An element with a molar mass of 27 g/mol forms a cubic unit cell with an edge length of 4.05 ×10^-8 cm. If the density of the element is 2.7 g/cm^-3. What will be the nature of a cubic unit cell?

The correct option is D

Both (A) and (B)

Explanation of the correct option:

(d) Step-1: Calculate the number of atoms in a unit cell (Z)

Given:

The molar mass of the given element (M) = 27 g mol^-1

Edge length, a = 4.05 × 10^-8 cm

Density, d = 2.7 g cm^-3

Formula used:

$Z=\frac{d\times a^3\times N_A}{M}$
where Z is the number of atoms in the unit cell

N_A is the Avogadro number.

Thus, on substituting values in the above formula

$Z=\frac{2.7\times {(4.05\times {10}^{-8})}^3\times 6.023\times {10}^{23}}{2.7}$

$Z=\frac{66.43\times 6.023}{100}=\frac{400}{100}=4$

Step-2: Find out the type of cell from the value of Z

Type of unit cell	Value of Z
Simple cubic lattice	1
Body-centered cubic (BCC)	2
Face centered cubic (FCC)	4

Since the number of atoms in the unit cell is 4, therefore the given cubic unit cell has a face-centered cubic (FCC).

Further, FCC and cubic-close packed (CCP) structures are the alternative terms for each other.

Explanation of the incorrect options:

(c) When Z=2, then the structure is BCC.

Hence, option (d) is correct