Solid State – JEE Main ChemistryPractice Questions & PYQs

In a face-centered cubic arrangement of an ionic crystal, the ions touch along the edge of the unit cell, meaning the edge length relates to the radii of the cation and anion through the expression $a = 2(r_{\text{cation}} + r_{\text{anion}})$. Given an edge length of 508 pm and a cation radius of 110 pm, we set up the equation
$508 = 2(110 + r_{\text{anion}})$
to solve for the unknown variable. Dividing the total edge length by two yields 254 pm as the sum of the cation and anion radii, and subtracting the cation radius of 110 pm from this result confirms the anion radius is 144 pm.

The percentage of free space in a crystal lattice is determined by subtracting the packing efficiency, which measures the volume occupied by spheres, from the total volume of the unit cell, as expressed by the equation $Free\ Space = 100\% - Packing\ Efficiency$. Since a cubic close packed (CCP) structure has a packing efficiency of 74% and a body-centered cubic (BCC) structure has an efficiency of 68%, calculating the difference for each reveals the respective void percentages. Consequently, the free space for CCP and BCC structures is 26% and 32% respectively.

In a face-centered cubic (fcc) crystal system, atoms are packed such that they make contact along the face diagonal of the unit cell. The length of this face diagonal, which geometrically measures $\sqrt{2}a$ for an edge length $a$, is equivalent to four times the atomic radius, expressed as $4r = \sqrt{2}a$. Given the edge length of $361 \pm$, we determine the radius by rearranging the equation to $r = \frac{\sqrt{2} \times 361}{4}$. Calculating this value provides a result of $127 \pm$.

In a cubic close-packed structure, the geometry dictates that the number of tetrahedral voids is exactly twice the count of the atoms forming the lattice itself. Given that element $Y$ occupies the ccp lattice sites, we define the unit cell as containing 4 atoms of $Y$. This structure inherently provides 8 tetrahedral voids, and since element $X$ fills $\frac{2}{3}$ of these spaces, the number of $X$ atoms per unit cell is calculated as $8 \times \frac{2}{3} = \frac{16}{3}$.

Comparing these quantities gives the atomic ratio of $X$ to $Y$ as $\frac{16}{3} : 4$. Multiplying both sides by 3 to clear the fraction yields a ratio of 16 to 12, which simplifies to 4 to 3, resulting in the chemical formula $X_4 Y_3$.

The total volume of atoms within a unit cell is the product of the number of atoms it contains and the volume of a single spherical atom. A face-centered cubic unit cell contains 4 atoms, and since each atom has a volume defined by $\frac{4}{3} \pi r^3$, the total volume is found by multiplying these values:

$4 \times \frac{4}{3} \pi r^3 = \frac{16}{3} \pi r^3$

This calculation determines the space occupied by all constituent atoms within the lattice.

Determining the stoichiometry of an ionic crystal depends on how many neighboring unit cells share each lattice site. Because corner positions are shared among 8 adjacent cubes, only 1/8 of each corner atom resides within a single unit cell, so the total contribution for $A$ ions is $8 \times \frac{1}{8} = 1$. Similarly, atoms placed on the centers of the faces are shared by 2 adjacent cells, meaning each contributes 1/2 to the unit cell, which leads to a total of $6 \times \frac{1}{2} = 3$ for $B$ ions. Given this 1:3 ratio, the empirical formula is $AB_3$.

The Schottky defect occurs when an equal number of cations and anions are missing from their designated lattice sites, a process that preserves the overall electrical neutrality of the ionic crystal. Because the diagram illustrates vacant positions resulting from the departure of an equivalent pair of oppositely charged ions, it confirms the presence of this specific irregularity. This type of defect is most commonly observed in ionic solids that exhibit a high coordination number and feature ions of similar sizes, such as $\mathrm{NaCl}$ and $\mathrm{KCl}$.

Calculating the number of unit cells in a specific mass of a crystal requires finding the total number of formula units present and dividing that value by the number of formula units contained in a single unit cell. The molar mass of sodium chloride is the sum of the atomic weights of sodium and chlorine, resulting in $23 + 35.5 = 58.5\ \mathrm{g/mol}$. Given the crystal mass of $1.00\ \mathrm{g}$, the total number of formula units is obtained by multiplying the number of moles by Avogadro's constant, which is $\frac{1.00}{58.5} \times 6.02 \times 10^{23}$. Since the NaCl lattice contains $4$ formula units per unit cell, the total number of unit cells is calculated as:

$\frac{1.00 \times 6.02 \times 10^{23}}{58.5 \times 4} = 2.57 \times 10^{21}$

The number of atoms within a unit cell is determined by the contribution of atoms at specific lattice positions shared with neighboring cells. In a Body-Centered Cubic (BCC) arrangement like that of sodium, atoms are located at the corners and the body center, resulting in a total count of $8 \times \frac{1}{8} + 1 = 2$ atoms. Conversely, a Face-Centered Cubic (FCC) structure like that of magnesium places atoms at the corners and the centers of the six faces, yielding a total count of $8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$ atoms. Therefore, these crystal structures contain 2 and 4 atoms per unit cell, respectively.