chemistry : class12 Fluorite Structure : Ionic solids : Solid State MCQ Questions and Answers
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Fluorite Structure
MCQ questions and answers.
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MCQ Questions
1. In the ideal fluorite structure of CaF2, which arrangement correctly describes the ions in the unit cell?
A.
Ca2+ ions form a body-centred cubic lattice and F− ions occupy octahedral sites
B.
Ca2+ ions form a face-centred cubic lattice and F− ions occupy all tetrahedral sites
C.
F− ions form a face-centred cubic lattice and Ca2+ ions occupy all tetrahedral sites
D.
Ca2+ and F− ions separately form two interpenetrating simple cubic lattices
Correct Answer:
Ca2+ ions form a face-centred cubic lattice and F− ions occupy all tetrahedral sites
Explanation:
Fluorite consists of a face-centred cubic array of Ca2+ ions, with F− ions occupying all eight tetrahedral voids associated with the conventional FCC unit cell.
2. What are the coordination numbers of Ca2+ and F−, respectively, in the fluorite structure?
A.
4 and 4
B.
6 and 4
C.
8 and 4
D.
8 and 8
Correct Answer:
8 and 4
Explanation:
Each Ca2+ ion is surrounded by eight F− ions, while each F− ion is tetrahedrally surrounded by four Ca2+ ions. Thus, the coordination numbers are 8:4.
3. How many formula units of CaF2 are present in one conventional unit cell of the fluorite structure?
A.
1
B.
2
C.
4
D.
8
Correct Answer:
4
Explanation:
The FCC arrangement contributes four Ca2+ ions per unit cell. All eight tetrahedral sites are occupied by F− ions, giving eight fluoride ions. Therefore, the cell contains Ca4F8, or four CaF2 formula units.
4. An FCC unit cell contains eight tetrahedral voids. If all of them are occupied by monovalent anions and the FCC lattice ions are divalent cations, what is the resulting fluorite-type formula?
A.
MX
B.
MX2
C.
M2X
D.
M2X3
Correct Answer:
MX2
Explanation:
An FCC unit cell contains four lattice cations and eight tetrahedral anions when all tetrahedral sites are occupied. The ratio is therefore 4:8, which simplifies to M:X = 1:2, giving MX2.
5. A compound with the fluorite structure has molar mass 78.0 g mol−1 and cubic unit-cell edge length 5.46 Å. Using Avogadro's constant as 6.02 × 10^23 mol−1, what is its approximate density?
A.
1.59 g cm−3
B.
2.38 g cm−3
C.
3.18 g cm−3
D.
6.36 g cm−3
Correct Answer:
3.18 g cm−3
Explanation:
The fluorite unit cell contains Z = 4 formula units. Thus, density = 4M/(N_Aa^3) = 4(78.0)/[6.02 × 10^23(5.46 × 10−8)^3] ≈ 3.18 g cm−3.
6. In the conventional fluorite unit cell, which statement about the tetrahedral sites is correct?
A.
There are four tetrahedral sites, and only two are occupied
B.
There are eight tetrahedral sites, and all eight are occupied by anions
C.
There are twelve tetrahedral sites, and four are occupied
D.
There are eight tetrahedral sites, and all eight are occupied by cations
Correct Answer:
There are eight tetrahedral sites, and all eight are occupied by anions
Explanation:
An FCC unit cell has eight tetrahedral voids. In the fluorite structure, every tetrahedral void is occupied by an F− ion, producing the characteristic 1:2 cation-to-anion ratio.
7. If the nearest Ca2+–F− contact in an ideal fluorite structure occurs along the body diagonal from a cube corner to a tetrahedral site at fractional coordinates (1/4, 1/4, 1/4), which relation is correct?
A.
r(Ca2+) + r(F−) = a/4
B.
r(Ca2+) + r(F−) = a/√2
C.
r(Ca2+) + r(F−) = √3a/4
D.
r(Ca2+) + r(F−) = √3a/2
Correct Answer:
r(Ca2+) + r(F−) = √3a/4
Explanation:
The distance from the cube corner (0,0,0) to (1/4,1/4,1/4) is √[(a/4)^2 + (a/4)^2 + (a/4)^2] = √3a/4. At contact, this distance equals the sum of the ionic radii.
8. For an ideal fluorite structure, r(Ca2+) = 0.100 nm and r(F−) = 0.133 nm. What is the approximate unit-cell edge length, assuming nearest Ca2+–F− ions are in contact along the body diagonal?
A.
0.269 nm
B.
0.404 nm
C.
0.539 nm
D.
0.932 nm
Correct Answer:
0.539 nm
Explanation:
Using r(Ca2+) + r(F−) = √3a/4, a = 4(0.100 + 0.133)/√3 nm = 0.538 nm, approximately 0.539 nm.
9. Which atom-counting result correctly represents the conventional unit cell of a fluorite-type compound MX2?
A.
One M atom from corner contributions and two X atoms from edge contributions
B.
Four M atoms from the FCC lattice and eight X atoms entirely inside the unit cell
C.
Four M atoms from body-centre positions and eight X atoms from face-centre positions
D.
Eight M atoms from tetrahedral sites and four X atoms from the FCC lattice
Correct Answer:
Four M atoms from the FCC lattice and eight X atoms entirely inside the unit cell
Explanation:
The FCC M sublattice contributes 8(1/8) + 6(1/2) = 4 atoms. The eight tetrahedral X sites lie within the conventional cell, contributing eight X atoms, so the cell composition is M4X8.
10. A crystal has a cubic unit cell in which the larger ions form an FCC lattice and the smaller ions occupy every tetrahedral void. Which combination is characteristic of the fluorite structure?
A.
Cation coordination number 4, anion coordination number 8, and formula M2X
B.
Cation coordination number 6, anion coordination number 6, and formula MX
C.
Cation coordination number 8, anion coordination number 4, and formula MX2
D.
Cation coordination number 12, anion coordination number 6, and formula M2X3
Correct Answer:
Cation coordination number 8, anion coordination number 4, and formula MX2
Explanation:
The defining local and stoichiometric features of fluorite are eightfold coordination of the FCC cations, fourfold tetrahedral coordination of the anions, and the MX2 composition resulting from complete occupation of tetrahedral sites.
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