Question

Chemistry question? Lattice energy

Consider the ionic compounds KF, NaCl, NaBr, and LiCl.

a) Use ionic radii to estimate the cation-anion distance for each compound.

b) Based on your answer to part (a), arrange these 4 compounds in order of decreasing lattice energy.

Answer 1

LiCl > KF > NaCl > NaBr

Explanation:

(a) Cation-anion distance

`ulbb("Compound"color(white)(ml)r_text(+)"/Å"color(white)(ll)r_text(-)"/Å"color(white)(ll)d_text(+-)"/Å")`
`color(white)(mm)"KF"color(white)(mmmI'll)1.52color(white)(m)1.19color(white)(mll)2.71`
`color(white)(mm)"NaCl"color(white)(mmmll)1.16color(white)(m)1.67color(white)(mll)2.83`
`color(white)(mm)"NaBr"color(white)(mmmll)1.16color(white)(m)1.82color(white)(mll)2.98`
`color(white)(mm)"LiCl"color(white)(mmmm)0.90color(white)(m)1.67color(white)(mll)2.57`

(b) Order of lattice energies

The lattice energy `Δ_text(lattice)H` of a salt is the energy required to break it apart and convert its ions into the gaseous state.

`"MX(s)" → "M"^"+""(g)" + "X"^"-""(g)"`

The attractive force `F` between oppositely charged ions is given by the formula

`color(blue)(bar(ul(|color(white)(a/a)F = "-"(kq_1q_2)/d^2color(white)(a/a)|)))" "`

where

`q_1` and `q_2` are the charges on the ions
`d` is the difference between them, and
`k` is a proportionality constant

For all of the above salts, `q_1` and `q_2` are +1 and -1, so the formula becomes

`F = "-"k/d^2`

Thus, the attractive force is inversely proportional to the square of the distance between the ions.

In other words, the greater the interionic distance, the smaller the lattice energy.

Thus, the lattice energy decreases in the order

`"LiCl > KF > NaCl > NaBr"`
`747color(white)(mll)786color(white)(mm)821color(white)(mml)853color(white)(l) "kJ·mol"^"-1"`

Just to confirm our predictions, I have listed the actual lattice energies below the formulas.