Solid State – JEE Main ChemistryPractice Questions & PYQs

Non-stoichiometric compounds like $A_{0.95} O$ arise from metal deficiency, which necessitates the presence of both $A^{2+}$ and $A^{3+}$ ions to maintain electrical neutrality within the crystal lattice. To compensate for the deficit of metal atoms relative to $O^{2-}$ ions, three $A^{2+}$ ions are replaced by two $A^{3+}$ ions, a process that naturally creates a vacant site for every pair of $A^{3+}$ ions introduced. Consequently, the lattice structure of this compound is characterized by the presence of these varying oxidation states alongside the anions and vacancies required to balance the overall ionic charge.

In a body-centered cubic (BCC) crystal structure, the atoms are in contact along the body diagonal of the unit cell, creating a geometric relationship defined by the equation $\sqrt{3} a = 4 r$, where $a$ is the edge length and $r$ is the atomic radius. Substituting the given unit cell edge length of $4 \text{\AA}$ into this expression results in $\sqrt{3} \times 4 = 4 r$, which simplifies directly to $r = \sqrt{3} \text{\AA}$. Using the value of $\sqrt{3} \approx 1.732$, the atomic radius is $1.732 \text{\AA}$. Expressing this value as $17.32 \times 10^{-1} \text{\AA}$ and rounding to the nearest integer leads to a final value of $17$.

The relationship between the unit cell edge length $a$ and the sphere radius $r$ is defined by the specific geometric diagonal path along which constituent spheres make physical contact. In a face-centered cubic structure, spheres touch along the face diagonal, which spans four sphere radii and has a length of $\sqrt{2}a$, yielding the equality $2\sqrt{2}r = a$. Similarly, atoms \in a body-centered cubic structure make contact along the body diagonal, which also spans four sphere radii and has a length of $\sqrt{3}a$, resulting \in the relation $4r = \sqrt{3}a$.

In a hexagonal close-packed structure, the density of voids is directly proportional to the number of constituent metal atoms, with each atom creating one octahedral void and two tetrahedral voids, summing to three voids per atom. Because there are three voids for every atom, the total number of voids in 0.02 moles of the metal is determined by multiplying the number of moles by Avogadro's number, $6.02 \times 10^{23}$, and then multiplying that product by three. The arithmetic $3 \times 0.02 \times 6.02 \times 10^{23}$ yields $0.3612 \times 10^{23}$, which is equivalent to $36.12 \times 10^{21}$. Approximating to the nearest integer results in 36.

The density of a crystal lattice is defined by the relationship $d = \frac{Z \times M}{N_A a^3}$, where $Z$ is the number of atoms per unit cell and $a$ is the edge length. Because the metal is identical in both scenarios, the molar mass and Avogadro's constant cancel out, leaving the density proportional to the ratio of $Z$ to $a^3$. The face-centered cubic lattice has $Z=4$ atoms per unit cell with an edge length of 2.0 Å, whereas the body-centered cubic lattice has $Z=2$ atoms per unit cell with an edge length of 2.5 Å. Comparing the densities involves calculating the ratio $\frac{4 / 2.0^3}{2 / 2.5^3}$, which simplifies to the expression $\frac{0.5}{0.128}$. This calculation results in approximately 3.90, and rounding to the nearest integer gives 4.

The density of a crystalline solid relates the microscopic properties of its unit cell to its macroscopic mass, represented by the relationship $d = \frac{Z \times M}{N_A \times a^3}$. In this formula, $d$ denotes the density, $Z$ is the number of atoms per unit cell, $M$ is the molar mass, $N_A$ is Avogadro's constant, and $a$ represents the length of the unit cell edge.

To calculate the number of atoms, one must first ensure that the edge length is in centimeters to match the units of density, giving $a = 300 \times 10^{-10}\ \mathrm{cm}$. Rearranging the density equation to isolate $Z$ yields $Z = \frac{d \times N_A \times a^3}{M}$. Substituting the known values into this equation results in $Z = \frac{3 \times 6.02 \times 10^{23} \times (300 \times 10^{-10})^3}{12}$. Simplifying the cube of the edge length produces $27 \times 10^{-24}\ \mathrm{cm}^3$, allowing the calculation to continue as $Z = \frac{3 \times 6.02 \times 10^{23} \times 27 \times 10^{-24}}{12}$. This simplifies further to $\frac{48.762}{12}$, which results in approximately $4.0635$, meaning the number of atoms present in the unit cell is $4$ when rounded to the nearest integer.

The 14 Bravais lattices determine which unit cell arrangements, such as primitive and centered types, are physically permissible for each of the seven crystal systems. Body-centered unit cells, symbolically represented by the lattice type $I$, cannot exist in all crystal structures due to their specific geometric constraints. Among the provided options, only the cubic, tetragonal, and orthorhombic systems possess the necessary symmetry to accommodate a body-centered arrangement. This confirms that there are exactly 3 crystal systems where this unit cell can be found.

The empirical formula of a crystal structure is determined by calculating the effective contribution of each atom to a single unit cell. Atoms of element $X$ are located on four alternate corners, each contributing $\frac{1}{8}$ of an atom to the unit cell, plus one atom at the body center that contributes fully, resulting in a total of $X = 4 \times \frac{1}{8} + 1 = 1.5$ atoms per unit cell. Element $Y$ is situated at one-third of the six total faces, and since each face atom is shared between two adjacent cells, the contribution is $Y = 6 \times \frac{1}{3} \times \frac{1}{2} = 1$ atom per unit cell. The resulting stoichiometric ratio is $X_{1.5} Y_1$, which simplifies to the empirical formula $X_3 Y_2$.

The density of a crystal links the mass of the unit cell to its physical volume, governed by the relationship $d = \frac{Z \times M}{N_A \times a^3}$. In this expression, $Z$ represents the number of formula units per unit cell, $M$ is the molar mass of the compound, $N_A$ is Avogadro's number, and $a$ is the edge length of the cubic unit cell.

To calculate $Z$, rearrange the formula to $Z = \frac{d \times N_A \times a^3}{M}$. The molar mass of FeO is 72 g/mol. Converting the edge length 5.0 Å to centimeters gives $5.0 \times 10^{-8}$ cm, making the volume $a^3$ equal to $125 \times 10^{-24}$ cm$^3$. Substituting the density of 4.0 g/cm$^3$, Avogadro's number of $6.0 \times 10^{23}$ mol$^{-1}$, and the calculated volume into the rearranged equation gives the expression $Z = \frac{4.0 \times 6.0 \times 10^{23} \times 125 \times 10^{-24}}{72}$. Simplifying the numerator yields $Z = \frac{300}{72}$, which results in approximately 4.166; rounding to the nearest integer gives 4.

The $\mathrm{CsCl}$ unit cell adopts a body-centered structure where $\mathrm{Cl}^{-}$ anions occupy the eight corners and a $\mathrm{Cs}^{+}$ cation resides at the body center. These ions come into direct contact along the body diagonal of the cube, which has a total geometric length of $\sqrt{3}a$. Since the distance from a corner of the cube to its center represents the sum of the ionic radii of the cation and the anion, this distance is equal to half the length of the body diagonal, which leads to the expression $r_{\mathrm{Cs}}^{+} + r_{\mathrm{Cl}}^{-} = \frac{\sqrt{3}}{2}a$.

In a cubic close-packed arrangement, the number of lattice atoms determines the total count of available interstitial sites. If element $Y$ crystallizes in a CCP structure, it provides $4$ atoms per unit cell, and the inherent geometry of this lattice dictates that there are twice as many tetrahedral voids as lattice atoms, resulting in $8$ tetrahedral voids. With element $X$ occupying exactly one-third of these voids, its count is $\frac{8}{3}$ per unit cell.

The stoichiometric relationship between the elements is expressed by the ratio of $\frac{8}{3}$ atoms of $X$ for every $4$ atoms of $Y$. Multiplying both sides of the ratio by $3$ to convert these values into whole numbers yields $8:12$, which simplifies to the smallest whole-number ratio of $2:3$. The resulting formula for the compound is $X_2Y_3$

In a cubic close packing arrangement, the anions occupy the lattice positions while the cations reside within the octahedral voids located at the edge centers and the body center of the unit cell. Geometrically, this configuration forces the cation and anion to touch along the edge of the unit cell, which means the edge length corresponds to twice the sum of the cation and anion radii. By substituting the provided ionic radii into this relationship, the calculation follows the expression $a = 2(r_{A^+} + r_{B^-})$, which equates to $2(102 + 181) = 566$ pm.

Imperfections in a solid crystal lattice are directly linked to changes in its density based on how the particle count per unit volume is affected. A vacancy defect involves the absence of particles from regular lattice sites, causing a reduction in mass while the volume remains constant, which necessarily decreases the overall density of the substance. In contrast, interstitial defects pack additional particles into lattice voids to increase density, and Frenkel defects maintain consistent density by merely shifting an existing particle into an interstitial position. Thus, asserting that a vacancy defect increases the density of the substance is fundamentally incorrect.

In an $A B$ crystal where $A$ atoms form a cubic close-packed structure and $B$ atoms occupy all octahedral sites, the unit cell initially contains 4 $A$ atoms and 4 $B$ atoms. When two atoms are removed from opposite face centers, these atoms belong to the $A$ lattice, reducing the effective count of $A$ atoms to $4 - (2 \times \frac{1}{2}) = 3$.

Because the octahedral voids occupied by $B$ atoms are located at the edge centers and the body center, the $B$ atoms remain entirely unaffected by this removal, maintaining their total count of 4. This results in a final stoichiometry of $A_{3} B_{4}$, which means the value of $x$ is 3.

The density of a crystalline solid depends on the mass of the atoms within its unit cell and the volume of that cell. For a face-centered cubic structure, there are $Z=4$ atoms per unit cell. The relationship between density, molar mass, and unit cell dimensions is given by the formula $\rho = \frac{Z M}{N_A a^3}$, where $\rho$ is the density, $M$ is the molar mass, $N_A$ is Avogadro's number, and $a$ is the edge length.

The expression can be rearranged to isolate the molar mass, yielding $M = \frac{\rho N_A a^3}{Z}$. By substituting the given parameters, the equation becomes $M = \frac{9.03 \times 6.02 \times 10^{23} \times (4.0 \times 10^{-8})^3}{4}$. Simplifying the cubic volume term results \in the expression $M = \frac{9.03 \times 6.02 \times 64 \times 10^{-1}}{4}$. These operations result in 86.9 g/mol, which is 87 g/mol when rounded to the nearest integer.

In a body-centered cubic unit cell, the structure effectively contains two atoms. The density of the crystal is defined by the mass of these atoms per unit cell volume, expressed by the equation $\rho = \frac{Z \times A}{V}$, where $\rho$ is the density, $Z$ is the number of atoms, $A$ is the mass of one atom, and $V$ is the volume of the unit cell.

With an edge length of $300 \times 10^{-10}$ cm, the unit cell volume is $(300 \times 10^{-10})^3$ cm$^3$. Given a density of $6.0$ g cm$^{-3}$ and $Z = 2$, we solve for the mass of a single atom using the equation $6.0 = \frac{2 \times A}{(300 \times 10^{-10})^3}$, which gives $A = 81 \times 10^{-24}$ g. The total number of atoms in $180$ g of the element is calculated by $\frac{180}{81 \times 10^{-24}}$, resulting in approximately $22.22 \times 10^{23}$ atoms. Rounding to the nearest integer, we find the answer is 22.

In an hexagonal close-packed unit cell, there are 6 atoms, and since the number of tetrahedral voids is twice the number of lattice particles, 12 such voids exist. Given that element Y occupies 2/3 of these voids, the number of Y atoms per unit cell is
$\frac{2}{3} \times 12 = 8$
The total number of particles in the lattice is the sum of 6 atoms of element X and 8 atoms of element Y, which equals 14. The percentage of element X is calculated as
$\frac{6}{14} \times 100 = 42.85$
which, when rounded to the nearest integer, equals 43.

The density of a crystal lattice relates its macroscopic mass to the microscopic arrangement of atoms within its unit cell through the expression $\rho = \frac{Z \times M}{a^3 \times N_A}$. Since the element possesses a body-centred cubic structure, the number of atoms per unit cell, Z, equals 2. Using the provided density of 5 g cm^-3 and the unit cell edge length of 200 pm, which converts to 200 x 10^-10 cm, we can rearrange the equation to isolate the molar mass M.

Substituting these values into $\frac{\rho \times a^3 \times N_A}{Z} = M$ allows us to solve for the molar mass, which is 12 g mol^-1. Because the molar mass represents the weight of Avogadro's number of atoms, a 300 g sample contains 300/12, or 25, times the Avogadro constant, resulting in a total of 25 $N_A$ atoms.

The principle of electroneutrality dictates that in any ionic crystal, the total positive charge must equal the total negative charge. For the non-stoichiometric compound $\mathrm{Fe}_{0.93}\mathrm{O}$, this means the net charge contributed by iron ions in their $+2$ and $+3$ oxidation states must balance the negative charge from the oxygen ions.

Assuming a sample contains $100$ oxide ions to simplify calculations, we have a total negative charge of $200$ and $93$ total iron ions. If $x$ represents the number of $\mathrm{Fe}^{2+}$ ions, the remaining $(93 - x)$ ions must be $\mathrm{Fe}^{3+}$ ions. Balancing the charges yields the equation $x(2) + (93 - x)(3) = 200$. Solving this results in $2x + 279 - 3x = 200$, leading to $x = 79$. The percentage of $\mathrm{Fe}^{2+}$ ions is calculated as $\frac{79}{93} \times 100$, which is approximately $85$.

The crystal structure of NaCl consists of a face-centered cubic arrangement where the number of formula units per unit cell is $Z=4$ and the shortest distance between $\mathrm{Na}^{+}$ and $\mathrm{Cl}^{-}$ ions is exactly half of the unit cell edge length, represented as $\frac{a}{2}$. By utilizing the density formula $\rho = \frac{Z \times M}{a^3 \times N_A}$ with the provided parameters $\rho = 43.1 \ \mathrm{g \ cm^{-3}}$, $Z=4$, $M=58.5 \ \mathrm{g \ mol^{-1}}$, and $N_A = 6.02 \times 10^{23} \ \mathrm{mol^{-1}}$, the expression for the edge length becomes $a^3 = \frac{4 \times 58.5}{43.1 \times 6.02 \times 10^{23}}$. This calculation results in $a^3 = 9 \times 10^{-24} \ \mathrm{cm^3}$, which corresponds to an edge length $a = 2.08 \times 10^{-8} \ \mathrm{cm}$ or $2.08 \times 10^{-10} \ \mathrm{m}$. Because the distance between the ions is $\frac{a}{2}$, the value is $1.04 \times 10^{-10} \ \mathrm{m}$, which rounds to 1 as the nearest integer.