Question

MnO has either the NaCl type structure or the CsCl type structure. The edge length of the MnO unit cell is 4.47×10^-8 cm, and the density of MnO is 5.28 g/cm³.

a. Does MnO crystallize in the NaCl type or the CsCl type structure?

b. Assuming that the ionic radius of oxygen is 140. pm, estimate the ionic radius of manganese.

c. Does the calculated cation-to-anion radius ratio for MnO substantiate your answer in part a? Explain.

Short answer

(a) MnO crystallizes in the NaCl type structure.

(b) Ionic radii of manganese is 84 pm.

(c)Because the calculated radius ratio falls within the prescribed limits, we expect Mn²⁺to occupy the octahedral holes formed by the cubic closest packed array of O^2-ions.

Step-by-step solution

Given information

The edge length of the MnO unit cell is 4.47×10^-8 cm

The density of MnO is 5.28 g/cm³.

Concept Introduction

Cubic cell

The cubic unit cell is known as the smallest repeating unit, and it applies when all angles are $\mathbf{90}\mathbf{°}$and also all lengths are equal, with each axis being defined by a coordinate.

The simple cubic unit cell is denoted by eight atoms marked with the actual cube.

Face-centered cubic cell

A face-centered cubic unit cell is a well-known structure that consists of atoms arranged in a cube in which each corner of the cube has a fraction of an atom with six additional full atoms positioned at the center of each cube face.

Solution

(a)

The NaCl unit cell has a face-centered cubic arrangement of the anions with cations in the octahedral holes. There are four NaCl formula units per unit cell, and since there is a 1:1 ratio of cations to anions in MnO, there would be four MnO formula units per unit cell assuming a NaCl-type structure. The CsCl unit cell has a simple cubic structure of anions with the cations in the cubic holes. There is one CsCl formula unit per unit cell, so there would be one MnO formula unit per unit cell if a CsCl structure is observed

$$\begin{gathered} \frac{\mathrm{Formula}\mathrm{of}\mathrm{MnO}}{\mathrm{Unit}\mathrm{Cell}}={\left(4.47\times {10}^{-8}\mathrm{cm}\right)}^3\times \frac{5.28\mathrm{gm}\mathrm{of}\mathrm{MnO}}{{\mathrm{cm}}^3}\times \frac{1\mathrm{mole}\mathrm{MnO}}{70.94\mathrm{gm}\mathrm{of}\mathrm{MnO}}\times 6.6023\times {10}^{23}\mathrm{formula}\mathrm{units}\mathrm{of}\mathrm{MnO} \\ \frac{\mathrm{Formula}\mathrm{of}\mathrm{MnO}}{\mathrm{Unit}\mathrm{Cell}}=4\mathrm{formula}\mathrm{units}\mathrm{of}\mathrm{MnO} \end{gathered}$$

From the calculation, MnO crystallizes in the NaCl type structure.

(b)

From the NaCl structure and assuming the ions touch each other, then L cube edge length= 2r_Mn²⁺ + 2r_O^2-

l=4.47×10^-8 cm=2r _Mn²⁺ +2(1.40×10^-8 cm).

r_Mn²⁺= 8.35 x 10^-8 cm = 84 pm

(c)

r_Mn²⁺ /r_O^2-=84pm/140pm=0.60

Octahedral holes should be filled when 0.414<r_Mn²⁺ /r_O^2-<0.732. Because the calculated radius ratio falls within the prescribed limits, we expect Mn²⁺to occupy the octahedral holes formed by the cubic closest packed array of O^2-ions.