Solid state practice · Easy

Option D (Glass) is correct because glass is an amorphous solid, lacking long-range order, and is therefore non-crystalline.

Option B (Super conductor) is correct because superconductors exhibit zero electrical resistance below a critical temperature, approaching 0 K.

Option A (1) is correct because in a cubic close-packed (CCP) structure, the number of octahedral voids is equal to the number of atoms in the unit cell.

Option D (MnO) is correct because MnO exhibits antiferromagnetic behavior due to the alignment of magnetic moments in opposite directions.

Option A (0.414) is correct because the ratio of cation to anion in an octahedral close packing is 0.414.

Option B (Quartz) is correct because quartz exhibits piezoelectricity, generating an electric charge under mechanical stress.

Option B (GaAs - Insulator) is the wrong match because GaAs is a semiconductor, not an insulator.

Option C (Molecular solid) is correct because iodine crystals are composed of I2 molecules held together by weak van der Waals forces.

Option C (Fcc and bcc) is correct because interchanging $\mathrm{N}{\mathrm{a}^ + }$ and $\mathrm{C}{\mathrm{l}^ - }$ in NaCl results in $\mathrm{N}{\mathrm{a}^ + }$ forming an fcc lattice and $\mathrm{C}{\mathrm{l}^ - }$ forming a bcc lattice.

Option B ($50\%$) is correct because in diamond, half of the tetrahedral voids are occupied by carbon atoms.

Option D is correct because NO, Fe3+, and Fe2+ all possess unpaired electrons, making them paramagnetic.

Option C (Metallic crystals) is correct because metallic crystals are known for their good electrical and thermal conductivity due to the presence of free electrons.

Option B (FCC) is correct because NaCl has a face-centered cubic arrangement of ions.

Option C (Barium titanate) is correct because barium titanate ($BaTiO_3$) is a well-known ferroelectric material.

Option D (In Schottky defect equal number of cations and anions are missing from their lattice sites) is correct because Schottky defects involve the absence of an equal number of cations and anions to maintain electrical neutrality.

Option C (Ge) is correct because Germanium (Ge) is a common semiconductor material.

Option B (Glass) is correct because glass is an amorphous solid that lacks long-range order and exhibits properties of a supercooled liquid, hence it's considered a pseudo solid.

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 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 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.

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

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 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 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 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.