Solid State – JEE Main ChemistryPractice Questions & PYQs

Substances that remain electrical insulators in both solid and molten states cannot be metallic or ionic, as these solids rely on mobile electrons or ions for conductivity in those conditions. While molecular solids are generally defined by their relatively low melting points due to weak intermolecular attractions, a hard material that combines an extremely high melting point with total electrical insulation must be a covalent network solid, characterized by a vast, rigid lattice of strong, directional chemical bonds.

In an fcc or ccp lattice, the ratio of octahedral voids to lattice sites is determined by comparing their respective total counts. Summing the 8 corner positions and 6 face-centered positions results in 14 total lattice sites, while the 12 edge-centered positions combined with the 1 body-centered position identify 13 octahedral voids. Calculating the ratio as $\frac{13}{14}$ yields approximately 0.928. Rounding this value to the nearest integer provides 1.

The rounding of sharp glass edges upon heating occurs because liquids naturally tend to minimize their surface area, a process driven by the property of surface tension rather than viscosity. Although the viscosity of glass does indeed decrease as temperature rises, this physical characteristic is independent of the surface tension-driven smoothing process. Consequently, while both statements are accurate descriptions of how glass behaves under heat, the decrease in viscosity is not the underlying cause for the edges losing their sharpness.

The atomic radius is an inherent physical property of an element and remains constant regardless of the crystal lattice geometry. When atoms arrange themselves in a body-centered cubic (BCC) structure, they make contact along the body diagonal, which establishes the relationship $4r = \sqrt{3} a_{bcc}$ where $r$ represents the radius and $a_{bcc}$ denotes the unit cell edge length. Conversely, \in a face-centered cubic (FCC) arrangement, the atoms touch along the face diagonal, defining the relationship $4r = \sqrt{2} a_{fcc}$ where $a_{fcc}$ represents the edge length of the FCC unit cell.

Since the radius $r$ is identical in both structures, we can equate these two expressions to solve for the unknown edge length. By setting the expressions equal, we obtain $\sqrt{3} a_{bcc} = \sqrt{2} a_{fcc}$ and by substituting the given BCC edge length of $27$ Å along with the provided values of $\sqrt{3} = 1.73$ and $\sqrt{2} = 1.41$, the equation becomes $1.73 \times 27 = 1.41 \times a_{fcc}$ Calculating the left side yields $46.71$, and dividing this product by $1.41$ results \in approximately $33.12$. Rounding this to the nearest integer gives an FCC edge length of $33$ Å.

In a hexagonal close-packed structure, the unit cell contains an effective count of 6 atoms. Geometric principles show this cell hosts 12 tetrahedral and 6 octahedral voids, resulting in 18 total voids per unit cell. Since there are 6 atoms per unit cell, there are 3 voids associated with every atom in the lattice.

To calculate the total voids in 0.581 g of gallium, first determine the number of moles by dividing the sample mass by the atomic mass, $\frac{0.581}{70}$. Multiplying this by Avogadro's constant, $6.02 \times 10^{23}$, gives the number of atoms. Multiplying this by 3 provides the total count. The calculation $\frac{0.581}{70} \times 6.02 \times 10^{23} \times 3$ yields $14.99 \times 10^{21}$, which rounds to $15 \times 10^{21}$.

In a cubic close packed lattice, the effective number of B anions is 4 per unit cell. The total number of octahedral voids in this arrangement is inherently equal to the number of constituent particles, meaning there are 4 such voids. Given that all octahedral sites are occupied by A cations, the compound contains 4 A atoms and 4 B atoms. The resulting formula is $A_4B_4$, which simplifies to $AB$, confirming that the value of x is 1.

Crystal systems are defined by the geometric relationship between their axial lengths and interfacial angles. The dimensions $a=2.5$, $b=3.0$, and $c=4.0$ satisfy the condition $a \neq b \neq c$. Additionally, with the angles $\alpha=90^{\circ}$, $\gamma=90^{\circ}$, and $\beta=120^{\circ}$, the crystal structure adheres to the relationship $\alpha=\gamma=90^{\circ} \neq \beta$. These specific parameters uniquely characterize the monoclinic crystal system.

The arrangement of atoms in an HCP unit cell dictates that the number of tetrahedral voids is exactly double the number of atoms present in the lattice. Because element $A$ forms the HCP structure, it contributes $6$ atoms per unit cell, which implies there are $2 \times 6 = 12$ tetrahedral voids available for occupation. Since element $M$ fills $\frac{2}{3}$ of these voids, we calculate the number of $M$ atoms as $\frac{2}{3} \times 12 = 8$. This creates a stoichiometric relationship of $M_8 A_6$, which reduces to the simplest whole-number ratio of $M_4 A_3$.

The coordination number is defined as the total count of nearest neighboring atoms that directly touch a central atom within a crystal lattice. In a body-centered cubic structure, an atom at the center is equidistant to eight surrounding corner atoms, specifically touching four atoms in the layer above and four in the layer below, with none in its own plane. Since these eight corner atoms represent the immediate points of contact for the central particle, the coordination number is 8.

The density of a crystalline solid is determined by the ratio of the mass of the particles within its unit cell to the total volume occupied by that cell. A face-centered cubic structure contains 4 atoms per unit cell, and the mass contribution of these atoms is derived from the molar mass divided by Avogadro's constant. Since the final density must be expressed in kilograms per cubic meter, the molar mass must be converted to kilograms per mole and the edge length of the unit cell must be converted to meters before performing the calculation.

The density is calculated using the formula $d = \frac{Z \times M}{a^{3} \times N_{A}}$ where $Z = 4$, $M = 63.54 × 10^{-3}$ kg/mol, $a = 3.596 × 10^{-10}$ m, and $N_{A} = 6.022 × 10^{23}$ mol$^{-1}$. Substituting these values into the equation, the volume is found by cubing the edge length, and the total mass per unit cell is determined by multiplying the molar mass by 4 and dividing by Avogadro's constant. This arithmetic results \in a density of 9076.26 kg/m$^{3}$, which rounds to 9077 kg/m$^{3}$.

A Frenkel defect arises when an ion is displaced from its original lattice position to an interstitial site, thereby generating both a vacancy and an interstitial defect within the crystal structure. Regarding coloration, ionic solids only exhibit this property due to the presence of F-centers, which are formed by metal excess defects involving trapped electrons in anionic vacancies, meaning Frenkel defects are unrelated to this specific phenomenon.

When a divalent $\mathrm{Sr}^{2+}$ ion replaces a monovalent $\mathrm{K}^+$ ion in a $\mathrm{KBr}$ crystal lattice, the principle of charge neutrality necessitates the removal of one $\mathrm{K}^+$ ion to compensate for the charge difference. This process creates exactly one cationic vacancy for every $\mathrm{Sr}^{2+}$ ion introduced, making the number of vacancies equal to the total number of dopant ions added.

The molar mass of $\mathrm{KBr}$ is $39.1 + 79.9 = 119\ \mathrm{g/mol}$, so a $1\ \mathrm{g}$ sample contains $\frac{1}{119}$ moles of the crystal. A doping concentration of $10^{-5}$ mole percent means that for every $100$ moles of $\mathrm{KBr}$, there are $10^{-5}$ moles of $\mathrm{SrBr}_2$. The amount of $\mathrm{Sr}^{2+}$ ions is thus $\frac{10^{-5}}{100}$ of the total moles of $\mathrm{KBr}$, which calculates to $\frac{1}{119} \times 10^{-7}$ moles. Multiplying this quantity by Avogadro's constant, $N_A = 6.023 \times 10^{23}$, provides the total number of vacancies: $\frac{6.023 \times 10^{23}}{119} \times 10^{-7} \approx 5.06 \times 10^{14}$. Rounding this value to the nearest integer results in $5$.

The structure of diamond follows a face-centered cubic arrangement where the fundamental lattice positions are supplemented by occupying specific interstitial sites. In an FCC unit cell, the atoms located at the corners and face centers contribute a total of 4 atoms to the lattice. Because an FCC structure inherently creates 8 tetrahedral voids per unit cell, filling half of these positions adds another 4 atoms to the total count. Combining these contributions, the total number of carbon atoms per unit cell is $4 + 4 = 8$.

Crystalline solids are defined by a highly organized, repeating geometric arrangement of constituent particles that persists over long distances, which confirms they possess long-range order. Amorphous solids differ significantly because their constituent particles are held together by varying bond strengths, causing these materials to gradually soften across a broad range of temperatures rather than melting sharply. Due to this lack of rigid structural regularity and their tendency to flow extremely slowly over time, amorphous solids are frequently categorized as pseudo solids.

The density of a crystalline solid is fundamentally tied to its unit cell geometry through the relationship between the mass of the particles within the unit cell and the total volume they occupy. For a cubic close-packed structure, which corresponds to a face-centered cubic arrangement, the unit cell contains 4 particles. The mass of these units is determined by the density of the substance and the dimensions of its unit cell.

The edge length of the unit cell must be converted from nanometers to centimeters, giving 0.4518 × 10⁻⁷ cm. The relationship between density ($\rho$), molar mass ($M$), Avogadro's number ($N_A$), and edge length ($a$) is given by:

$\rho = \frac{Z \times M}{N_A \times a^3}$

Rearranging the formula to isolate the molar mass yields $M = \frac{\rho \times N_A \times a^3}{Z}$. Substituting the values 7.62 g cm⁻³, 6.022 × 10²³ mol⁻¹, and the cubed edge length into the equation results in:

$M = \frac{7.62 \times 6.022 \times 10^{23} \times (0.4518 \times 10^{-7})^3}{4}$

Calculating this expression leads to a value of approximately 105.8 g mol⁻¹, which rounds to the nearest integer of 106.

The occurrence of Frenkel and Schottky defects depends primarily on the relative sizes of the constituent ions, with Frenkel defects favoring crystals with a large disparity in ionic size and Schottky defects favoring those with similar ionic radii. Silver bromide (AgBr) is a unique exception because it possesses an intermediate radius ratio that allows its lattice to accommodate both types of defects simultaneously. While compounds like zinc sulfide (ZnS) are restricted to exhibiting Frenkel defects, substances such as potassium bromide (KBr) and cesium chloride (CsCl) are typically characterized by Schottky defects alone.

The density of a crystal lattice is determined by the mass of the particles packed within its unit cell volume. For a body-centered cubic structure, there are two molecules per unit cell, which allows us to relate the given density, the length of the cell edge, and the molar mass using the formula $d = \frac{Z \times M}{N_A \times a^3}$.

Converting the edge length to centimeters, we have $a = 300\text{ pm} = 3 \times 10^{-8}\text{ cm}$. Substituting the known values of density $d = 6.17\text{ g cm}^{-3}$, edge length $a = 3 \times 10^{-8}\text{ cm}$, and unit cell particles $Z = 2$ into the equation gives $6.17 = \frac{2 \times M}{N_A \times (3 \times 10^{-8})^3}$. Solving this expression for the molar mass $M$ yields a value of approximately $50\text{ g mol}^{-1}$. The number of molecules in a $200\text{ g}$ sample is found by dividing the total mass by the molar mass and multiplying by the Avogadro constant, resulting in $\frac{200}{50} \times N_A$, which simplifies to $4 N_A$.

The density of a crystalline lattice is defined by the formula $\rho = \frac{zM}{N_A a^3}$, where $z$ represents the number of atoms per unit cell, $M$ is the molar mass, $N_A$ is Avogadro's number, and $a$ is the edge length. Substituting the known values of 2.7 x 10^3 kg m^-3 for density, 2.7 x 10^-2 kg mol^-1 for molar mass, 405 x 10^-12 m for edge length, and 6 x 10^23 mol^-1 for Avogadro's number allows for the calculation of $z$. Solving this equation results in a value of 4, indicating that the element crystallizes in a face-centered cubic structure.

In a face-centered cubic structure, the atoms touch along the face diagonal, establishing the geometric relationship $r = \frac{\sqrt{2}a}{4}$ for the atomic radius $r$. Substituting the value 405 x 10^-12 m for the edge length $a$ into this formula leads to a calculated atomic radius of approximately 143 x 10^-12 m.

In a face-centred cubic lattice, octahedral voids are located at the center of each edge and the body center of the unit cell. The nearest octahedral voids, positioned at an edge center and the body center, are separated by a displacement of half the edge length along two perpendicular axes. This geometry forms a right triangle, leading to the calculation of the distance as $\sqrt{ \left( \frac{a}{2} \right)^2 + \left( \frac{a}{2} \right)^2 }$. Evaluating this expression directly yields $\frac{a}{\sqrt{2}}$.

The stoichiometry of an ionic crystal is determined by the requirement of electrical neutrality, where the total positive charge from cations must balance the total negative charge from anions within the crystal lattice. In a cubic close-packed structure containing 4 oxide ions, there are 4 octahedral voids and 8 tetrahedral voids available. Since the cation $M_1$ occupies 50% of the octahedral voids, it provides 2 ions per unit cell, and since $M_2$ occupies 12.5% of the tetrahedral voids, it provides 1 ion per unit cell, resulting in the empirical formula $(M_1)_2(M_2)O_4$. Because the 4 oxide ions contribute a total charge of -8, the cations must collectively provide a charge of +8 to maintain stability. Evaluating the potential oxidation states, only the combination where $M_1$ has a charge of +2 and $M_2$ has a charge of +4 satisfies this requirement, as shown by the following balance:

$2(+2) + 1(+4) = 8$