Ferrous oxide has a cubic structure. The length of edge of the unit cell is 5 Å and the density of the oxide is 4.0 ~g ~cm^ -3 . Then the number of Fe^ 2+ and O^ 2- ions present in each unit cell will be

Correct answer: B. four Fe^ 2+ and four O^ 2- aligned & d=(Z × M)/(N_A × V) & Z=(d × N_A × V)/(M) aligned As a=5 Å=5 × 10^ -8 ~cm V=(5 × 10^ -8 )^3 ~cm^3 Molecular mass, M=56+16=72 ~g aligned Z & = 4 × 6.023 × 10^ 23 ×(5 × 10^ -8 )^3 72 & =4.16 4 aligned Four molecules of ferrous oxide, FeO per unit cell, it means that 4 Fe^ 2+ and 4 O^ 2- ions are present in each unit cell.

JIPMER 2015

Exam: JIPMER
Year: 2015
Subject: Chemistry
Chapter: Solid State

JIPMER 2015 · Chemistry · Solid State

Ferrous oxide has a cubic structure. The length of edge of the unit...

Ferrous oxide has a cubic structure. The length of edge of the unit cell is

5 Å

and the density of the oxide is

4.0 \mathrm{~g} \mathrm{~cm}^{-3}

. Then the number of

\mathrm{Fe}^{2+}

and

\mathrm{O}^{2-}

ions present in each unit cell will be

Options

Afour $\mathrm{Fe}^{2+}$ and two $\mathrm{O}^{2-}$
Bfour $\mathrm{Fe}^{2+}$ and four $\mathrm{O}^{2-}$
Ctwo Fe ${ }^{2+}$ and four $\mathrm{O}^{2-}$
Dtwo Fe ${ }^{2+}$ and two O₂-

Correct answer

B. four $\mathrm{Fe}^{2+}$ and four $\mathrm{O}^{2-}$

Step-by-step solution

\begin{aligned} & \because d=\frac{Z \times M}{N_A \times V} \\ & \therefore Z=\frac{d \times N_A \times V}{M}\end{aligned}

a=5 Å=5 \times 10^{-8} \mathrm{~cm} \therefore V=\left(5 \times 10^{-8}\right)^3 \mathrm{~cm}^3

Molecular mass,

M=56+16=72 \mathrm{~g}

\begin{aligned} Z & =\frac{4 \times 6.023 \times 10^{23} \times\left(5 \times 10^{-8}\right)^3}{72} \\ & =4.16 \approx 4\end{aligned}

Four molecules of ferrous oxide, FeO per unit cell, it means that

4 \mathrm{Fe}^{2+} and 4 \mathrm{O}^{2-}$ ions are present in each unit cell.

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