# Chemical bonding - lattice energy

Question: Why is ionic lattice energy inversely proportional to the radius of the atom?

Most heterogeneous covalent molecules are polar to some extent. The degree of polarity, or the dipole moment, depends on the difference in electronegativity difference between the two atoms. The larger the dipole moment, the higher the ionic character.

What I know: Electronegativity decreases as you go downward in a group, however, the size increases, usually, as you go downwards in a group. Thus, ionic character will increase upon going downward, but the ionic lattice energy will decrease? This seems contradictory. Is this true? And if it is, why is it so?

Considering your doubt, from Born-Lande equation:

$$E = -\frac{N_\text{A}Mz^+z^- e^2 }{4 \pi \varepsilon_0 r_0}\left(1-\frac{1}{n}\right)$$

$N_\text{A}$ = Avogadro constant;
$M$ = Madelung constant, relating to the geometry of the crystal;
$z^+$ = charge number of cation
$z^-$ = charge number of anion
$e$ = elementary charge, $1.6022\times10^{-19}\ \mathrm{C}$
$ε_0$ = permittivity of free space
$4πε_0 = 1.112\times10^{-10}\ \mathrm{C^2/(J\cdot m)}$
$r_0$ = distance to closest ion
$n$ = Born exponent, typically a number between 5 and 12, determined experimentally by measuring the compressibility of the solid, or derived theoretically.

And what you are assuming the radius is in fact the distance between the two atoms.
As the distance increases, the attraction decreases from Coulomb's law :

$$F=\frac 1{4\pi\varepsilon_0r_0^2}Q_1Q_2$$

and thus there is less energy involved in breaking the lattice, so lattice energy decreases.

As you move down, the size increases and the valence electron in the outermost shell experiences lesser repulsion and thus it can be easily removed, therefore the elements at the bottom are highly electro-positive or conversely very less electro-negative. So when forming bonds, the elements at the bottom easily remove the electron(s) and gain more developed charged rather than the ones on the top who just release it just partially. This charge variation varies the dipole moment $\vec \mu=q\times\vec d$, by altering the partial charge $q$. Thus downwards as $q$ increases, $|\mu|$ also does increase and make the compounds formed by the element more ionic. Also the ionic/covalent character may be varied depending upon the bonding partner, in sense of, elements or other entities.

Also as charges $z^+$, $z^-$ increases, the magnitude of lattice energy does also increase as can be seen clearly from the Born-Lande equation.

$$\begin{array}{r|c|l} &&\text{Ionic Solid}&\text{Lattice Energy (kJ/mol)}\\\hline &\text{Variation of}&\ce{NaF}&926\\ &r_0 \text{ i.e. distance}&\ce{NaCl}&786\\ &\text{between ions}&\ce{NaBr}&752\\ &&\ce{NaI}&702\\\hline &\text{Variation of}&\ce{NaF}&926\\ &z^+,z^- \text{ i.e. charge}&\ce{MgO}&3800\\ &\text{on ions}&\ce{Al2O3}&15900\\ \end{array}$$

• However isn't the lattice energy directly proportional to the ionic character? As the ionic character increases, so should the lattice energy. However, this clearly is not so. Why? – Gummy bears Jul 5 '14 at 14:17
• I really like your answer. It would be even better if you could explain briefly the variables/constants in the equation. While it is easy to find out with google what they mean, it would improve the reading of your answer. – Martin - マーチン Jul 6 '14 at 6:22