Ideas: 2 Chemistry JEE Solid State (2005

Multiple Choice Questions

1. The correct statement(s) regarding defects in solids is (are) [JEE 2009 P-I]

(a) Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion

(b) Frenkel defect is a dislocation defect

(d) Schottky defects have no effect on the physical properties of solids

Answer: (a, c)

Explanation: Frenkel defect occurs when an ion is dislocated to an interstitial site (b is also correct, but as per options/key provided, a, c). Trapping of an electron in an anion vacancy leads to F-centre formation.

2. The packing efficiency of the two-dimensional square unit cell shown below is: [JEE 2010 P-II]

(a) 39.27 %

(b) 68.02%

(d) 78.54%

Answer: (d)

Explanation: $a = 2\sqrt{2}r$. $\text{P.F.} = \frac{2 \times \pi r^2}{(2\sqrt{2}r)^2} = \frac{2\pi r^2}{8r^2} = \frac{\pi}{4} = 0.7854 = 78.54\%$.

3. The arrangement of ions around $\text{A}^+$ ion in solid AX is given in the figure (not drawn to scale). If the radius of $\text{X}^-$ is 250 pm, the radius of $\text{A}^+$ is: [JEE Adv. 2013 P-I]

(a) 104 pm

(b) 125 pm

(d) 57 pm

Answer: (a)

Explanation: According to the given figure, $\text{A}^+$ is present in the octahedral void of $\text{X}^-$. The limiting radius ratio in octahedral void is related to the radius of sphere as:

$$r_{\text{void}} = 0.414 r_{\text{sphere}}$$

$$r_{\text{A}^+} = 0.414 r_{\text{X}^-} = 0.414 \times 250\text{ pm} = 103.5\text{ pm} \approx 104\text{ pm}$$

4. Sodium metal crystallises in a body centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium atom is approximately: [JEE Main 2015]

(a) 1.86 Å

(b) 3.22 Å

(d) 0.93 Å

Answer: (a)

Explanation: For BCC unit cell:

$$\sqrt{3}a = 4r$$

$$r = \frac{\sqrt{3}}{4} a = \frac{\sqrt{3}}{4} \times 4.29\text{ \AA} = 1.857\text{ \AA} \approx 1.86\text{ \AA}$$

5. If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with m fraction of octahedral holes occupied by aluminium ions and n fraction of tetrahedral holes occupied by magnesium ions, m and n, respectively, are: [JEE Adv. 2015 P-I]

(a) $\frac{1}{2}, \frac{1}{8}$

(b) $1, \frac{1}{4}$

(d) $\frac{1}{4}, \frac{1}{8}$

Answer: (a)

Explanation: In ccp lattice:

Number of $\text{O}$ atoms $\longrightarrow 4$

Number of Octahedral voids $\longrightarrow 4$

Number of tetrahedral voids $\longrightarrow 8$

Number of $\text{Al}^{3+} = 4 \times m$

Number of $\text{Mg}^{2+} = 8 \times n$

Due to charge neutrality: $4(-2) + 4m(+3) + 8n(+2) = 0 \implies m = \frac{1}{2}\text{ and } n = \frac{1}{8}$.

6. The correct statement(s) for cubic close packed (ccp) three dimensional structure is (are) [JEE Adv. 2016 P-II]

(a) The number of the nearest neighbours of an atom present in the topmost layer is 12

(b) The efficiency of atom packing is 74%

(d) The unit cell edge length is $2\sqrt{2}$ times the radius of the atom

Answer: (b, c, d)

Explanation: (a) For any atom in top most layer, coordination number is not 12 since there is no layer above top most layer, (b) Fact, (c) Fact, (d) $\sqrt{2}a = 4R \implies a = 2\sqrt{2}R$.

7. A metal crystallises in a face centered cubic structure. If the edge length of its unit cell is 'a', the closest approach between two atoms in metallic crystal will be [JEE Main 2017]

(a) $\frac{a}{\sqrt{2}}$

(b) $\sqrt{2} a$

(d) $2 a$

Answer: (a)

Explanation: In FCC unit cell atoms are contact along face diagonal:

$$\sqrt{2}a = 4R$$

$$\text{Closest distance } (2R) = \frac{\sqrt{2}a}{2} = \frac{a}{\sqrt{2}}$$

8. Which type 'defect' has the presence of cations in the interstitial sites? [JEE Main 2018]

(a) Metal deficiency defect

(b) Schottky defect

(d) Frenkel defect

Answer: (d)

Explanation: In Frenkel defect, cation is dislocated from its normal lattice site to an interstitial site.

9. Element 'B' forms ccp structure and 'A' occupies half of the octahedral voids, while oxygen atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is: [JEE Main 2019]

(a) $\text{A}_2\text{BO}_4$

(b) $\text{A}_2\text{B}_2\text{O}$

(d) $\text{AB}_2\text{O}_4$

Answer: (d)

Explanation: Number of B atoms $= Z = 4$. Number of A atoms $= \frac{1}{2} \times 4 = 2$. Number of O atoms $= 8$.

Formula: $\text{A}_2\text{B}_4\text{O}_8 \equiv \text{AB}_2\text{O}_4$.

10. The statement that is incorrect about the interstitial compounds is: [JEE Main 2019]

(a) They have high melting points

(b) They are chemically reactive

(d) They are very hard

Answer: (b)

Explanation: Generally interstitial compounds are chemically inert.

11. Consider the bcc unit cells of the solids 1 and 2 with the position of atoms as shown below. The radius of atom B is twice that of atom A. The unit cell edge length is 50% more in solid 2 than in 1. What is the approximate packing efficiency in solid 2? [JEE Main 2019]

(a) 45%

(b) 65%

(d) 75%

Answer: (c)

Explanation:

$$\text{P.F.} = \frac{z_{\text{eff, A}} \times \frac{4}{3}\pi r_{\text{A}}^3 + z_{\text{eff, B}} \times \frac{4}{3}\pi r_{\text{B}}^3}{a^3}$$

$$2(r_{\text{A}} + r_{\text{B}}) = \sqrt{3}a \implies 2(r_{\text{A}} + 2r_{\text{A}}) = \sqrt{3}a \implies 6r_{\text{A}} = \sqrt{3}a$$

$$\text{P.F.} = \frac{1 \times \frac{4}{3}\pi r_{\text{A}}^3 + 1 \times \frac{4}{3}\pi (8r_{\text{A}}^3)}{\left(\frac{6r_{\text{A}}}{\sqrt{3}}\right)^3} = \frac{12\pi r_{\text{A}}^3}{\frac{216}{3\sqrt{3}} r_{\text{A}}^3} = \frac{36\sqrt{3}\pi}{216} = \frac{\sqrt{3}\pi}{6} \approx 0.90$$

$$\text{Packing efficiency} \approx 90\%$$

12. At 100°C, copper (Cu) has FCC unit cell structure with cell edge length of x Å. What is the approximate density of Cu (in $\text{g cm}^{-3}$) at this temperature? [Atomic Mass of Cu = 63.55u] [JEE Main 2019]

(a) $\frac{105}{x^3}$

(b) $\frac{211}{x^3}$

(d) $\frac{422}{x^3}$

Answer: (d)

Explanation: FCC unit cell $Z = 4$, $N_A = 6 \times 10^{23}$

$$d = \frac{63.55 \times 4}{6 \times 10^{23} \times (x \times 10^{-8})^3} = \frac{254.2}{6 \times 10^{-1} x^3} = \frac{423.33}{x^3} \approx \frac{422}{x^3}$$

13. Which primitive unit cell has unequal edge lengths ($a \neq b \neq c$) and all axial angles different from 90°? [JEE Main 2019]

(a) Tetragonal

(b) Hexagonal

(d) Triclinic

Answer: (d)

Explanation: In triclinic unit cell $a \neq b \neq c$ and $\alpha \neq \beta \neq \gamma \neq 90^\circ$.

14. A compound of formula $\text{A}_2\text{B}_3$ has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms: [JEE Main 2019]

(a) hcp lattice-A, $\frac{2}{3}$ tetrahedral voids-B

(b) hcp lattice-B, $\frac{1}{3}$ tetrahedral voids-A

(d) hcp lattice-A, $\frac{1}{3}$ tetrahedral voids-B

Answer: (b)

Explanation: $\text{A}_2\text{B}_3$ has HCP lattice. If B forms HCP lattice (6 atoms), then tetrahedral voids $= 12$. To get formula $\text{A}_2\text{B}_3$, number of A atoms must be 4. Fraction of tetrahedral voids occupied by A $= \frac{4}{12} = \frac{1}{3}$.

15. A solid having density of $9 \times 10^3\text{ kg m}^{-3}$ forms face centred cubic crystals of edge length $200\sqrt{2}\text{ pm}$. What is the molar mass of the solid? [JEE Main 2019] (Avogadro constant $\cong 6 \times 10^{23}\text{ mol}^{-1}$, $\pi \cong 3$)

(a) $0.0216\text{ kg mol}^{-1}$

(b) $0.0305\text{ kg mol}^{-1}$

(d) $0.0432\text{ kg mol}^{-1}$

Answer: (b)

Explanation: $d = \frac{Z \times M}{N_A \times a^3} \implies M = \frac{d \times N_A \times a^3}{Z}$

$$M = \frac{9000 \times 6 \times 10^{23} \times (200\sqrt{2} \times 10^{-12})^3}{4} = 0.0305\text{ kg mol}^{-1}$$

16. The radius of the largest sphere which fits properly at the centre of the edge of body centred cubic unit cell is: (Edge length is represented by 'a'): [JEE Main 2019]

(a) 0.134 a

(b) 0.027 a

(d) 0.047 a

Answer: (c)

Explanation: $a = 2(R + r) \implies R + r = \frac{a}{2}$. Also for BCC, $\sqrt{3}a = 4R \implies R = \frac{\sqrt{3}}{4}a$.

$$r = \frac{a}{2} - R = \frac{a}{2} - \frac{\sqrt{3}}{4}a = a\left(0.5 - 0.433\right) = 0.067a$$

Matrix Match/Column Match

17. Match the crystal system/unit cells mentioned in Column I with their characteristic features mentioned in Column II. [IIT JEE 2007 P-II]

Column I	Column II
(A) Simple cubic and face-centred cubic	1. have these cell parameters $a = b = c$ and $\alpha = \beta = \gamma$
(B) Cubic and rhombohedral	2. are two crystal systems
(C) Cubic and tetragonal	3. have only two crystallography angles of 90°
(D) Hexagonal and monoclinic	4. belong to same crystal system

(a) A$\rightarrow$1,4; B$\rightarrow$ 1,2; C$\rightarrow$2; D$\rightarrow$2,3

(b) A$\rightarrow$1,2; B$\rightarrow$ 1,3; C$\rightarrow$4; D$\rightarrow$2,4

(d) A$\rightarrow$1,3; B$\rightarrow$ 1,2; C$\rightarrow$4; D$\rightarrow$2,3

Answer: (a)

Explanation: Simple cubic and face-centred cubic both belong to the cubic system ($a=b=c, \alpha=\beta=\gamma=90^\circ$). Cubic and rhombohedral both have $a=b=c$ and $\alpha=\beta=\gamma$, and are two crystal systems.

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Paragraph for Question Nos. 18 to 20

In hexagonal systems of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them. A space-filling model of this structure, called hexagonal close-packed (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of these three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be 'r'. [IIT JEE 2008 P-II]

18. The number of atoms on this HCP unit cell is:

(a) 4

(b) 6

(d) 17

Answer: (b)

Explanation: Total effective number of atoms in HCP unit cell $= 12 \times \frac{1}{6} + 2 \times \frac{1}{2} + 3 = 2 + 1 + 3 = 6$.

19. The volume of this HCP unit cell is:

(a) $24\sqrt{2}r^3$

(b) $16\sqrt{2}r^3$

(d) $\frac{64r^3}{3\sqrt{3}}$

Answer: (a)

Explanation: $\text{Height} = 4r\sqrt{\frac{2}{3}}$, $\text{Base Area} = 6 \times \frac{\sqrt{3}}{4}(2r)^2 = 6\sqrt{3}r^2$. $\text{Volume} = \text{height} \times \text{base area} = 24\sqrt{2}r^3$.

20. The empty space in this HCP unit cell is:

(a) 74%

(b) 47.6%

(d) 26%

Answer: (d)

Explanation: Packing fraction $= 74\%$; Empty space $= 100\% - 74\% = 26\%$.

Integer and Subjective

21. An element crystallises in FCC lattice have edge length 400 pm. Calculate the maximum diameter of atom which can be placed in interstitial site without distorting the structure in pm. [JEE 2005]

Answer: 117

Explanation: For maximum diameter of atom in interstitial site, octahedral void is considered:

$$\frac{r_{\text{void}}}{R_{\text{host}}} = 0.414$$

For FCC, $4R = \sqrt{2}a \implies R = \frac{\sqrt{2} \times 400}{4} = 100\sqrt{2}\text{ pm} = 141.4\text{ pm}$.

$$r_{\text{void}} = 0.414 \times 141.4\text{ pm} = 58.54\text{ pm}$$

$$\text{Diameter} = 2 \times 58.54\text{ pm} \approx 117\text{ pm}$$

22. The edge length of unit cell of a metal having molecular weight 75 g/mol is 5 Å which crystallises in cubic lattice. If the density is 2 g/cc then find the radius of metal atom. Give the answer in pm [JEE 2006]

Answer: 216

Explanation: $d = \frac{Z \times M}{N_A \times a^3} \implies 2 = \frac{Z \times 75}{6.023 \times 10^{23} \times (5 \times 10^{-8})^3} \implies Z = 2$ (BCC lattice).

For BCC lattice, $4r = \sqrt{3}a \implies r = \frac{\sqrt{3}}{4} \times 500\text{ pm} \approx 216.5\text{ pm}$.

23. The number of hexagonal faces that are present in a truncated octahedron is: [JEE 2011 P-II]

Answer: 8

Explanation: A truncated octahedron has 14 total faces consisting of 8 hexagonal faces and 6 square faces.

24. A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of 400 pm. If the density of the substance in the crystal is $8\text{ g cm}^{-3}$, then the number of atoms present in 256 g of the crystal is $N \times 10^{24}$. The value of N is: [JEE Adv. 2017 P-I]

Answer: 2

Explanation: Volume of crystal $= \frac{256\text{ g}}{8\text{ g cm}^{-3}} = 32\text{ cm}^3$.

Number of unit cells $= \frac{32}{(400 \times 10^{-10})^3} = \frac{32}{64 \times 10^{-24}} = 0.5 \times 10^{24}$.

In FCC, 4 atoms per unit cell $\implies \text{Number of atoms} = 4 \times 0.5 \times 10^{24} = 2 \times 10^{24} \implies N = 2$.

25. Consider an ionic solid MX with NaCl structure. Construct a new structure (Z) whose unit cell is constructed from the unit cell of MX following the sequential instructions given below. Neglect the charge balance. [JEE Adv. 2018 P-I]

(i) Remove all the anions (X) except the central one

(ii) Replace all the face centered cations (M) by anions (X)

(iii) Remove all the corner cations (M)

(iv) Replace the central anion (X) with cation (M)

The value of $\left(\frac{\text{Number of anions}}{\text{Number of cations}}\right)$ in Z is ______.

Answer: 3

Explanation: In original NaCl structure, $\text{M}^+$ occupies FCC lattice positions (corners + face centers = 4) and $\text{X}^-$ occupies octahedral voids (edge centers + body center = 4).

After steps (i) to (iv):

Final number of cations in a unit cell $= 1$

Final number of anions in a unit cell $= 3$

$$\text{Ratio} = \frac{\text{Number of anions}}{\text{Number of cations}} = \frac{3}{1} = 3.00$$