chemistry : class12 Edge Length–Radius Relations : Crystalline solids : Solid State MCQ Questions and Answers
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Edge Length–Radius Relations
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MCQ Questions
1. A face-centred cubic crystal has an edge length of 508 pm. What is the atomic radius of the constituent atom?
A.
155.0 pm
B.
179.6 pm
C.
203.2 pm
D.
220.3 pm
Correct Answer:
179.6 pm
Explanation:
For an FCC unit cell, atoms touch along the face diagonal, so 4r = √2a. Hence, r = a/(2√2) = 508/(2√2) = 179.6 pm.
2. The atomic radius of a body-centred cubic metal is 125 pm. What is the edge length of its unit cell?
A.
216.5 pm
B.
250.0 pm
C.
288.7 pm
D.
353.6 pm
Correct Answer:
288.7 pm
Explanation:
In a BCC unit cell, atoms touch along the body diagonal, giving √3a = 4r. Therefore, a = 4r/√3 = 4(125)/√3 = 288.7 pm.
3. For cubic unit cells having the same edge length, what is the ratio of the atomic radius in a BCC structure to that in an FCC structure?
A.
0.816
B.
1.000
C.
1.225
D.
1.414
Correct Answer:
1.225
Explanation:
For BCC, r = √3a/4, while for FCC, r = √2a/4. Their ratio is (√3a/4)/(√2a/4) = √(3/2) = 1.225.
4. A cubic crystal has an edge length of 400 pm and an atomic radius of approximately 173.2 pm. Which type of cubic unit cell is consistent with these dimensions?
A.
Simple cubic
B.
Body-centred cubic
C.
Face-centred cubic
D.
A structure with atoms touching only along an edge and a face diagonal
Correct Answer:
Body-centred cubic
Explanation:
For BCC, r = √3a/4 = √3(400)/4 = 173.2 pm. The given radius therefore satisfies the BCC edge-length relation.
5. For the same edge length, by what percentage is the maximum atomic radius in an FCC unit cell smaller than that in a simple cubic unit cell?
A.
18.4%
B.
25.0%
C.
29.3%
D.
35.4%
Correct Answer:
29.3%
Explanation:
For simple cubic, rSC = a/2. For FCC, rFCC = a/(2√2). The percentage decrease is [1 − 1/√2] × 100 = 29.3%.
6. In an FCC crystal, the nearest-neighbour distance is 0.286 nm. What is the edge length of the unit cell?
A.
0.202 nm
B.
0.286 nm
C.
0.404 nm
D.
0.572 nm
Correct Answer:
0.404 nm
Explanation:
In an FCC unit cell, the nearest-neighbour distance is a/√2, because atoms touch along the face diagonal. Thus, a = √2(0.286) = 0.404 nm.
7. A cubic unit cell has an edge length of 0.500 nm, and the atoms touch along its body diagonal. What is the atomic radius?
A.
0.125 nm
B.
0.177 nm
C.
0.217 nm
D.
0.250 nm
Correct Answer:
0.217 nm
Explanation:
Touching along the body diagonal identifies a BCC relation: 4r = √3a. Therefore, r = √3(0.500)/4 = 0.2165 nm, approximately 0.217 nm.
8. A cubic crystal has an atomic-radius-to-edge-length ratio of 0.354. Which edge-length–radius relation does this indicate?
A.
r = a/2
B.
r = √3a/4
C.
r = √2a/4
D.
r = a/√3
Correct Answer:
r = √2a/4
Explanation:
The value √2/4 = 0.3536 corresponds to an FCC unit cell, in which atoms touch along the face diagonal.
9. A BCC metal has an atomic radius of 0.160 nm. What edge length would the same-radius atoms require in an FCC arrangement?
A.
0.277 nm
B.
0.320 nm
C.
0.369 nm
D.
0.453 nm
Correct Answer:
0.453 nm
Explanation:
For an FCC unit cell, a = 2√2r. Substituting r = 0.160 nm gives a = 2√2(0.160) = 0.4525 nm, or approximately 0.453 nm.
10. An FCC crystal has an edge length of 0.400 nm. What is the distance between neighbouring atomic centres along a face diagonal?
A.
0.141 nm
B.
0.200 nm
C.
0.283 nm
D.
0.400 nm
Correct Answer:
0.283 nm
Explanation:
The face diagonal of a cubic cell is √2a. In an FCC structure, this diagonal contains four radii, so the nearest-neighbour centre-to-centre distance is 2r = a/√2 = 0.400/√2 = 0.283 nm.
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