Solid state practice · Mixed difficulty

Option B ($2\mathrm{Z}$) is correct because the number of tetrahedral voids in a closest-packed structure is twice the number of atoms in the unit cell.

Option B ($\mathrm{NaW}{\mathrm{O}_3}$) is correct because it represents the correct ratio of Na, W, and O atoms based on their positions in the crystal lattice.

Option D (The fraction of the total volume occupied by the atoms in a primitive cell is 0.48) is correct because the packing fraction in a primitive cell is 0.52, not 0.48, and diamond has 8 carbon atoms in its unit cell.

Option D (${MnO}$) is correct because manganese(II) oxide (${MnO}$) exhibits antiferromagnetic properties due to the antiparallel alignment of magnetic moments.

Option A (${A}_3 {B}_4$) is correct because removing two A atoms from the corners changes the ratio of A to B in the unit cell.

Option A is correct because NaCl (sodium chloride) is a classic example of an ionic solid, formed by the electrostatic attraction between Na+ and Cl- ions.

Option B ($p -$ type semiconductor) is correct because doping silicon with boron, a group 13 element, creates holes, resulting in a p-type semiconductor.

The correct answer is A (Paramagnetic) because $Fe_3O_4$ (magnetite) exhibits ferrimagnetism, which is a form of paramagnetism, due to the unequal alignment of magnetic moments in its crystal structure.

Option D (23) is correct because a cubic crystal possesses 23 elements of symmetry.

Option A (1) is correct because in an FCC structure, the number of octahedral sites is equal to the number of atoms in the unit cell, which is 4. Since there are 4 spheres, there is 1 octahedral site per sphere (4/4 = 1).

Option C (${r}=0.414 {R}$) is correct because the relationship between the radius of an octahedral void (r) and the radius of the atom in close packing (R) is given by r = 0.414R.

Option A ($1.857\mathrm{ \times }{10^{( - 8)}}cm$) is correct because the radius of the sodium atom in a BCC lattice is calculated using the formula $r = \frac{{\sqrt 3 a}}{4}$, where a is the cell edge.

Option D ($0.414-0.732$) is correct because the radius ratio for an octahedral arrangement of ions falls within this range.

Option C (8) is correct because in the fluorite structure, each $C{a^{2 + }}$ ion is coordinated by 8 $F^-$ ions.

Option D (Diamond) is correct because diamond, being an insulator, has the highest energy gap compared to aluminum, silver, and germanium.

Option B (0.255 to 0.414) is correct because the limiting radius ratio for tetrahedral shape falls within this range.

Option A (12) is correct because in a hexagonal close-packed (hcp) structure, each atom is surrounded by 12 nearest neighbors.

Option A (1) is correct because a simple cubic unit cell contains only 1 atom per unit cell.

Option A (24.16 × 10²³) is correct because in a bcc structure, each unit cell contains 2 atoms, so the total number of atoms is twice the number of unit cells.

Option C (Tetrahedral voids) is correct because in an antifluorite structure, the cations occupy the tetrahedral voids, while the anions form a face-centered cubic lattice.

Option C (Void space in hcp mode of packing is $32 \%$.) is correct because the void space in hcp is 26%, not 32%.

Option C (6) is correct because a radius ratio in the range of 0.414 - 0.732 corresponds to a coordination number of 6, which typically results in an octahedral geometry.

Option A is correct because in NaCl crystal, each Na+ ion is indeed touching 6 Cl- ions, and these Cl- ions do not touch each other due to the radius ratio being greater than 0.414, which is the correct explanation.

Option A (ZnS) is correct because Zinc Sulfide (ZnS) exhibits Frenkel defect due to the significant difference in size between the Zn²⁺ and S²⁻ ions.

Option C (Aluminium) is correct because doping germanium with a group 13 element like aluminum creates a p-type semiconductor.

Option D (5.48 Å) is correct because it is calculated using the bcc structure formula, density, and atomic weight to find the edge length and then the body diagonal.

Option A (Yellow) is correct because ${ZnO}$ when heated, loses oxygen and forms a metal excess defect, leading to the formation of F-centers which absorb light and appear yellow.

Option D (1) is correct because in a simple cubic unit cell, there is one atom at each of the eight corners, and each corner atom contributes 1/8 to the unit cell, resulting in a total of 1 atom per unit cell (8 x 1/8 = 1).

Option A (3.72 pm) is correct because it is calculated using the formula for the shortest interionic distance in a BCC structure, which is $\frac{{\sqrt{3}a}}{{2}}$, where a is the edge length.

Correct Answer + Core Reason: Option A (Frenkel defect) is correct because a Frenkel defect involves an ion leaving its lattice site and occupying an interstitial position.