Test: Numericals on Unit Cells Dimensions - Chemistry MCQ

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.

Given,
Edge length of FCC crystal (a_FCC) = 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= ρFCC/ρBCC = (Z_FCC x a³_BCC) / (Z_BCC x a³_FCC)
1.5 = (4 x (a_BCC)³) / ( 2 x (4 x 10^-10)₃)
(a_BCC)³ = (1.5 x 2 x 64 x 10^-30)/ 4 = 48 x 10^-30
Therefore a_BCC = 3.63 Å
Difference in Unit Cell Length = 4.0 – 3.63 = 0.37 Å.

Explanation: Given,
Interionic radius (r) = 55 pm
Edge length (a) =?
For BCC, r =  x a
Or a = x r= 4 x 55/1.732 = 127 pm.

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%.

Given,
Edge length of the Gold unit cell (a) = 0.407 x 10^-7m
For FCC unit cell, the atomic radius (r) = a/(2√2)
= 0.407 x 10^-9/(2√2)
= 0.144 nm.

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.

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.

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.

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= 
Or a=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.

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.