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A compound is formed by two elements: Chemistry | JEE Main

A compound is formed by two elements

\text{X}

and

\text{Y}

. The element

\text{Y}

forms cubic close packed arrangement and those of element

\text{X}

occupy one third of the tetrahedral voids. What is the formula of the compound?

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\text{X}_2\text{Y}_3

\text{X}_3\text{Y}

\text{X}_1\text{Y}_2

\text{XY}_3

Step-by-step solutionView

Correct answer

For $N$ atoms of element $\text{Y}$ in a ccp lattice, there are $2N$ tetrahedral voids, meaning $\text{X}$ occupies $\frac{1}{3}(2N) = \frac{2}{3}N$ sites, yielding the formula $\text{X}_2\text{Y}_3$ .

Option analysis

Why each option works or fails

A ·

\text{X}_2\text{Y}_3

None. In a ccp lattice with $N$ atoms of $\text{Y}$ , the number of tetrahedral voids is $2N$ . Since $\text{X}$ occupies $\frac{1}{3}$ of these voids, the number of $\text{X}$ atoms is $\frac{2}{3}N$ . The ratio $\text{X}:\text{Y} = \frac{2}{3}:1 = 2:3$ , giving the formula $\text{X}_2\text{Y}_3$ .

B ·

\text{X}_3\text{Y}

Inverting the ratio or confusing the occupancy fraction $\frac{1}{3}$ directly as the number of $\text{Y}$ atoms per $\text{X}$ atom. Multiply the total number of tetrahedral voids ( $2N$ ) by the occupancy fraction ( $\frac{1}{3}$ ) to find the quantity of $\text{X}$ , then express the mole ratio $\text{X}:\text{Y}$ in simplest integers.

C ·

\text{X}_1\text{Y}_2

Assuming that a ccp lattice has only $N$ tetrahedral voids instead of $2N$ , or miscalculating $\frac{2}{3}:1$ as $1:2$ . Remember that the number of tetrahedral voids in any close-packed lattice is always twice the number of lattice atoms ( $2N$ ).

D ·

\text{XY}_3

Assuming the number of tetrahedral voids equals the number of lattice atoms ( $N$ ) and taking $\frac{1}{3}N$ for $\text{X}$ , leading to an incorrect ratio $\text{X}:\text{Y} = 1:3$ . Use $2N$ for the number of tetrahedral voids in a ccp lattice formed by $N$ atoms.

Reviewed route

Solution

StepWorking

01given

Element $\text{Y}$ forms a cubic close packed (ccp) lattice. Element $\text{X}$ occupies $\frac{1}{3}$ of the tetrahedral voids.

02find

Find the stoichiometric formula of the compound formed by $\text{X}$ and $\text{Y}$ .

03strategise

In a ccp (fcc) unit cell containing $N$ atoms of $\text{Y}$ , the number of tetrahedral voids is $2N$ . If $N = 4$ , tetrahedral voids $= 8$ . Number of $\text{X}$ atoms is $\frac{1}{3} \times 8 = \frac{8}{3}$ . Then simplify the ratio $\text{X} : \text{Y}$ .

04execute

Number of $\text{Y}$ atoms $= 4$ . Number of $\text{X}$ atoms $= \frac{1}{3} \times 8 = \frac{8}{3}$ . Ratio $\text{X} : \text{Y} = \frac{8}{3} : 4 = 8 : 12 = 2 : 3$ . Thus, formula is $\text{X}_2\text{Y}_3$ .

✓verify

Ratio $\text{X} : \text{Y} = 2 : 3$ . For 3 atoms of $\text{Y}$ , tetrahedral voids $= 6$ . One-third of 6 is 2 atoms of $\text{X}$ . Perfectly consistent.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Why are there 8 tetrahedral voids for 4 lattice points in ccp?

A ccp unit cell can be divided into 8 miniature cubes. Each miniature cube has 4 alternating corners occupied by lattice atoms, creating exactly 1 tetrahedral void at its center. Thus, $8 \times 1 = 8$ tetrahedral voids per unit cell.

Answer

For $N$ atoms of element $\text{Y}$ in a ccp lattice, there are $2N$ tetrahedral voids, meaning $\text{X}$ occupies $\frac{1}{3}(2N) = \frac{2}{3}N$ sites, yielding the formula $\text{X}_2\text{Y}_3$ .

Why each option works or fails

A · correct: $\text{X}_2\text{Y}_3$ - None. In a ccp lattice with $N$ atoms of $\text{Y}$ , the number of tetrahedral voids is $2N$ . Since $\text{X}$ occupies $\frac{1}{3}$ of these voids, the number of $\text{X}$ atoms is $\frac{2}{3}N$ . The ratio $\text{X}:\text{Y} = \frac{2}{3}:1 = 2:3$ , giving the formula $\text{X}_2\text{Y}_3$ .
B: $\text{X}_3\text{Y}$ - Inverting the ratio or confusing the occupancy fraction $\frac{1}{3}$ directly as the number of $\text{Y}$ atoms per $\text{X}$ atom. Multiply the total number of tetrahedral voids ( $2N$ ) by the occupancy fraction ( $\frac{1}{3}$ ) to find the quantity of $\text{X}$ , then express the mole ratio $\text{X}:\text{Y}$ in simplest integers.
C: $\text{X}_1\text{Y}_2$ - Assuming that a ccp lattice has only $N$ tetrahedral voids instead of $2N$ , or miscalculating $\frac{2}{3}:1$ as $1:2$ . Remember that the number of tetrahedral voids in any close-packed lattice is always twice the number of lattice atoms ( $2N$ ).
D: $\text{XY}_3$ - Assuming the number of tetrahedral voids equals the number of lattice atoms ( $N$ ) and taking $\frac{1}{3}N$ for $\text{X}$ , leading to an incorrect ratio $\text{X}:\text{Y} = 1:3$ . Use $2N$ for the number of tetrahedral voids in a ccp lattice formed by $N$ atoms.

Step-by-step solution

given: Element $\text{Y}$ forms a cubic close packed (ccp) lattice. Element $\text{X}$ occupies $\frac{1}{3}$ of the tetrahedral voids.
find: Find the stoichiometric formula of the compound formed by $\text{X}$ and $\text{Y}$ .
strategise: In a ccp (fcc) unit cell containing $N$ atoms of $\text{Y}$ , the number of tetrahedral voids is $2N$ . If $N = 4$ , tetrahedral voids $= 8$ . Number of $\text{X}$ atoms is $\frac{1}{3} \times 8 = \frac{8}{3}$ . Then simplify the ratio $\text{X} : \text{Y}$ .
execute: Number of $\text{Y}$ atoms $= 4$ . Number of $\text{X}$ atoms $= \frac{1}{3} \times 8 = \frac{8}{3}$ . Ratio $\text{X} : \text{Y} = \frac{8}{3} : 4 = 8 : 12 = 2 : 3$ . Thus, formula is $\text{X}_2\text{Y}_3$ .
verify: Ratio $\text{X} : \text{Y} = 2 : 3$ . For 3 atoms of $\text{Y}$ , tetrahedral voids $= 6$ . One-third of 6 is 2 atoms of $\text{X}$ . Perfectly consistent.

A compound is formed by two elements: Chemistry | JEE Main

For NNN atoms of element Y\text{Y}Y in a ccp lattice, there are 2N2N2N tetrahedral voids, meaning X\text{X}X occupies 13(2N)=23N\frac{1}{3}(2N) = \frac{2}{3}N31​(2N)=32​N sites, yielding the formula X2Y3\text{X}_2\text{Y}_3X2​Y3​.

01given

02find

03strategise

04execute

✓verify

Students also ask

For $N$ atoms of element $\text{Y}$ in a ccp lattice, there are $2N$ tetrahedral voids, meaning $\text{X}$ occupies $\frac{1}{3}(2N) = \frac{2}{3}N$ sites, yielding the formula $\text{X}_2\text{Y}_3$ .