Chapter 16, Problem 74E

(a)

Interpretation Introduction

Interpretation:

The crystalline structure of MnO whether it is NaCl type or CsCl type needs to be identified.

Concept Introduction :

Simple cubic, face-centered cubic and body centered cubic are the kinds of cubic cells explained on the basis of arrangement of spheres. All of them contains the identical volume thus the packing efficiency of these can be differentiated by calculating the number of spheres which best suited in the unit cell.

Answer to Problem 74E

MnO crystallizes in the NaCl type

Explanation of Solution

The NaCl unit cell contains a face-centered cubic arrangement of cations and anions. There are 4 NaCl units present per unit cell. The CsCl unit cell contains a simple cubic structure of anions with the cations in the cubic holes.There are 1CsCl units present per unit cell.

Therefore, we have to estimate how many units of MnO are present in the unit cell.

By using below formula, we can calculate:

  $\begin{array}{l}\frac{\text{Formula units of MnO}}{\text{Unit cell}}\text{= }\left(\text{4}{\text{.47×10}}^{\text{-8}}\text{cm}\right)\text{ × }\frac{\text{5}\text{.28gMnO}}{{\text{cm}}^{\text{3}}}\text{ × }\frac{\text{1molMnO}}{\text{70}\text{.94gMnO}}\\ \text{ = 6}{\text{.64×10}}^{\text{-24}}\text{mol MnO×}\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{formula units MnO}}{\text{mol MnO}}\\ \text{ =4}\text{.00 formula units MnO}\end{array}$

Due to the 4 formula units of MnO per unit cell, MnO crystallizes in the NaCl type.

(b)

Interpretation Introduction

Interpretation:

The ionic radius of manganese in MnO needs to be calculated.

Concept Introduction :

Simple cubic, face-centered cubic and body centered cubic are the kinds of cubic cells explained on the basis of arrangement of spheres. All of them contains the identical volume thus the packing efficiency of these can be differentiated by calculating the number of spheres which best suited in the unit cell.

Answer to Problem 74E

The radius of manganese is 84 pm.

Explanation of Solution

As we know that MnO crystallizes in the NaCl type, the edge length of Face-centered cubic is related to the ionic radii as shown below:

  $\begin{array}{l}{\text{a = 2r}}^{\text{+}}{\text{+2r}}^{\text{-}}\\ {\text{r}}^{\text{+}}{\text{+r}}^{\text{-}}\text{=}\frac{\text{a}}{\text{2}}\end{array}$

Radius of oxygen = r- = 140 pm

Radius of Mn = r⁺= ?

a = $4.47\times {10}^{-8}\text{cm}$

thus, putting all the values in the above formula:

  $\begin{array}{l}\text{4}{\text{.47×10}}^{\text{-8}}{\text{cm = 2r}}_{{\text{Mn}}^{\text{2+}}}\text{+2}\left(\text{140pm×}\frac{{\text{10}}^{\text{-10}}}{\text{1pm}}\right)\\ \text{4}{\text{.47×10}}^{\text{-8}}\text{cm = 2}\left({\text{r}}_{{\text{Mn}}^{\text{2+}}}\text{+}\left(\text{140}\bcancel{\text{pm}}\text{×}\frac{{\text{10}}^{\text{-10}}}{\text{1}\bcancel{\text{pm}}}\right)\right)\\ {\text{r}}_{{\text{Mn}}^{\text{2+}}}\text{=}\left(\frac{\text{4}{\text{.47×10}}^{\text{-8}}\text{cm}}{\text{2}}\right)\text{-}\left(\text{1}{\text{.40×10}}^{\text{-8}}\text{cm}\right)\\ {\text{r}}_{{\text{Mn}}^{\text{2+}}}\text{= 2}{\text{.235×10}}^{\text{-8}}\text{cm-1}{\text{.40×10}}^{\text{-8}}\text{cm}\end{array}$

  $\begin{array}{l}\text{ = 0}{\text{.835×10}}^{\text{-8}}\text{cm}\\ \text{ = 83}{\text{.5×10}}^{\text{-10}}\text{cm}\\ \text{ = 83}\text{.5pm}\approx \text{84pm}\end{array}$

Thus, the radius of manganese is 84 pm.

(c)

Interpretation Introduction

Interpretation:

Explanations require for the cation-to-anion ratio for MnO.

Concept Introduction :

Simple cubic, face-centered cubic and body centered cubic are the kinds of cubic cells explained on the basis of arrangement of spheres. All of them contains the identical volume thus the packing efficiency of these can be differentiated by calculating the number of spheres which best suited in the unit cell.

Answer to Problem 74E

The radius ratio value is 0.60.

Explanation of Solution

The cation-to-anion ratio can be calculated as below:

  $\begin{array}{l}\frac{{\text{r}}_{{\text{Mn}}^{\text{2+}}}}{{\text{r}}_{{\text{O}}^{\text{2-}}}}\text{=}\frac{\text{84pm}}{\text{140pm}}\\ \text{= 0}\text{.60}\end{array}$

We know from the data given in textbook, that octahedral holes should be filled when the radius ratio is $\text{0}{\text{.414 < r}}^{\text{+}}{\text{/r}}^{\text{-}}\text{< 0}\text{.732}$ . as the calculated value of radius ratio is in between the given value we can predict that Mn²⁺ occupy the octahedral holes formed by the cubic closest packed array of O^2- ions.