# Class 12 Chemistry MCQ – Solid State – Calculations Involving Unit Cells Dimensions

This set of Class 12 Chemistry Chapter 1 Multiple Choice Questions & Answers (MCQs) focuses on “Solid State – Calculations Involving Unit Cells Dimensions”.

1. Aluminium crystallises in a face-centred cubic lattice. The edge length of the unit cell of aluminium is 4.05 x 10-10m. What is the density of aluminium? (Atomic mass of Al=27)
a) 2700 kg m-3
b) 3000 kg m-3
c) 2400 kg m-3
d) 2100 kg m-3

Explanation: Given,
Atomic mass (M)=27 amu
For FCC structure, Z=4
Avogadro’s number (N0) = 6.02 x 1023
Edge length of the Al unit cell (a)= 4.05 x 10-10m
The density of aluminium (ρ) = (Z x M)/(a3 X N0)
= (4 x 27)/((4.05 x 10-10) 3 x 6.02 x 1023)
= 2700 kg m-3.

2. Gold (atomic mass 197 u) crystallises in a face-centred unit cell. What is its atomic radius if the edge length of the gold unit cell is 0.407 x 10-9m?
a) 0.115 nm
b) 0.144 nm
c) 0.235 nm
d) 0.156 nm

Explanation: Given,
Edge length of the Gold unit cell (a) = 0.407 x 10-7m
For FCC unit cell, the atomic radius (r) = a/(2$$\sqrt{2}$$)
= 0.407 x 10-9/(2$$\sqrt{2}$$)
= 0.144 nm.

3. A metal X has a BCC structure with nearest neighbor distance 365.9 pm. What is metal X if its density is 1.0016 g cm-3?
a) Aluminum
b) Magnesium
c) Sodium
d) Potassium

Explanation: Given,
Nearest neighbor distance (d) = 365.9 pm
Density (ρ) = 1.51 g cm-3
For the BCC structure, nearest neighbor distance (d) is related to the edge length (a) as d=$$\Big(\frac{\sqrt{3}}{2}\Big)$$ x a
Or a=$$\Big(\frac{2}{\sqrt{3}}\Big)$$ x d = 2/1.732 x 365.9 = 422.5 pm
For BCC structure, Z=2
We know, (ρ) = (Z x M)/(a3 x N0)
Or M = (ρ x a3 x N0)/Z
= (1.0016 x 106 x (422.5 x 10-12)3 x 6.02 x 1023)/2
= 23 amu
Therefore, the given metal X is Sodium.

4. Lithium forms a BCC lattice with an edge length of 350 pm. The experimental density of lithium is 0.53 g cm-3. What is the percentage of missing lithium atoms? (Atomic mass of Lithium = 7 amu)
a) 97.7%
b) 95.4%
c) 4.6%
d) 2.3%

Explanation: Given,
Edge length (a) = 350 pm = 3.5 x 10-8 cm
Atomic mass (M) = 7 amu
Avogadro’s number (N0) = 6.02 x 1023
Density (ρ) = (Z x M)/(a3 x N0)
= (2 x 7)/((3.5 x 10-8)3 x 6.02 x 1023)
= 0.542 g cm-3
% of lithium atoms occupied = (Experimental density/Theoretical density) x 100
= 0.53/0.542 x 100
= 97.7%
% of unoccupied lattice sites = 100 – 97.7
= 2.3%.

5. An element with cell edge of 288 pm has a density of 7.2 g cm-3. What type of structure does the element have if it’s atomic mass M=51.8 g mol-1?
a) Body-Centred Cubic (BCC)
b) Face-Centred Cubic (FCC)
c) Simple Cubic
d) Hexagonal Closed Packing (HCP)

Explanation: Given,
Edge length (a) = 288 pm
Density (ρ) = 7.2 g cm-3
Atomic mass (M) = 51.8 g mol-1
Avogadro’s number (N0) = 6.02 x 1023
We know, (ρ) = (Z x M)/(a3 x N0)
Or Z =(ρ x a3 x N0)/M = (7.2 x (288 x 10-10) 3 x 6.02 x 1023)/51.8
Z = 2
Therefore, the element has Body-Centred Cubic (BCC) type of structure.
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6. The radius of an atom of an element is 55 pm. What is the edge length of the unit cell if it is body-centred cubic?
a) 144.6 pm
b) 163.4 pm
c) 127.0 pm
d) 123.5 pm

Explanation: Given,
Interionic radius (r) = 55 pm
Edge length (a) =?
For BCC, r = $$\Big(\frac{\sqrt{3}}{4}\Big)$$ x a
Or a = $$\Big(\frac{4}{\sqrt{3}}\Big)$$ x r= 4 x 55/1.732 = 127 pm.

7. An element of density 8.0 g/cm3 forms an FCC lattice with unit cell edge of 300 pm. Calculate the number of atoms present in 0.5kg of the element.
a) 95 x 1023 atoms
b) 93.59 x 1023 atoms
c) 92.59 x 1023 atoms
d) 91.38 x 1023 atoms

Explanation: Given,
Density (ρ) = 8.0 g/cm3
For FCC structure, Z = 4
Avogadro’s number (N0) = 6.02 x 1023
Edge length of the unit cell (a) = 300 x 10-10 cm
The density of the element (ρ) = (Z x M)/ (a3 x N0)
Therefore, the Molar Mass (M) = (ρ x a3 x N0)/(Z)
= (8.0 x 6.02 x 1023 x 27.0 x 10-24) /4
= 32.508 g.
Therefore, 32.508 g of the element contains 6.02 x 1023 atoms.
500 g of the element would contain = (6.02 x 1023 x 500)/ 32.508 = 92.59 x 1023 atoms.

8. A metal crystallizes into two cubic phases BCC and FCC. The ratio of densities of FCC and BCC is equal to 1.5. Calculate the difference between the unit cell lengths of the FCC and BCC crystals if the edge length of the FCC crystal is equal to 4.0 Å.
a) 0.5 Å
b) 0.37 Å
c) 0. 28 Å
d) 0.73 Å

Explanation: Given,
Edge length of FCC crystal (aFCC) = 4.0 Å
For FCC structure, Z = 4
For BCC structure, Z=2
Avogadro’s number (N0) = 6.02 x 1023
The density of a crystal (ρ)=(Z x M)/(a3 x N0)
Therefore, the ratio of Densities= ρFCCBCC = (ZFCC x a3BCC) / (ZBCC x a3FCC)
1.5 = (4 x (aBCC)3) / ( 2 x (4 x 10-10)3)
(aBCC)3 = (1.5 x 2 x 64 x 10-30)/ 4 = 48 x 10-30
Therefore aBCC = 3.63 Å
Difference in Unit Cell Length = 4.0 – 3.63 = 0.37 Å.

9. If the radius of a Chloride ion is 0.154 nm, then what is the maximum size of a cation that can fit in each of its octahedral voids?
a) 1.15 x 10-1 nm
b) 1.21 x 10-1 nm
c) 1.18 x 10-1 nm
d) 1.13 x 10-1 nm

Explanation: Given,
Radius of Chloride ion (r) = 0.154 nm
Let radius of cation = r+
For Octahedral Voids, r+ / r = 0.732 (for maximum value of the size of the cation)
Therefore, r+ = 0.732 x 0.154 = 1.13 x 10-1 nm.

10. Rubidium Chloride (RbCl) has NaCl like structure at normal pressures. If the radius of the Chloride ion is 1.54 Å, what is the unit cell edge length for RbCl? (Assuming anion-anion contact)
a) 4.25 Å
b) 4.78 Å
c) 4.32 Å
d) 5.14 Å

Explanation: Given,
Radius of Chloride ion (r) = 0.154 nm
Distance between the centres of the Chloride ions = 2 x 0.154 = 0.308 nm
Let the edge length of cube = a
Distance between Rb+ and Cl ions = a/2
Therefore, the distance between Cl ions = (2 x (a/2)2)1/2
0.308 = (2 x (a/2)2)1/2
0.094864 = 2 x (a/2)2
0.047432 = (a/2)2
0.218 = (a/2)
a = 0.432 nm = 4.32 Å.

Sanfoundry Global Education & Learning Series – Chemistry – Class 12.

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