Chapter 4, Problem 4.19QA

Interpretation Introduction

To find:

The most negative lattice energy substance out of (a) KCl, (b) TiO², (c) BaCl₂, (d) KI

Answer to Problem 4.19QA

Solution:

TiO₂ has the most negative lattice energy.

Explanation of Solution

1) Concept: Using the formula below, we can calculate the lattice energy. Here, ${\mathrm{Q}}_1$ and ${\mathrm{Q}}_2$ are charges of cation and anion respectively and d is the distance between nuclei of cation and anion.

Ionization reactions for all four compounds:

$\left(\mathrm{a}\right)\mathrm{K}\mathrm{C}\mathrm{l}\to {\mathrm{K}}^{+}+{\mathrm{C}\mathrm{l}}^{-}$

$\left(\mathrm{b}\right)\mathrm{T}\mathrm{i}{\mathrm{O}}_2\to \mathrm{T}{\mathrm{i}}^{+2}+2\mathrm{C}{\mathrm{l}}^{-}$

$\left(\mathrm{c}\right)\mathrm{B}\mathrm{a}\mathrm{C}{\mathrm{l}}_2\to \mathrm{B}{\mathrm{a}}^{+2}+2\mathrm{C}{\mathrm{l}}^{-}$

$\left(\mathrm{d}\right)\mathrm{K}\mathrm{I}\to {\mathrm{K}}^{+}+{\mathrm{I}}^{-}$

2) Formula:

$\mathrm{ }{\mathrm{E}}_{\mathrm{e}\mathrm{l}}=2.31\times {10}^{-19}\mathrm{ }\mathrm{J}.\mathrm{n}\mathrm{m}\mathrm{ }\left[\frac{{\mathrm{Q}}_1\times {\mathrm{Q}}_2}{\mathrm{d}}\right]$

3) Given information: Ionic radii of ions are:

Ions	Ionic radius
K+	138 pm
Cl-	181 pm
Ti+2	100 pm
O-2	140 pm
Ba+2	149 pm
I-	206 pm

4) Calculations: The distance between nuclei of cation and anion is calculated as

$\left(\mathrm{a}\right)\mathrm{d}=({\mathrm{K}}^{+}\mathrm{ }\mathrm{ }\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{i}\mathrm{c}\mathrm{ }\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{i}\mathrm{u}\mathrm{s}+{\mathrm{C}\mathrm{l}}^{-}\mathrm{ }\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{i}\mathrm{c}\mathrm{ }\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{i}\mathrm{u}\mathrm{s})$

$\mathrm{d}=(138\mathrm{ }\mathrm{p}\mathrm{m}+181\mathrm{ }\mathrm{p}\mathrm{m})$

$\mathrm{d}=319\mathrm{ }\mathrm{p}\mathrm{m}$

$319\mathrm{ }\mathrm{p}\mathrm{m}\mathrm{ }\times \frac{1\mathrm{n}\mathrm{m}}{1000\mathrm{p}\mathrm{m}}=0.319\mathrm{ }\mathrm{n}\mathrm{m}$

And so, for the rest of compounds:$\left(\mathrm{b}\right)\mathrm{ }0.240\mathrm{ }\mathrm{n}\mathrm{m},\mathrm{ }\left(\mathrm{c}\right)\mathrm{ }0.330\mathrm{ }\mathrm{n}\mathrm{m},\mathrm{ }\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }\left(\mathrm{d}\right)\mathrm{ }0.344\mathrm{ }\mathrm{n}\mathrm{m}$

Now product of charges for all compounds

$\left(a\right)1\times \left(+1\right)\times 1\times \left(-1\right)=-1$

$\left(b\right)1\times \left(+2\right)\times 2\left(-1\right)=-4$

$\left(c\right)1\times \left(+2\right)\times 2\left(-1\right)= -4$

$\left(a\right)1\times \left(+1\right)\times 1\times \left(-1\right)=-1$

$\left[\frac{{\mathrm{Q}}_1\times {\mathrm{Q}}_2}{\mathrm{d}}\right]\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{ }\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{ }\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{u}\mathrm{d}\mathrm{s}:$

$\left(a\right)-\frac{1}{0.19}=-5.26$

$\left(b\right)-\frac{4}{0.24}=-16.67$

$\left(c\right)-\frac{4}{0.33}=-12.12$

$\left(d\right)-\frac{1}{0.344}=-2.91$

When the factor $\left[\frac{{\mathrm{Q}}_1\times {\mathrm{Q}}_2}{\mathrm{d}}\right]$ is large, it will lead to the most negative lattice energy. It is large for TiO_2. Therefore, TiO₂ has the most negative lattice energy.

Conclusion:

TiO₂ has the most negative lattice energy because the product of the charges is the highest.