Question

For a hypothetical element, the force of attraction between a monovalent anion and a divalent cation is 6.02 * 10-9N. If the ionic radius of the cation is 0.040 nm, what is the anion radius?

Answer 1

Anion radius; $r_{a}$ = 0.351 nm

We are given;

Force of attraction: $F_{A}$ = 6.02 × 10⁻⁹ N

Radius of cation; $r_{c}$ = 0.04 nm = 4 × 10⁻¹¹ m

Formula for the force of attraction is given by;

$F_{A}$ = $\frac{1}{4\pi \epsilon_{0} }$ × $\frac{|Z_{1}|*|Z_{2}|*e^{2}}{r^{2}}$

where;

ε₀ is a constant = 8.85 10⁻¹² N/m

Z₁ is charge on divalent cation = 2

Z₂ is charge on anion = -2

e is charge on an electron = 1.6 × 10⁻¹⁹ C

r = $r_{a}$ + $r_{c}$

where, $r_{a}$ is radius of anion

Thus;

r = = $r_{a}$ + (4 × 10⁻¹¹) m

Plugging in the relevant values, we have;

6.02 × 10⁻⁹ = $\frac{1}{4\pi * 8.85 * 10^{-12} }$ × $\frac{2 * 2 * (1.6 * 10^{-19})^{2} }{(r_{a} + (4 * 10^{-11})){2} }$

Simplifying this gives;

$r_{a}$ = 3.51 × 10⁻¹⁰ m = 0.351 nm

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