Chapter 9, Problem 80E

MnO has either the NaCI type structure or the CsCI type structure (see Exercise 69). 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 or the CsCl type structure?

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

Interpretation Introduction

Interpretation:

• The crystalline structure of $\text{MnO}$ whether it is $\text{NaCl}$ type or $\text{CsCl}$ type has to be identified.

• The ionic radius of Manganese in $\text{MnO}$ has to be determined.

Concept Introduction:

In packing of atoms in a crystal structure, the atoms are imagined as spheres. The two major types of close packing of the spheres in the crystal are – hexagonal close packing and cubic close packing. Cubic close packing forms two types of lattices – body – centered lattice and face – centered lattice.

In body-centered cubic unit cell, each of the six corners is occupied by every single atom. Center of the cube is occupied by one atom. Each atom in the corner is shared by eight unit cells and a single atom in the center of the cube remains unshared. Thus the number of atoms per unit cell in BCC unit cell is,

                      $\text{8}\text{×}\frac{\text{1}}{\text{8}}\text{atoms}\text{in}\text{corners}\text{+}\text{1}\text{atom}\text{at}\text{the}\text{center}\text{=}\text{1}\text{+}\text{1}\text{=}\text{2}\text{atoms}$

             $\begin{array}{l}\text{ The edge length of one unit cell is given by}\\ \text{a}\text{=}\frac{\text{4R}}{\sqrt{\text{3}}}\\ \text{where a}\text{=}\text{edge length of unit cell}\\ \text{R}\text{=}\text{radius}\text{of}\text{atom}\text{.}\\ \end{array}$

In face-centered cubic unit cell, each of the six corners is occupied by every single atom. Each face of the cube is occupied by one atom. Each atom in the corner is shared by eight unit cells and each atom in the face is shared by two unit cells. Thus the number of atoms per unit cell in FCC unit cell is,

                $\text{8}\text{×}\frac{\text{1}}{\text{8}}\text{atoms}\text{in}\text{corners}\text{+}\text{6}\text{×}\frac{\text{1}}{\text{2}}\text{atoms}\text{in}\text{faces}\text{=}\text{1}\text{+}\text{3}\text{=}\text{4}\text{atoms}$

                          $\begin{array}{l}\text{The edge length of a unit cell is given by}\\ \text{l}\text{=}\text{2R}\sqrt{\text{2}}\\ \text{where a}\text{=}\text{edge}\text{length of unit cell}\\ \text{R}\text{=}\text{radius}\\ \end{array}$

The edge length of the unit cell containing anion and cation is related to ionic radii as,

$\begin{array}{l}\text{l}\text{=}{\text{2r}}_{{\text{C}}^{\text{+}}}\text{+}{\text{2r}}_{{\text{A}}^{\text{-}}}\\ \text{where}{\text{r}}_{{\text{C}}^{\text{+}}}\text{and}{\text{r}}_{{\text{A}}^{\text{-}}}\text{are}\text{ionic}\text{radius}\text{of}\text{cation}\text{and}\text{anion}\text{respectively}\text{.}\end{array}$

Answer to Problem 80E

Answer

$\text{MnO}$ has $\text{NaCl}$ type structure.

Ionic radius of Manganese ion in $\text{MnO}$ is calculated as 83.5 pm.

Explanation of Solution

Explanation

Record the given data.

Given data: edge length of $\text{MnO}$ unit cell is $\text{4}\text{.47}\text{×}{\text{10}}^{\text{-8}}\text{cm}$

                     Density of $\text{MnO}$ is $\text{5}{\text{.28 g/cm}}^3$

                     Ionic radius of Oxygen is $\text{140}\text{pm}$

The edge length of a $\text{MnO}$ unit cell ‘l’ is given. Density of $\text{MnO}$ is given. The ionic radius of Oxygen is given.

Determine the mass of a unit cell of $\text{MnO}$ crystalline solid.

      $\begin{array}{l}\text{Density}\text{=}\frac{\text{mass}}{\text{volume}}\\ \text{5}\text{.28}{\text{g/cm}}^{\text{3}}=\frac{\text{mass}}{{\text{l}}^{\text{3}}}\\ \text{mass}\text{=}{\text{l}}^{\text{3}}\text{×}\text{5}\text{.28}{\text{g/cm}}^{\text{3}}\\ \text{=}{\left(\text{4}\text{.47}\text{×}{\text{10}}^{\text{-8}}\text{cm}\right)}^3\times \text{5}\text{.28}{\text{g/cm}}^{\text{3}}\\ =4.715\times {10}^{-22}\text{g}\\ \end{array}$

The edge length of the unit cell is known. Using that value the density of the unit cell is calculated using the equation $\text{Density}\text{=}\frac{\text{mass}}{\text{volume}}$ where cubic value of the edge length gives volume of the unit cell.

Determine the number of formula units per $\text{MnO}$ unit cell.

Let ‘n’ be the number of formula units per $\text{MnO}$ unit cell.

Then, $\text{mass}\text{of}\text{a}\text{unit}\text{cell}\text{=}\text{Number}\text{of}\text{formula}\text{units}\text{×}\text{average}\text{mass}\text{of}\text{one}\text{formula}\text{unit}\mathrm{......}\left(1\right)$

$\begin{array}{l}\text{Average mass of one MnO}\text{unit}\text{=}\frac{\text{molar}\text{mass}\text{of}\text{MnO}}{\text{Avogadro}\text{number}}\\ \text{=}\frac{\text{70}\text{.94}\text{g}}{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}}\\ \text{=}\text{1}\text{.18}\text{×}{\text{10}}^{\text{-22}}\text{g}\mathrm{......}\left(2\right)\end{array}$

Substitute (2) in (1). Therefore,

                 $\begin{array}{l}4.715\times {10}^{-22}\text{g}\text{=}\text{n}\text{×}\text{11}\text{.8}\text{×}{\text{10}}^{\text{-23}}g\\ \text{n}=\frac{4.715\times {10}^{-22}\text{g}}{\text{1}\text{.18}\text{×}{\text{10}}^{\text{-22}}g}=3.99\approx 4\end{array}$

The mass of a unit cell is calculated in the previous step. The average mass of one formula unit is determined using mass of a unit cell. The number of formula units per unit cell is calculated from this. The value obtained is 4. The crystalline structure of $\text{NaCl}$ is face -centered cubic. It has 4 units of $\text{NaCl}$ per unit cell. So the crystalline structure of $\text{MnO}$ also $\text{NaCl}$ type that is face – centered cubic.

Determine the ionic radius of Manganese ion in $\text{MnO}$

                 $\begin{array}{l}\text{l}\text{=}{\text{2r}}_{{\text{Mn}}^{\text{2+}}}\text{+}{\text{2r}}_{{\text{O}}^{\text{2-}}}\\ {\text{r}}_{{\text{O}}^{\text{2-}}}\text{=}\text{140}\text{pm}\text{=}\text{140}\text{×}{\text{10}}^{\text{-12}}\text{m}\text{=}\text{1}\text{.40}\text{×}{\text{10}}^{\text{-8}}\text{cm}\\ \text{l}\text{=}\text{4}\text{.47}\text{×}{\text{10}}^{\text{-8}}\text{cm}\\ {\text{2r}}_{{\text{Mn}}^{\text{2+}}}\text{=}\text{l}\text{-}{\text{2r}}_{{\text{O}}^{\text{2-}}}\text{=}\text{4}\text{.47}\text{×}{\text{10}}^{\text{-8}}\text{cm}-2\left(\text{1}\text{.40}\text{×}{\text{10}}^{\text{-8}}\text{cm}\right)\\ {\text{r}}_{{\text{Mn}}^{\text{2+}}}=\frac{1.67\text{×}{\text{10}}^{\text{-8}}\text{cm}}{2}=\text{0}\text{.835}\text{×}{\text{10}}^{\text{-8}}\text{cm}\text{=}\text{83}\text{.5}\text{pm}\end{array}$

The edge length of the unit cell is related to the ionic radius of ions of a ionic compound as $\text{l}\text{=}{\text{2r}}_{{\text{C}}^{\text{+}}}\text{+}{\text{2r}}_{{\text{A}}^{\text{-}}}\text{where}{\text{r}}_{{\text{C}}^{\text{+}}}\text{and}{\text{r}}_{{\text{A}}^{\text{-}}}\text{are}\text{ionic}\text{radius}\text{of}\text{cation}\text{and}\text{anion}\text{respectively}\text{.}$

Using the above formula the ionic radius of Manganese is calculated.

Conclusion

Conclusion

The crystalline structure of $\text{MnO}$ is ascertained and the ionic radius of Manganese is calculated.