JEE Chemistry The Solid State MCQs Set 18

Answer: (c) 124 pm

Explanation:
1. In bcc the atoms touch along the body diagonal, so \( 4r=\sqrt{3}\,a \).
2. \( r=\dfrac{\sqrt{3}\times 286}{4}=123.8\approx 124 \) pm.

Teacher's Note:
1. The body diagonal is \( \sqrt{3}a \) and carries two radii plus the whole atom at the centre, i.e. 4r.
2. Do not confuse pm with Å: 124 pm is 1.24 Å.

Answer: (a) 4.5 Å

Explanation:
1. Nearest neighbours in bcc touch along the body diagonal: distance = \( \dfrac{\sqrt{3}}{2}a \).
2. \( =0.866\times5.2=4.5 \) Å.

Teacher's Note:
1. Half the body diagonal runs from a corner to the central atom.
2. The next-nearest neighbours lie at distance a (along the edge).

Answer: (d) \( \dfrac{23}{3} \) amu/Å\(^3\)

Explanation:
1. Simple cubic: Z = 1.
2. Density = \( \dfrac{Z\,M}{a^3}=\dfrac{1\times207}{27}=\dfrac{23}{3} \) amu/Å\(^3\).

Teacher's Note:
1. Using atomic mass in amu and edge in Å gives density directly in amu/Å\(^3\).
2. Multiply by 1.66 to convert to g/cm\(^3\).

Answer: (b) \( d_e < d_{fd} < d_{bd} \)

Explanation:
1. Between successive lattice points: along the edge a, along the face diagonal \( \sqrt{2}a \) (the face has no centre atom in bcc), along the body diagonal \( \sqrt{3}a \) (that is the distance between corner atoms).
2. So \( a<\sqrt{2}a<\sqrt{3}a \).

Teacher's Note:
1. Check that the body-centre atom does not lie on the face diagonal; it lies on the body diagonal.
2. The question compares distances between corner-to-corner points.

Answer: (c) Less than 6 (three)

Explanation:
1. An hcp unit cell has 12 corner atoms (1/6 each = 2), 2 face-centre atoms (1/2 each = 1) and 3 atoms inside the cell.
2. Only the 3 atoms of the middle layer are completely inside the cell, which is fewer than 6.

Teacher's Note:
1. The total of 6 atoms per cell is 2 + 1 + 3.
2. Do not confuse atoms per cell with atoms that lie wholly inside it.

Answer: (b) \( 8\sqrt{\dfrac{2}{3}}\,r \)

Explanation:
1. In ABAB stacking the spacing between adjacent layers is \( h=\sqrt{\dfrac{8}{3}}\,r=2\sqrt{\dfrac{2}{3}}\,r \).
2. Layers 1 and 5 are four spacings apart: \( 4h=8\sqrt{\dfrac{2}{3}}\,r \).

Teacher's Note:
1. Layers 1, 3, 5 are identical (A); layers 2 and 4 are B.
2. The height of the unit cell, c = 2h, equals the distance between layers 1 and 3.

Answer: (a) \( 24\sqrt{2}\,r^3 \)

Explanation:
1. Base area = 6 equilateral triangles of side a = 2r: \( 6\times\dfrac{\sqrt{3}}{4}(2r)^2=6\sqrt{3}\,r^2 \).
2. Height \( c=2\sqrt{\dfrac{8}{3}}\,r \).
3. \( V=6\sqrt{3}r^2\times 2\sqrt{\dfrac{8}{3}}r=24\sqrt{2}\,r^3 \).

Teacher's Note:
1. Packing fraction check: 6 spheres of volume \( \tfrac43\pi r^3 \) over \( 24\sqrt2 r^3 \) gives 0.74.
2. The hexagonal base has 6 triangles, not 3.

Answer: (d) \( 2\sqrt{\dfrac{2}{3}}\,a \)

Explanation:
1. The layer spacing is the height of a regular tetrahedron of edge a: \( h=\sqrt{\dfrac{2}{3}}\,a \).
2. The unit cell contains two layer spacings: \( c=2h=2\sqrt{\dfrac{2}{3}}\,a \).

Teacher's Note:
1. With a = 2r this is \( c=4\sqrt{2/3}\,r \).
2. c/a = 1.633 for ideal hcp.

Answer: (c) 3.62 Å

Explanation:
1. In fcc the atoms touch along the face diagonal: \( \sqrt{2}a=4r \).
2. \( a=2\sqrt{2}\,r=2\times1.414\times1.28=3.62 \) Å.

Teacher's Note:
1. Face diagonal = 4r because it carries two corner radii and one full face-centre atom.
2. Compare bcc: \( a=4r/\sqrt{3} \).

Answer: (a) ABBAABBA

Explanation:
1. In a close-packed lattice a layer is placed over the hollows of the layer below, so two consecutive layers can never be identical.
2. ABBAABBA has two B layers in succession, so the second B sits directly over the first B and the spheres would not rest in hollows.

Teacher's Note:
1. Any sequence with no two adjacent letters equal is a valid close packing (hcp, ccp or mixed).
2. ABAB is hcp and ABCABC is ccp.

Answer: (d) \( \ce{C2A3} \)

Explanation:
1. Octahedral voids = number of anions. For n anions, cations = \( \tfrac{2}{3}n \).
2. C : A = 2/3 : 1 = 2 : 3.

Teacher's Note:
1. Tetrahedral voids are twice the number of anions; octahedral voids equal the number of anions.
2. Al2O3 (corundum) is an example of this type.

Answer: (b) 102.5 pm (minimum) and 333.3 pm (maximum)

Explanation:
1. Octahedral void: \( r_+/r_- \) lies between 0.414 and 0.732. Tetrahedral void: between 0.225 and 0.414.
2. Smallest anion: \( r_-=75/0.732=102.5 \) pm (octahedral limit).
3. Largest anion: \( r_-=75/0.225=333.3 \) pm (tetrahedral limit).

Teacher's Note:
1. A larger radius ratio needs a smaller anion.
2. The combined range of the two voids is 0.225 to 0.732.

Answer: (c) 6

Explanation:
1. \( r_+/r_-=126/216=0.583 \).
2. 0.414 to 0.732 corresponds to an octahedral void, i.e. coordination number 6.

Teacher's Note:
1. Ranges: 0.155-0.225 (3), 0.225-0.414 (4), 0.414-0.732 (6), 0.732-1 (8).
2. Actual AgI is zinc blende (CN 4) at room temperature, but the rule predicts 6.

Answer: (b) Vacant

Explanation:
1. In NaCl the \( \ce{Na+} \) ions occupy all the octahedral voids of the ccp \( \ce{Cl-} \) lattice.
2. The eight tetrahedral voids per cell are empty.

Teacher's Note:
1. Zinc blende uses half the tetrahedral voids; fluorite-type uses all of them.
2. Only octahedral voids are filled in rock salt.

Answer: (a) Interchanging the positions of the positive and negative ions

Explanation:
1. In fluorite (\( \ce{CaF2} \)) the cations form the ccp lattice and the anions fill all tetrahedral voids.
2. In antifluorite (\( \ce{Na2O} \)) the anions form the ccp lattice and the cations fill all tetrahedral voids, so the roles are exchanged.

Teacher's Note:
1. Coordination numbers: fluorite 8:4, antifluorite 4:8.
2. Examples of antifluorite: \( \ce{Li2O, Na2O, K2O} \).

Answer: (d) 4

Explanation:
1. In ZnS the \( \ce{S^{2-}} \) ions form a ccp lattice and \( \ce{Zn^{2+}} \) ions occupy half of the tetrahedral voids.
2. Each Zn is surrounded tetrahedrally by 4 S ions, and each S by 4 Zn ions.

Teacher's Note:
1. Radius ratio of ZnS is about 0.40 (0.225 - 0.414), tetrahedral.
2. Zinc blende has 4 formula units per cell.

Answer: (c) The strontium ions are in a face-centred cubic arrangement

Explanation:
1. In the fluorite structure the cations form a ccp (fcc) lattice and the anions fill all eight tetrahedral voids.
2. So each Sr is surrounded by 8 Cl (cube) and each Cl by 4 Sr (tetrahedron).

Teacher's Note:
1. Coordination numbers are 8 for the cation and 4 for the anion.
2. The two wrong 'centre' statements have the coordination numbers reversed.

Answer: (a) A occupies one-eighth of the tetrahedral holes and B occupies one-half of the octahedral holes

Explanation:
1. Per 4 \( \ce{O^{2-}} \) ions there are 8 tetrahedral and 4 octahedral voids.
2. \( \ce{AB2O4} \) has 1 A and 2 B per 4 O.
3. A: 1 of 8 tetrahedral = 1/8. B: 2 of 4 octahedral = 1/2.

Teacher's Note:
1. Normal spinel example: \( \ce{MgAl2O4} \) (Mg tetrahedral, Al octahedral).
2. Always use 2N tetrahedral and N octahedral voids for N close-packed ions.

Answer: (d) An fcc arrangement with 50% of the tetrahedral holes occupied

Explanation:
1. Diamond has 8 atoms per unit cell: 4 from the fcc lattice and 4 in alternate tetrahedral voids.
2. 4 of the 8 tetrahedral voids are filled, i.e. 50%.

Teacher's Note:
1. Zinc blende is the same structure with Zn and S as different atoms.
2. Each carbon is tetrahedrally bonded, coordination number 4.

Answer: (b) Frenkel defect

Explanation:
1. A Frenkel defect is a vacancy plus an interstitial ion of the same species (dislocation defect).
2. Density is unchanged because no ion leaves the crystal.

Teacher's Note:
1. Frenkel defects occur when the cation is much smaller than the anion (AgCl, AgBr, ZnS).
2. A Schottky defect removes equal numbers of cations and anions from the crystal.

Answer: (c) AgCl shows Frenkel defect and \( \ce{CaF2} \) shows Schottky defect

Explanation:
1. \( \ce{Ag+} \) is small and easily moves to an interstitial site, so AgCl has cation Frenkel defects.
2. \( \ce{CaF2} \) has a high coordination number and large cations, so equal numbers of ions are missing: Schottky defect.

Teacher's Note:
1. Schottky defect lowers density; Frenkel defect does not change it.
2. AgBr shows both defects.