EAMCET PROBLEMS WITH SOLUTIONS FROM SOLIDSTATE CHAPTER PART-3
Example 36. If silver iodide crystallizes in a zinc blende structure with I- ions forming the lattice then calculate fraction of the tetrahedral voids occupied by Ag+ ions.
Solution: In AgI, if there are nI- ions, there will be nAg+ ions. As I- ions form the lattice, number of tetrahedral voids = 2n. As there are nAg+ ions to occupy these voids, therefore fraction of tetrahedral voids occupied by Ag+ ions = n/2n = ½ = 50%.
Example 37. The co-ordination number of a metal crystallising in hcp structure is
(A) 12 (B) 10
(C) 8 (D) 6
Solution: (A)
Example 38. Out of NaCl and CsCl, which one is more stable and why?
Solution: CsCl is more stable than NaCl. This is because higher the coordination number, greater are the forces of attraction between the cations and the anions in the close-packed arrangement. As CsCl has co-ordination number of 8 : 8 while NaCl has a coordination number of 6 : 6, therefore CsCl is more stable.
Example 39. Compare the structure of zinc blend (ZnS) with that of diamond.
Solution: The structure of zinc blend (ZnS) is similar to that of diamond. In diamond each carbon atom is linked to four other atoms tetrahedrally. Similarly, in zinc blende, each Zn+ ion is surrounded by 4s2- ions tetrahedrally and each S2- ion is surrounded by 4Zn2+ ions tetrahedrally. Thus if all Zn2+ ions and S2- ions in zinc blende are replaced by carbon atoms, if gives rise to structure of diamond.
Solution: (B)
Example 40. Compute the percentage void space per unit volume of unit cell in zinc-fluoride structure.
Solution: Since anions occupy fcc positions and half of the tetrahedral holes are occupied by cations.
Since there are four anions and 8 tetrahedral holes per unit cell, the fraction of volume occupied by spheres/unit volume of the unit cell is
= 4 × (4/3 πra3) + 1/2 × 8 × (4/3πrc3) = 16√2πra3= π/3√2 {1 + (rc/ra)3}
for tetrahedral holes,
rc / ra = 0.225 = π/3√2 {1 + (0.225)3} = 0.7496
∴ Void volume = 1 – 0.7496 = 0.2504/unit volume of unit cell
% void space = 25.04%
Example 41. Select the correct statements:-
(A) For CsCl unit cell (edge-length = a), rc + ra = 3√2 a
(B) For NaCl unit cell (edge-length =), rc + ra = l/2
(C) The void space in a b.c.c. unit cell is 0.68
(D) The void space % in a face-centered unit cell is 26%
Solution: In bcc structure are r+ + r– = √3 / 2 a
Hence, for CsCl, rC + ra = √3 / 2 a
∴ (A)
Since, NaCl crystallises in fcc structure
∴ 2rC + 2ra = edge length of the unit cell
Hence, rC + ra = l/2
∴ (B)
Since packing fraction of a bcc unit cell is 0.68
∴ void space = 1–0.68 = 0.32
∴ (C)
In fcc unit cell PF = 74%
∴ VF = 100–74 = 26 \(D)
Example 42. Addition of CdCl2 to AgCl yields solid solutions, where the divalent cations Cd+2occupy the Ag+ sites which one of the following statements is true
(A) The no. of cationic vacancies is one half of the no. of that of divalent ions added.
(B) The no. of cationic vacancies is one half of the no. of that of divalent ions added.
(C) The no. of anionic vacancies is equal in no. to that of divalent ions added.
(D) No cationic or anionic vacancies are produced.
Solution: To maintain the electrical neutrality, the no. of cationic vacancies is equal to the no. of divalent ions added.
Example 43. CsCl has bcc structure with at the centre and ion at each corner. If rCs+ = 1.69Å and rCl– = 1.81Å, what is the edge length “a” of the cube?
(A) 3.50Å (B) 3.80Å
(C) 4.04Å (D) 4.50Å
Solution: (C) Assuming the closest approach between and ions, the internuclear separation is one-half of the cubic diagonal i.e.
.69 + 1.81 = 3.50 = a√3 / 2
∴ a = 2 × 3.5 / √3 = 4.04Å
Example 44. Out of SiO2(s), Si(s), NaCl(s) and Br2(l) which is the best electrical conductor?
Solution: Si(s) because only this is a semi-conductor, while others SiO2(s), NaCl(s) and Br2(l) are insulators.
Example 45. Explain:
(a) The basis of similarities and difference between metallic and ionic crystals.
(b) Unit cell is not simply a cube of 4Na+ ions and 4Cl- ions.
(c) Can a cube consisting of Na+ and Cl- ions at alternate corners serve as satisfactory unit cell for the sodium chloride lattice?
(d) Ionic solids are hard and brittle.
Solution: (a) Similarities:
(i) Both involve electrostatic forces of attraction.
(ii) Both are non-directional.
Differences: Ionic bond is a strong bond due to electrostatic forces of attraction while metallic bond may be weak or strong depending upon the kernels.
(b) Unit cell of NaCl has fcc arrangement of Cl- ions and Na+ ions are present at the edge centres and one at the body-cnetre.
Thus there are 14Cl- ions and 13Na+ ions in the unit cell. However their net contribution towards the unit cell is 4Na+ and 4Cl- ions.
(c) Yes because its repetition in different directions produces the complete space lattice.
(d) Ionic solids are hard because there are strong electrostatic forces of attraction. However they are brittle because the bond is non-directional.
Example 46. Analysis shows that nickel oxide has the formula Ni0.98O. What fractions of the nickel exist as Ni2+ and Ni3+ ions?
Solution: 98 Ni atoms are associated with 100 O-atom. Out of 98 Ni atoms, suppose Ni present as Ni2+ = x
Then Ni present as Ni3+ = 98 – x
Total charge on xNi2+ and (98 – x)Ni3+ should be equal to charge n 100O2-. Hence
x × 2 + (98 – x) × 3 = 100 × 2
or 2x + 294 – 3x = 200
or x = 94
∴ Fraction of Ni present as Ni2+ = 94 / 98 × 100 = 96%
∴ Fraction of Ni present as Ni3+ = 4 / 98 × 100 = 4%
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