Solid State – JEE Main ChemistryPractice Questions & PYQs

The thickness of a molecular monolayer is determined by the relationship $Volume = Area \times height$. Since 0.27 g of fatty acid is dissolved in 100 cm³, the 10 cm³ sample contains 0.027 g of the fatty acid. The surface area of the circular watch glass is calculated using the formula $Area = \pi r^2$, which with a radius of 10 cm and $\pi = 3$ yields $Area = 3 \times 10^2 = 300\,\mathrm{cm}^2$. The volume of the fatty acid layer is found using $Volume = \frac{mass}{density} = \frac{0.027}{0.9} = 0.03\,\mathrm{cm}^3$. Finally, the height is determined by the equation $height = \frac{Volume}{Area} = \frac{0.03}{300} = 10^{-4}\,\mathrm{cm}$, which is equivalent to $h = 10^{-6}\,\mathrm{m}$.

Packing efficiency represents the fraction of a unit cell's volume occupied by the spheres contained within it. In a body-centered cubic arrangement consisting of two types of atoms, A and B, the spheres make contact along the body diagonal of the cube. The length of this diagonal, $\sqrt{3} a$, is equal to the sum of the diameters of the two spheres, or $2r + 4r = 6r$, allowing the edge length $a$ to be expressed as $2\sqrt{3}r$.

The total volume occupied by the two spheres is $\frac{4}{3} \pi r^3 + \frac{4}{3} \pi (2r)^3$, which simplifies to $12 \pi r^3$. Given the unit cell volume as $a^3 = (2\sqrt{3}r)^3 = 24\sqrt{3}r^3$, the packing efficiency is determined by dividing the total atomic volume by the cell volume. This calculation yields a value of approximately 0.90, indicating that 90% of the unit cell space is occupied by the atoms.

In a face-centered cubic lattice, the structure can be visualized as being composed of eight smaller sub-cubes, each having an edge length of $\frac{a}{2}$. Tetrahedral voids are positioned at the centers of these sub-cubes, specifically located along the body diagonals at a distance from the corners. Because these tetrahedral voids lie at the corners of adjacent sub-cubes, the physical distance between the centers of any two nearest tetrahedral voids is identical to the side length of these divided regions, resulting in a value of $\frac{a}{2}$.

The molar mass ($M$) of a crystalline solid is calculated using the relation $\rho = \frac{Z M}{N_A a^3}$, where $\rho$ is the density, $Z$ is the number of atoms per unit cell, $N_A$ is the Avogadro constant, and $a$ is the edge length. For a face-centred cubic (FCC) crystal, $Z = 4$. Given the edge length $a = 200\sqrt{2} \text{ pm} = 200\sqrt{2} \times 10^{-12} \text{ m}$, the volume of the unit cell is $a^3 = (200\sqrt{2} \times 10^{-12})^3 = 16\sqrt{2} \times 10^{-30} \text{ m}^3$.

Rearranging the formula to solve for molar mass gives $M = \frac{\rho N_A a^3}{Z}$. Substituting the given values, $M = \frac{9 \times 10^3 \times 6 \times 10^{23} \times 16\sqrt{2} \times 10^{-30}}{4}$. Simplifying this expression, $M = \frac{864\sqrt{2} \times 10^{-4}}{4} = 216\sqrt{2} \times 10^{-4}$. Using $\sqrt{2} \approx 1.414$, the molar mass is approximately $0.0305 \text{ kg mol}^{-1}$.

The total number of effective atoms in a unit cell depends on the fractional contribution of atoms located at specific positions, as corner atoms are shared by eight cells, face atoms by two, and body-centered atoms belong exclusively to a single unit cell. For a simple cubic structure, the effective number of atoms is calculated as $Z_{\mathrm{SC}} = 8 \times \frac{1}{8} = 1$. In a body-centered cubic structure, we add the single central atom to the contribution from the corners to find $Z_{\mathrm{BCC}} = 1 + 1 = 2$. For a face-centered cubic structure, we combine the contribution from the eight corners and the six face-centered atoms as $Z_{\mathrm{FCC}} = 1 + 6 \times \frac{1}{2} = 4$. Comparing these effective counts gives a ratio of 1 : 2 : 4 for the simple cubic, body-centered cubic, and face-centered cubic structures respectively.

In any crystal lattice, the stoichiometry of a compound is determined by the ratio of atoms occupying specific interstitial voids relative to the atoms at the lattice points. A cubic close-packed (ccp) arrangement contains 4 atoms of element B per unit cell. The total number of octahedral voids in this structure is equal to the number of lattice points, providing 4 available sites, so element A occupying half of these contributes 2 atoms. Because the number of tetrahedral voids is twice the number of lattice points, there are 8 such sites, all of which are occupied by oxygen atoms. This results in an atomic ratio represented by the formula $A_2 B_4 O_8$, which simplifies to $AB_2O_4$.

The total surface area of a monolayer is established by the sum of the footprints of all individual polar head groups present. We first determine the total quantity of surfactant in the system by multiplying the concentration $10^{-3} \, \text{M}$ by the volume $10 \, \text{mL}$, which results in $10^{-5} \, \text{moles}$, or $6 \times 10^{18}$ individual polar heads. The monolayer covers an area of $0.24 \, \text{cm}^2$, which converts to $2.4 \times 10^{-5} \, \text{m}^2$. Approximating each head as a cube with edge length $a$, the total area is the number of heads multiplied by the square of the edge length, leading to the relation $a^2 \times 6 \times 10^{18} = 2.4 \times 10^{-5}$. Simplifying this expression gives $a^2 = 4 \times 10^{-24} \, \text{m}^2$, and taking the square root reveals an edge length $a = 2 \times 10^{-12} \, \text{m}$, or $2.0 \, \text{pm}$.

The density of a crystal lattice is determined by dividing the total mass of the atoms within the unit cell by the volume of that cell. For a copper face-centered cubic structure, the number of atoms per unit cell is $Z = 4$, and the atomic mass is 63.5 g per mole. To perform the calculation in consistent units, the edge length is defined by $a = x \times 10^{-8}$ cm, while Avogadro's number is $N_A = 6 \times 10^{23}$ per mole.

Applying these values to the density formula $d = \frac{Z \cdot M}{N_A \cdot a^3}$ yields the expression $d = \frac{4 \cdot 63.5}{(6 \times 10^{23}) \cdot (x \times 10^{-8})^3}$. The denominator simplifies to $6 \times 10^{23} \cdot x^3 \cdot 10^{-24}$, which evaluates to $0.6 \cdot x^3$. Dividing the numerator of 254 by this value provides the final density result of $\frac{422}{x^3}$ g/cm^3.

Crystal systems are defined by the geometric relationships between their edge lengths ($a$, $b$, and $c$) and axial angles ($\alpha$, $\beta$, and $\gamma$). The triclinic system distinguishes itself as the most asymmetric primitive unit cell, requiring that $a \neq b \neq c$ and $\alpha \neq \beta \neq \gamma \neq 90^\circ$. This specific configuration ensures that no two edges or angles are equivalent, providing the least amount of symmetry among all lattice types.

In a body-centered cubic unit cell, the largest sphere positioned at the midpoint of an edge is constrained by the two corner atoms defining that edge. Since the distance from a corner to the edge center is half the edge length, we have the geometric constraint $R + r = \frac{a}{2}$, where $R$ is the radius of the corner atoms and $r$ is the radius of the sphere fitting at the edge center.

In a BCC arrangement, the atoms touch each other along the body diagonal, which corresponds to the relationship $4R = a\sqrt{3}$, or $R = \frac{a\sqrt{3}}{4}$. Substituting this value of $R$ into the first relationship allows us to isolate $r$ as $r = \frac{a}{2} - \frac{a\sqrt{3}}{4}$. Simplifying this expression to $r = a \left(\frac{2 - \sqrt{3}}{4}\right)$ and calculating the numerical value results in $r = 0.067 a$.

In any hexagonal close-packed lattice containing $n$ atoms, there are exactly $2n$ tetrahedral voids. To determine the arrangement for the compound $A_2B_3$, we test which atom forms the lattice to satisfy the required 2:3 atomic ratio. Assuming B forms the lattice, there are $n$ atoms of B, which necessitates $\frac{2}{3}n$ atoms of A to match the formula. Placing these A atoms into the $2n$ available tetrahedral voids results in an occupancy fraction defined by $\frac{\frac{2}{3}n}{2n}$, confirming that A must occupy $\frac{1}{3}$ of the tetrahedral voids.

The crystal structure of ionic compounds is determined by the radius ratio of the cation to the anion, denoted as $r_+/r_-$, which dictates the coordination number and packing geometry of the lattice. While LiCl, NaCl, and RbCl crystallize in the rock salt structure with a 6:6 coordination number, CsCl adopts the cesium chloride structure characterized by an 8:8 coordination number. This distinct packing arrangement in CsCl is required to accommodate the significantly larger ionic radius of the $Cs^+$ cation compared to the other alkali metal ions listed.

In ionic crystals, a Frenkel defect occurs when a smaller ion, typically a cation, is displaced from its normal lattice position and relocates into a vacant interstitial space within the crystal structure. This shift maintains the overall electrical neutrality of the solid because the total number of ions remains unchanged, even though the internal arrangement is locally altered. Because the cation essentially vacates its original lattice site to occupy a different void, the defining characteristic of this specific imperfection is the presence of a cation in an interstitial site.

Antiferromagnetic substances are characterized by the arrangement of atomic magnetic moments in an anti-parallel orientation of equal magnitude, which leads to the complete cancellation of the net magnetic moment. This structural alignment is schematically represented by alternating arrows pointing in opposite directions, following an up-down-up-down pattern. This characteristic distinguishes antiferromagnetic materials from ferromagnetic substances, where moments align in parallel, and ferrimagnetic substances, where moments align in unequal anti-parallel numbers.

In a face-centered cubic lattice, the atoms are arranged such that they touch one another along the diagonal of each square face of the unit cell. Given an edge length of $a$, the geometric length of this face diagonal is calculated using the Pythagorean theorem as $\sqrt{a^2 + a^2}$, resulting in $\sqrt{2}a$. This path spans four atomic radii, represented by the equation
$\sqrt{2}a = 4R$
where $R$ is the atomic radius. The distance of closest approach between two adjacent atoms is equivalent to their diameter, which is $2R$. By dividing both sides of the previous equation by two, we find that this distance is equal to $\frac{\sqrt{2}a}{2}$, which simplifies to $\frac{a}{\sqrt{2}}$.

In a body-centered cubic structure, atoms touch along the body diagonal, creating a specific geometric relationship between the cube edge length and the atomic radius. Because this diagonal spans two atomic radii at the corners plus the full diameter of the central atom, the relationship is defined as:

$4r = a\sqrt{3}$

The atomic radius $r$ is derived from the edge length $a$ of 4.29 Å by rearranging this expression:

$r = \frac{a\sqrt{3}}{4}$

Substituting the numerical values yields 1.857 Å, which simplifies to approximately 1.86 Å.

In the cesium chloride crystal structure, which adopts a body-centered cubic arrangement, the cesium ion is positioned at the center of the unit cell while chloride ions occupy the eight corners. These ions are in contact along the body diagonal, which spans the interior of the cube and is geometrically defined as $\sqrt{3}a$. Because this line of contact encompasses a chloride ion at the corner and the central cesium ion, the distance between the center of the cesium ion and the center of the chloride ion corresponds to half of the total body diagonal. This geometric relationship establishes that the sum of the radii of the cesium and chloride ions is equal to $\frac{\sqrt{3}a}{2}$.

Covalent crystals, often referred to as network solids, are defined by an infinite three-dimensional lattice where atoms are interconnected by strong, continuous covalent bonds throughout the entire material. Silicon uniquely forms such a rigid structure, whereas Sulphur, Phosphorous, and Iodine exist as discrete molecular units-specifically $S_8$, $P_4$, and $I_2$ respectively-which are classified as molecular solids because they are held together by significantly weaker intermolecular forces rather than a giant covalent framework.

In a body-centered cubic arrangement, atoms make direct contact along the body diagonal, establishing a geometric relationship where the length of this diagonal equals four times the atomic radius. This geometry is defined by the equation:
$\sqrt{3} a = 4r$
Rearranging the formula to isolate the atomic radius $r$ provides:
$r = \frac{\sqrt{3}}{4} a$
Substituting the provided side length of 351 pm into this expression leads to the following calculation:
$r = \frac{\sqrt{3}}{4} \times 351 = 152 \text{ pm}$

In a face-centered cubic unit cell, the stoichiometry depends on the fractional contribution of each lattice site to the interior of the structure. Atoms positioned at the corners are shared by eight adjacent unit cells, meaning each corner contributes $1/8$ of an atom, while atoms sitting on the faces are shared between two cells, resulting in a contribution of $1/2$ per atom.

To determine the count for A, we multiply the 8 corner positions by their contribution, which results in 1 atom. For atom B, we identify that while a standard face-centered cubic structure possesses 6 face positions, the loss of one leaves 5 atoms. Multiplying these 5 atoms by their contribution of $1/2$ yields an effective value of $2.5$ atoms of B. Combining these results gives a ratio of 1 atom of A for every $2.5$ atoms of B, which translates to the integer formula $A_2 B_5$.