chemistry : class12 ZnS Crystal Structure : Ionic solids : Solid State MCQ Questions and Answers
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ZnS Crystal Structure
MCQ questions and answers.
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MCQ Questions
1. In the zinc blende form of ZnS, which description of the crystal arrangement is correct?
A.
S2− ions form a face-centred cubic lattice, and Zn2+ ions occupy all octahedral holes
B.
S2− ions form a face-centred cubic lattice, and Zn2+ ions occupy half of the tetrahedral holes
C.
Zn2+ ions form a body-centred cubic lattice, and S2− ions occupy all cubic holes
D.
Zn2+ and S2− ions form separate primitive cubic lattices interpenetrating each other
Correct Answer:
S2− ions form a face-centred cubic lattice, and Zn2+ ions occupy half of the tetrahedral holes
Explanation:
Zinc blende is based on a cubic close-packed, or face-centred cubic, arrangement of S2− ions. Zn2+ ions occupy half of the available tetrahedral voids, producing the 1:1 ZnS stoichiometry.
2. How many Zn2+ ions and S2− ions are effectively contained in one unit cell of cubic zinc blende ZnS?
A.
Zn2+: 4; S2−: 4
B.
Zn2+: 2; S2−: 4
C.
Zn2+: 4; S2−: 8
D.
Zn2+: 8; S2−: 4
Correct Answer:
Zn2+: 4; S2−: 4
Explanation:
The S2− ions occupy the face-centred cubic positions, contributing 4 ions per unit cell. Half of the 8 tetrahedral holes are occupied by Zn2+, also giving 4 ions, so the unit cell contains four ZnS formula units.
3. What is the coordination number of each Zn2+ ion and each S2− ion in the zinc blende structure?
A.
Zn2+: 4, S2−: 6
B.
Zn2+: 6, S2−: 4
C.
Zn2+: 4, S2−: 4
D.
Zn2+: 6, S2−: 6
Correct Answer:
Zn2+: 4, S2−: 4
Explanation:
Each Zn2+ ion is surrounded tetrahedrally by four S2− ions, and each S2− ion is tetrahedrally surrounded by four Zn2+ ions. Therefore, the coordination is 4:4.
4. A cubic zinc blende unit cell has lattice parameter a. The nearest-neighbour Zn2+–S2− distance is
A.
a
B.
a√2/2
C.
a√3/4
D.
a√3/2
Correct Answer:
a√3/4
Explanation:
In zinc blende, a Zn2+ ion and its nearest S2− neighbour are separated along one-fourth of a body diagonal of the cubic unit cell. Since the body diagonal is a√3, the nearest-neighbour distance is a√3/4.
5. If rZn and rS are the ionic radii of Zn2+ and S2−, respectively, which relation applies when the ions are treated as touching along the nearest-neighbour direction in zinc blende?
A.
rZn + rS = a√3/4
B.
rZn + rS = a√2/2
C.
rZn + rS = a/2
D.
rZn + rS = a√3/2
Correct Answer:
rZn + rS = a√3/4
Explanation:
The nearest Zn2+–S2− separation in cubic zinc blende is a√3/4. If the ions touch along this direction, their radii sum must equal this separation.
6. The anions in cubic zinc blende occupy an fcc arrangement. If all tetrahedral holes of this fcc anion lattice were filled by Zn2+ ions, the resulting composition would be
A.
ZnS
B.
Zn2S
C.
ZnS2
D.
Zn2S3
Correct Answer:
Zn2S
Explanation:
An fcc unit cell contains 4 S2− ions and has 8 tetrahedral holes. Filling all of them would place 8 Zn2+ ions around 4 S2− ions, giving Zn2S. ZnS results because only half of the tetrahedral holes are occupied.
7. Which crystallographic change converts the ideal cubic zinc blende arrangement into the hexagonal wurtzite arrangement while retaining tetrahedral Zn2+ and S2− coordination?
A.
Replacing half-filled tetrahedral sites by fully filled octahedral sites
B.
Changing the close-packed anion stacking sequence from ABCABC to ABAB
C.
Changing the coordination number from 4 to 6
D.
Removing all Zn2+ ions from the tetrahedral holes
Correct Answer:
Changing the close-packed anion stacking sequence from ABCABC to ABAB
Explanation:
Zinc blende has cubic close packing of anions with ABCABC stacking, whereas wurtzite has hexagonal close packing with ABAB stacking. In both structures, the ions retain tetrahedral coordination.
8. For cubic zinc blende ZnS, the number of unoccupied tetrahedral holes per unit cell is
A.
2
B.
4
C.
6
D.
8
Correct Answer:
4
Explanation:
An fcc unit cell containing four S2− ions has twice as many tetrahedral holes, namely eight. Since four are occupied by Zn2+ ions in ZnS, four tetrahedral holes remain vacant.
9. Which statement best explains why cubic zinc blende has the formula ZnS rather than ZnS2 when Zn2+ ions occupy tetrahedral holes?
A.
An fcc anion lattice contains twice as many anions as tetrahedral holes
B.
Only half of the tetrahedral holes are occupied, giving equal numbers of Zn2+ and S2− ions
C.
Tetrahedral holes can accommodate only monovalent cations
D.
Each Zn2+ ion occupies two tetrahedral holes simultaneously
Correct Answer:
Only half of the tetrahedral holes are occupied, giving equal numbers of Zn2+ and S2− ions
Explanation:
The fcc S2− framework has eight tetrahedral holes per unit cell but only four are occupied by Zn2+. Thus, the numbers of Zn2+ and S2− ions are equal, giving the formula ZnS.
10. Assuming Zn2+ and S2− ions touch only along the nearest-neighbour direction, which expression gives the cubic lattice parameter of zinc blende in terms of their ionic radii?
A.
a = 2(rZn + rS)
B.
a = √2(rZn + rS)
C.
a = 4(rZn + rS)/√3
D.
a = 4√3(rZn + rS)
Correct Answer:
a = 4(rZn + rS)/√3
Explanation:
The nearest-neighbour distance is a√3/4 and equals rZn + rS when the ions touch. Rearranging gives a = 4(rZn + rS)/√3.
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