Solid state practice · Easy

Option C (Molecular) is correct because molecular crystals are held together by weak intermolecular forces, resulting in low melting points and softness.

Option C (693 pm) is correct because the body diagonal of a cube is related to the edge length by $\sqrt{3}a$.

Option C (CsF) is correct because Cs has the largest cation size and F has the smallest anion size among the given options, leading to the highest cation to anion size ratio.

Option D (Body centred cubic) is correct because a radius ratio of 0.93 falls within the range (0.732 - 1.000) that corresponds to a body-centered cubic structure.

Option B (Face centred cubic) is correct because the crystalline structure of NaCl is a face-centered cubic (FCC) lattice.

Option B (Greater then 0.73) is correct because for an AX ionic crystal to exist in a bcc structure, the radius ratio must be greater than 0.73.

Option D (All of these) is correct because ${\mathrm{V}_2}{\mathrm{O}_3}$, $\mathrm{V}{\mathrm{O}_2}$, and $\mathrm{T}{\mathrm{i}_2}{\mathrm{O}_3}$ all exhibit metal-insulator transitions, behaving as semiconductors.

Option C is correct because a matchbox has three unequal sides (length, width, and height) and all angles are 90 degrees, which corresponds to orthorhombic geometry.

Option D ($\frac{{\sqrt 3 a}}{4}$) is correct because it represents half the distance between the body center (octahedral void) and a corner (tetrahedral void) in an FCC lattice.

Option A ($NaCl$) is correct because Schottky defects are commonly observed in highly ionic compounds with a small difference in cation and anion size.

Option D ($\frac{\pi }{6}$) is correct because the packing fraction for a simple cubic cell is calculated as the volume occupied by the atoms divided by the total volume of the unit cell, which simplifies to $\frac{\pi }{6}$.

Option C (Do not undergo clean cleavage when cut with knife) is correct because amorphous solids lack long-range order, resulting in irregular fracture patterns when cut.

Option C is correct because iodine crystals are formed by weak van der Waals forces between $I_2$ molecules, making them molecular solids.

Option B ($\mathrm{Zns}$) is correct because in ZnS, sulfide ions form a face-centered cubic lattice, and zinc ions occupy alternate tetrahedral voids.

Option D (Triclinic) is correct because the triclinic crystal system has unequal axes ($a \neq b \neq c$) and unequal angles ($\alpha \neq \beta \neq \gamma \neq 90^0$).

Option A is correct because in a ccp lattice, A atoms occupy the lattice points, and with the formula $AB_2$, B must occupy all the tetrahedral voids. The coordination number of an atom in a tetrahedral void is 4.

Option B (liquid crystals) is correct because liquid crystals exhibit a turbid liquid phase before clearing upon heating.

Option C (Electrons) is correct because the hardness of metals increases with the number of delocalized electrons involved in metallic bonding.

Option B (Li) is correct because Lithium is a metal and metals are generally good conductors of heat and electricity.

Option C ($32\%$) is correct because the packing efficiency in a BCC lattice is $68\%$, leaving $32\%$ as vacant space.

Option D ($\frac{4}{{\sqrt 2 }}r$) is correct because in a face-centered cubic (FCC) unit cell, the atoms touch along the face diagonal, and the relationship between the edge length (a) and the atomic radius (r) is $a = \frac{4r}{\sqrt{2}}$.

Option A ($HCP\,\,and\,\,CCP$) is correct because $ABAB.....$ packing is characteristic of Hexagonal Close Packing (HCP) and $ABCABC.....$ packing is characteristic of Cubic Close Packing (CCP).

Option A (0.1547) is correct because it represents the limiting radius ratio for a coordination number of 3, where the cation touches three anions that also touch each other.

Option D (Glass) is correct because glass lacks a long-range ordered structure, making it a non-crystalline solid.

Option D (616 pm) is correct because the shortest separation in an FCC lattice is half the face diagonal, which is calculated as $\frac{{\sqrt{2}}}{{2}}a$.

Option C ($241.5\,pm$) is correct because the radius ratio for NaCl structure falls within a specific range, allowing calculation of the anion radius.

Option D ($\frac{\pi }{6}$) is correct because the packing fraction for a simple cubic cell is calculated as the volume occupied by the atoms divided by the total volume of the unit cell, which simplifies to $\frac{\pi }{6}$.

Option B ($B$) is correct because doping silicon with Boron (B), a group 13 element, creates a p-type semiconductor due to the introduction of holes.

Option B ($\mathrm{Fe}{\mathrm{O}_{0.98}}$ has non stoichiometric metal deficiency defect) is incorrect because while FeO can have a metal deficiency defect, the statement implies it *always* does, which isn't true; it depends on the specific composition.

Option B (Electrons in atom) is correct because the free electrons in metals absorb and re-emit light, making them excellent reflectors.