Solid State

Number of particles $=5 \mathrm{~N}_{\mathrm{A}}$

Number of THV $=2 \times$ number of particles, for close packing

$\begin{aligned} & =2 \times 5 \mathrm{~N}_{\mathrm{A}} \\ & =10 \mathrm{~N}_{\mathrm{A}} \\ & =10 \times 6.023 \times 10^{23} \\ & =6.023 \times 10^{24} \end{aligned}$

For CCP (FCC)

$\begin{aligned} & 4 \mathrm{r}=\sqrt{2} \mathrm{a} \\ & \mathrm{a}=\frac{4 \mathrm{r}}{\sqrt{2}} \\ & \mathrm{a}=2 \sqrt{2} \mathrm{r} \end{aligned}$

$\mathrm{x+y=5}$

$\to$ Edge centred octahedral void is shared between four unit cells

$\to$ Per unit cell contribution is 1/4

$\bullet$ Frenkel defect : It is shown by ionic solids. The smaller ion (usually cation) is dislocated from its normal site to an interstitial site. It does not change the density of the solid.

$\bullet$ Schottky defect : It is a vacancy defect in ionic solids. It decreases the density of substance.

$\bullet$ Vacancy defect : When some of the lattice sites are vacant, the crystal is said to have vacancy defect. This results in decrease in density of the substance.

$\bullet$ Interstitial defect : When some constituent particles occupy an interstitial site, the crystal is said to have interstitial defect. This defect increases the density of the substance.

$\bullet$ Reason statement is true but not correct explanation as it is defining Frenkel defect in which ion does not leave crystal.

Diamond :

$\bullet$ sp³ hybridised carbon atom

$\bullet$ Covalent solid

$\bullet$ 3-D structure

$\bullet$ Cannot be used as dry lubricant

Graphite :

$\bullet$ sp² hybridised carbon atom

$\bullet$ Covalent solid

$\bullet$ 3-D structure

$\bullet$ Used as dry lubricant

$d = {{ZM} \over {{N_A}{{(a)}^3}}}$

Z = 4(FCC), d = 8.92 g cm^$-$3, N_A = 6.023 $\times$ 10²³, a = 3.608 $\times$ 10^$-$8 cm

$M = {{d{N_A}{{(a)}^3}} \over Z}$

$ = {{8.92 \times 6.023 \times {{10}^{23}} \times {{(3.608 \times {{10}^{ - 8}})}^3}} \over 4}$

$ = {{8.92 \times 6.023 \times {{10}^{23}} \times 46.97 \times {{10}^{ - 24}}} \over 4} = {{2523.47 \times {{10}^{ - 1}}} \over 4}$

$ = 630.8 \times {10^{ - 1}} = 63.08 \simeq 63.1\,u$

To determine the formula of the compound formed by elements X and Y based on their positions in a cubic structure, we need to count the number of atoms per unit cell for each type of atom and then simplify the ratio.

For element X:

• Corners of a cube: Each corner atom is shared by 8 adjacent cubes, so each corner contributes 1/8 of an atom to the cube. Since there are 8 corners in a cube, the total contribution from corners is $8 \times \frac{1}{8} = 1$ atom of X per unit cell.

For element Y:

• Face-centers of a cube: Each face-centered atom is shared by 2 adjacent cubes, so each face-center contributes 1/2 of an atom to the cube. Since there are 6 faces on a cube, the total contribution from face-centers is $6 \times \frac{1}{2} = 3$ atoms of Y per unit cell.

Therefore, within one unit cell, we have 1 atom of X and 3 atoms of Y, giving us the simplest ratio of $X:Y = 1:3$. Hence, the formula of the compound is $XY_3$.

So, the correct answer is:

Option A: XY₃

In a hexagonal close-packed (hcp) crystal structure, there are 2 tetrahedral voids for every atom present in the structure.

Given :

Number of moles of hcp crystal structure = 0.5 mole

Avogadro's number ($N_A$) = $6.02 \times 10^{23}$ atoms/mol

First, calculate the total number of atoms in 0.5 mole of hcp crystal structure :

$ 0.5 \text{ mole} \times 6.02 \times 10^{23} \text{ atoms/mole} = 3.01 \times 10^{23} \text{ atoms} $

Since there are 2 tetrahedral voids per atom, the total number of tetrahedral voids can be calculated as :

$ 2 \times 3.01 \times 10^{23} = 6.02 \times 10^{23} $

Thus, the number of tetrahedral voids in 0.5 mole of hcp crystal structure is :

$6.02 \times 10^{23}$

Option D is the correct answer.

Packing efficiency for:

Hexagonal close packing $(\mathrm{hcp})=74 \%$

Face-centered cubic close packing $(\mathrm{fcc})=74 \%$

Body centered cubic close packing (bcc) $=68 \%$

and simple cubic $(\mathrm{sc})=52 \%$

Thus, correct order of packing efficiency is

$\mathrm{hcp}=\mathrm{fcc}>\mathrm{bcc}>\mathrm{sc}$

Hence, option (a) is the correct.

Zinc oxide which is white in colour on heating loses oxygen and turns yellow.

$\mathrm{ZnO} \xrightarrow{\text { Heating }} \mathrm{Zn}^{2+}+\frac{1}{2} \mathrm{O}_2+2 e^{-}$

It is due to metal excess defect in which excess ions (cation and electron) occupy interstitial site. A negative ion missing from its lattice site leaving a hole which is occupied by an extra electron to maintain electrical neutrality. Thus, anionic site is called $F$-centre.

Hence, both assertion and reason are correct but reason is not the correct explanation of assertion.

The network solid among the options provided is Option C: Diamond.

A network solid is a chemical compound wherein the atoms are bonded covalently in a continuous network extending throughout the material. In such solids, there are no discrete molecules; the bonding continues throughout the crystal. This results in very hard substances with high melting points and poor conductivity.

Explanation:

Diamond: In diamond, each carbon atom is covalently bonded to four other carbon atoms in a tetrahedral structure, forming a three-dimensional network. This extensive covalent bonding throughout the crystal provides diamond its hardness and high melting point.

$\mathrm{SO}_2$ (solid): Sulfur dioxide forms discrete molecules held together by van der Waals forces, not a network solid structure.

$\mathrm{I}_2$: Iodine consists of discrete $\mathrm{I}_2$ molecules held together by van der Waals forces, not covalent bonds in a network solid structure.

$\mathrm{H}_2 \mathrm{O}$ (ice): Ice is a molecular solid where water molecules are held together by hydrogen bonds, not covalently bonded in a network fashion.

Therefore, the continuous tetrahedral carbon arrangement in diamond is a perfect example of a network solid.

Given,

$\rho(\text { density })=8.55 \mathrm{~g} \mathrm{~cm}^{-3}$

Atomic mass of niobium $=93 \mathrm{u}$

As we know that,

$\text { Density }(\rho)=\frac{M Z}{a^3 N_A}$ ..... (i)

where, $M=$ atomic mass, $Z=$ number of unit cell

$a=$ edge length, $N_A=$ Avogadro's number

For bcc, $Z=2$

Putting the values in equation (i), we get

$8.55=\frac{93 \times 2}{a^3 \times 6.02 \times 10^{23}}$

$\begin{aligned} \text{or} \quad a^3 & =3.61 \times 10^{-23} \mathrm{~cm}^3 \\ \text{or} \quad a & =3.306 \times 10^{-8} \mathrm{~cm} \end{aligned}$

For body centred cube,

$\text { radius }=\frac{\sqrt{3}}{4} a$

or, $r_{\text {bcc }}=\frac{\sqrt{3}}{4} \times 3.306 \times 10^{-8}$

or, $r_{\mathrm{bcc}} =1.43 \times 10^{-8} \mathrm{~cm}=1.43 \times 10^{-10} \mathrm{~m}=143~\mathrm{pm}$

In a unit cell,

$\mathrm{W}$ atoms at the corner $=\frac{1}{8} \times 8=1$

$\mathrm{O}$ atoms at the centre of edges $=\frac{1}{4} \times 12=3$

$\mathrm{Na}$ atoms at the centre of the cube $=1$

$\therefore \quad \mathrm{W}: \mathrm{O}: \mathrm{Na}=1: 3: 1$

Hence, formula of compound is $\mathrm{NaWO}_3$.