Solid state practice · Medium

Option C (The internuclear distance ${r_\mathrm{c}} + {r_\mathrm{a}}$ is greater for $\mathrm{RbBr}$ than for $\mathrm{NaF}$) is correct because a smaller internuclear distance in NaF leads to stronger electrostatic forces and thus a higher melting point.

Option B ($63 \mathrm{~amu}$) is correct because the molecular mass can be calculated using the density, edge length, Avogadro's number, and the number of atoms per FCC unit cell.

Option D (Octahedral) is correct because an ionic radius ratio of 0.52 falls within the range (0.414 - 0.732) that corresponds to an octahedral arrangement.

Option A ($\frac{a}{2}\mathrm{cm}$) is correct because in a $NaCl$ type structure, the distance between nearest cation and anion is half of the cell edge.

Option D (Orthorhombic) is correct because the orthorhombic crystal system has the maximum number of Bravais lattices, which is 4.

Option A is correct because amorphous solids like $B_2O_3$ resemble liquids in their disordered structure, and this disorder is due to the random arrangement of B-O bonds, making it non-crystalline.

Option C ($3.72A \circ$) is correct because it's the calculated shortest interionic distance in a BCC structure.

Option B (Cubic) is correct because cubic voids are formed by the arrangement of eight spheres, making them the largest among the given options.

Option C (4) is correct because in a diamond structure, each carbon atom contributes a specific fraction to the unit cell, resulting in a total of 4 effective carbon atoms.

Option D ($1 \times {10^{20}}$) is correct because it is calculated by dividing the total mass by the mass of one unit cell and accounting for unit conversions.

Option A is correct because quartz is a crystalline form of silica ($SiO_2$).

Option A ($27\,g/mol$) is correct because the molar mass can be calculated from the density, edge length, and the number of atoms per unit cell for an FCC lattice.

Option D (6, 12, 8) is correct because the coordination numbers for simple cubic, FCC, and BCC lattices are 6, 12, and 8, respectively.

Option C is correct because Bragg's Law (nλ = 2dsinθ) is used to calculate the interplanar spacing (d), and the calculation yields approximately 0.905 Å.

Option C (155.9 pm) is correct because it's the calculated radius for a BCC structure with the given edge length.

Option C ((A) is correct but (R) is not correct) is correct because FCC is more closely packed than BCC, but the packing fraction of FCC is not double that of BCC.

Option C (Silicon) is correct because amorphous silicon is commonly used as a photovoltaic material to convert sunlight into electricity.

Option B (Between 0.73 and 0.41) is correct. The correct range is between 0.41 and 0.73 because a coordination number of 6 corresponds to an octahedral geometry, and the radius ratio for octahedral geometry falls within that range.

Option D (All are correct answer) is correct because Zinc is added to the Ag-Pb alloy, then cooled, forming immiscible layers X and Y, where X contains Zn and Ag, and Y contains Pb and Ag with less silver than X.

Option D ($100 \%$) is correct because in a simple cubic unit cell, atoms are present only at the corners, and the edge is fully occupied by the atoms.

Option A (The shift of a positive ion from its normal lattice site to an interstitial site) is correct because this is the defining characteristic of a Frenkel defect.

Option A ($ABC$) is correct because if C forms CCP, A occupies all octahedral voids, and B occupies half the tetrahedral voids, the ratio of A:B:C is 1:1:1.