Problem 140 - A Complete Resource Book in Chemistry for JEE Main

Question

In a compound, atoms of element \(\mathrm{Y}\) from ccp lattice and those of element \(\mathrm{X}\) occupy \(2 / 3^{\text {nd }}\) oftetrahedral voids. The formula of the compound will be (a) \(\mathrm{X}_{4} \mathrm{Y}_{3}\) (b) \(\mathrm{X}_{2} \mathrm{Y}_{3}\) (c) \(\mathrm{X}_{2} \mathrm{Y}\) (d) \(\mathrm{X}_{3} \mathrm{Y}_{4}\)

Step-by-Step Solution

Answer

The formula of the compound is \(X_4Y_3\) (option a).

Step 1: Determine the number of atoms in the ccp lattice

In a cubic close-packed (ccp) structure, there is 1 atom per unit cell. Since element \(Y\) forms the ccp lattice, in a given volume, the number of \(Y\) atoms is taken as 1.

Step 2: Calculate the number of tetrahedral voids

For every atom in a ccp lattice, there are 2 tetrahedral voids. Thus, for 1 atom of \(Y\), there are 2 tetrahedral voids available.

Step 3: Determine the occupancy of tetrahedral voids by element X

The problem states that atoms of element \(X\) occupy \(\frac{2}{3}\) of the tetrahedral voids. Thus, of the 2 available voids, \(\frac{2}{3} \times 2 = \frac{4}{3}\) atoms of \(X\) occupy the voids.

Step 4: Find the ratio of X and Y

We have 1 atom of \(Y\) and \(\frac{4}{3}\) atoms of \(X\). Therefore, the ratio of \(X:Y\) is \(\frac{4}{3}:1\). This can be converted to whole numbers for stoichiometric purposes as \(4:3\).

Step 5: Write the chemical formula

The ratio \(4:3\) signifies that for every 3 atoms of \(Y\), there are 4 atoms of \(X\). Thus, the chemical formula is \(X_4Y_3\).

Key Concepts

Cubic Close-Packed Structure

In the world of chemistry, many substances are organized into structured patterns known as lattices. One of the most efficient and densely packed structures is the cubic close-packed (ccp) arrangement. This structure can be visualized as layers of spheres (atoms or ions) stacked in a repeating ABCABC... pattern. This clever arrangement maximizes packing efficiency, which means more atoms are packed into a given volume.
For every cubic unit cell in the structure, there is effectively 1 atom per unit. This is because the ccp structure contains 8 atoms at the corners of the cube and 6 atoms in the faces of the cube, summing up to an equivalent of 4 atoms, but due to the shared nature of these atoms within multiple cells, it simplifies to 1 atom per unit cell for stoichiometric calculations.
This structure also leads to the formation of different types of interstitial sites or voids. Among these are the tetrahedral and octahedral voids, which play a crucial role in understanding the behavior of other elements that might be introduced into the lattice.

Tetrahedral Voids

When atoms in the ccp lattice are arranged close together, they leave tiny gaps or spaces in the structure. These spaces are known as voids. The specific type of void that arises in the spaces between four atoms and forms a pyramid shape is known as a tetrahedral void. In a ccp structure, for each atom, there are two such tetrahedral voids available.
These voids are crucial because they determine how other atoms or ions can fit into the lattice. If another element is introduced into the crystal, it might occupy these voids, affecting both the physical and chemical properties of the compound. For example, smaller atoms or ions may nestle into these tetrahedral voids.
In our exercise scenario, element \(X\) occupies \(\frac{2}{3}\) of the tetrahedral voids for every 'Y' atom. Thus, if there are two voids for every atom of \(Y\), then \(\frac{2}{3} \times 2 = \frac{4}{3}\) portions of these voids are filled by \(X\). This fraction is essential in calculating the exact stoichiometry of the mixed compound.

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that refers to the quantitative relationship between reactants and products in a chemical reaction. In simpler terms, it's about measuring the exact proportions of elements that make up compounds.
In the context of lattices and interstitial voids, stoichiometry informs us how different elements in a compound relate numerically. For this exercise, we have the elements \(X\) and \(Y\) in a calculated ratio. We determined this ratio by understanding that each \(Y\) atom in the ccp lattice provides 2 tetrahedral voids, and element \(X\) fills \(\frac{4}{3}\) of these voids per unit cell.
Thus, from the ratio \(\frac{4}{3}:1\) of \(X:Y\), we derive the whole number stoichiometric ratio of \(4:3\). This transformation gives us the simplest whole number formula for the compound, which in our case is \(X_4Y_3\). Stoichiometry ultimately helps in the determination of the empirical formula of chemical compounds by such quantitative analysis.