean that individual ions are divided. At x = 0.75, each formula unit has normalised counts of 1/4 divalent M and 1/2 trivalent M. Scaling to four formula units gives one divalent M ion, two trivalent M ions, eight trivalent Y ions and sixteen oxide ions.

Why is option C, 0.67, wrong in the metal-deficient oxide question?

The value 0.67 comes from assigning a +3 charge to every M ion and solving 3x + 6 = 8. That ignores the stated one-third divalent population. Counting both oxidation states separately gives x = 0.75, so this is a method error rather than a rounding issue.

Properties JEE 2024: Metal-Deficient Oxide Solution

What is the answer to the Properties JEE 2024 metal-deficient oxide question?

How much positive charge must the M ions supply?

How do you turn the one-third fraction into ion counts and find x?

How can fractional ion counts satisfy both the given fraction and neutrality?

Why does treating every M ion as trivalent give option C?

How do you solve two related mixed-valence practice questions?

What is x if half the M ions are divalent?

What fraction of M is divalent when its coefficient is 0.9?

Frequently asked questions

What is the answer to the Properties JEE 2024 metal-deficient oxide question?

Why is the number of divalent M ions x/3 and not 1/3?

How can an oxide formula have fractional ion counts?

Why is option C, 0.67, wrong in the metal-deficient oxide question?

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