Test: Packing Efficiency (Old NCERT) - NEET MCQ

For fcc,
2(r₊ + r_-) = a
2(110 + r_-) = 508
r_- = 508/2 - 110 = 144 pm

For fcc arrangement , distance of nearest neighbour (d) is

= 609.6 pm

2(r₊ + r_-) = a
2(2_Na⁺ + r_Cl^-) = 564
2_Na⁺  = 564/2 - 181 = 101 pm

In NaCl crystal , Edge length = 2 x distance between Na⁺ and Cl⁻
=2 x 265 = 530 pm

a = (r₊ + r_-) = 2(95 + 181) = 552 pm

For fcc,
r = √2 / 4 x a = √2 / 4 x 361
= 
= 127 pm

a = 2√2 r
Volume of the cell = a3 = (2√2 r)³ =  16√2r³
No. of shperes in fcc = 8 x 1/8 + 6 x 1/2 = 4
Volume of the four spheres = 4 x 4/3πr³
= 16πr³

Distance between nearest neighbours, d = AD/2
In right angled △ABC,AC² = AB² + BC² 
AC² = a² + a²  or AC = √2a
Now in right angled △ADC,
AD² = AC² + DC²
∴ d = √3a / 2
Radius , r = d/2 = √3/4 a

In bcc lattice, r = √3/4
r = 
= 124.27 pm

For simple cube,
Radius(r) = a / 2 [a = edge length]
Volume of the atom = 
Packing fraction = 
Volume of the sphere (atoms) in an unit cell.
For simple cubic, Z = 1 atom
Packing fraction =