Chapter 18, Problem 18.126QA

Interpretation Introduction

To find:

a) The density of $MgSr$.

b) If the atoms touch each other in the body diagonal of the unit cell.

c) The reason why we cannot determine the atom present in the cubic hole.

d) The arrangement of the conduction band and valence bands.

Answer to Problem 18.126QA

Solution:

a) The density of $MgSr$ is $3.13 g/c{m}^3$.

b) Yes, the atoms of $Mg$ and $Sr$ will touch each other along the body diagonal.

c) The formula of the alloy does not allow us to differentiate the arrangement of  $Mg$ and $Sr$ atoms in the holes because the number of atoms of $Mg$ present in the cubic voids is the same as that of the number of atoms of $Sr$ present in the cubic voids in the alloy of $MgSr$.

d) $MgSr$ alloy will have an overlapping of conduction and valence bands.

Explanation of Solution

1) Concept:

The $CsCl$ has the cubic unit cell where the chloride ions sit in the corners of the lattice and the cesium occupies the hole in the center of the unit cell. Both the atoms have a coordination number of eight.

In the unit cell of $MgSr$, let us consider that the magnesium is occupying the cubic void in the unit cell and the lattice is made up of strontium ions. The edge length of the unit cell has been provided which is used for determination of the volume. The molar mass of the unit cell can be calculated and hence the density.

2) Formulae:

i) $V_{unit cell}=l^3 c{m}^3$

ii) $density= \raisebox{1ex}{${mass}_{unit cell}$}\!\left/ \!\raisebox{-1ex}{$V_{unit cell}$}\right.$

3) Given:

i) Edge length of the unit cell $l=390 pm=390 pm \times \frac{1.0 \times {10}^{-10}cm}{1 pm }=3.90 \times {10}^{-8} cm$

ii) Molar mass of $Mg=24.305 g/mol$

iii) Molar mass of $Sr=87.62 g/mol$

iv) Avogadro’s number $1 mol =6.023 \times {10}^{23}atoms$

4) Calculations:

a) Calculate the density of MgSr

Volume of unit cell

$V_{unit cell}=l^3 c{m}^3= {\left(3.90 \times {10}^{-8}\right)}^3 c{m}^3=5.93 \times {10}^{-23} c{m}^3$

Molar mass of the $MgSr$ can be calculated as

$\frac{1 atom}{6.022 \times {10}^{23}atoms/mol}\times \frac{24.305 g}{1 mol}+\frac{1 atom}{6.022 \times {10}^{23}atoms/mol}\times \frac{87.62 g}{1 mol}$

$mass ofMgSr=4.04 \times {10}^{-23}g+1.45\times {10}^{-22} g$

$mass of MgSr=1.86 \times {10}^{-22} g$

Density of the $MgSr$ alloy

$density, d= \raisebox{1ex}{${mass}_{unit cell}$}\!\left/ \!\raisebox{-1ex}{$V_{unit cell}$}\right.$

$d= \frac{1.86 \times {10}^{-22} g}{5.93 \times {10}^{-23} c{m}^3}$

$\therefore d=3.13 g/c{m}^3$

Hence, the density of the $MgSr$ alloy is $3.13 g/c{m}^3$.

b) The atomic radius of magnesium is $160 pm$ and that of strontium is $215 pm$. If we take a unit cell in which the magnesium ions are forming the lattice and the strontium ions are present in the cubic void, the atoms will touch each other as the $Sr$ atoms are present in the body diagonal of the unit cell. It can be shown by the following calculation. $MgSr$ is in cubic unit cell which is similar to the bcc unit cell, and the edge length of the unit cell is related to the radius of the atoms by the relation $4r= \sqrt{3}l$. From this relation, we see that $4r= \sqrt{3}\times 390 pm=675.48 pm$. We can theoretically calculate the value of $4r as 4r= 2r_{Mg}+2r_{Sr}=2\times 160 pm+2\times 215 pm= 750 pm$. Since this theoretical value is greater than the calculated value, therefore, the atoms will touch each other in the body diagonal of the unit cell.

c)  In the cubic unit cell, the cations and anions are present in the cubic voids in the alternate unit cells. That is in one-unit cell, $Mg$ atom is present in the void and $Sr$ atoms are making the lattice, and in the next unit cell,  $Sr$ atoms are sitting in the voids and $Mg$ atom makes the lattice. Because of this arrangement, the number of $Mg$ and $Sr$ occupying the voids is the same, and hence, we cannot comment on the exact position based on the formula of the alloy.

d)  Since both $Mg$ and $Sr$ belong to the same group, it will not have a partially filled valence band because of the group electronic configuration $n{s}^2$, both will have an empty conduction band.  But in the case of $Sr$, there will be an overlap of the conduction and valence bands which is responsible for the conduction of electricity. The overlap of the empty conduction band and filled valence band is more prominent in the case of $Sr$ than $Mg$ because the conduction band is broader in case of $Sr$ which helps in easy overlap of the bands.

Conclusion:

The density of the alloy is calculated from the edge length and the mass of the unit cell. The body diagonal is calculated from the atomic radii.