# NEET Chemistry The Solid State Questions Solved

BOARD

Iron(II) oxide has a cubic structure and each unit cell has a size of 5$\stackrel{\circ }{A}$. If density of this oxide is 4 g cm-3, calculate the number of Fe2+ and O- ions present in each unit cell.

(Atomic mass of Fe = 56, O = 16, NA = 6.023$×$1023 and  1$\stackrel{\circ }{A}$ = 10-8 cm)

Concept Videos :-

#9 | FCC Unit Cell: Packing Fraction
#20 | Rank of HP Unit Cell
#21 | Packing Fraction of HP Unit Cell
#24 | Density of Unit Cell

Concept Questions :-

Density/Formula/packing fraction/ Semiconductors

Given : $\rho$ = 4 g cm-3

a = 5A = 5$×$10-8 cm M = 72 g/mol, Z = ?

Using the formula for cubic crystals

$\rho$ = $\frac{Z×M}{{a}^{3}×{N}_{0}}$ $⇒$ Z = $\frac{\rho ×{a}^{3}×{N}_{0}}{M}$

$⇒$ Z =

$\therefore$ Z = $\frac{3011×{10}^{-1}}{72}$ = 4.18 $\approx$4

There are four formula units of FeO present per unit cell. Hence it has face-centred cubic lattice where each Fe+ and O2- are four in number.

Difficulty Level:

Crack NEET with Online Course - Free Trial (Offer Valid Till September 24, 2019)