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Chemistry · real student question

The attractive force between an Mg²⁺ ion and an S²⁻ ion is 1.49 x 10^-8 N. The radius of the anion is 0.184 nm and the radius of the cation is 0.065 nm. Taking the Born exponent n = 8, calculate the net potential energy when the two ions are just touching.

Question

The attractive force between an Mg2+\mathrm{Mg^{2+}} ion and an S2−\mathrm{S^{2-}} ion is 1.49×10−8 N1.49\times 10^{-8}\ \mathrm{N}. The anion radius is 0.184 nm0.184\ \mathrm{nm} and the cation radius is 0.065 nm0.065\ \mathrm{nm}.

Taking the Born exponent n=8n=8, calculate the net potential energy EnetE_{\text{net}} when the two ions are just touching.

Step-by-step solution

  1. Find the separation at contact. "Just touching" means the centre-to-centre distance is the sum of the ionic radii, and this is also the equilibrium separation r0r_0:

    r0=0.065+0.184=0.249 nm=2.49×10−10 m.r_0=0.065+0.184=0.249\ \mathrm{nm}=2.49\times 10^{-10}\ \mathrm{m}.

  2. Extract the attraction constant from the given force. The Coulomb attraction has the form F=Ar2F=\dfrac{A}{r^2}, where A=k ∣z1z2∣e21A=\dfrac{k\,|z_1 z_2|e^2}{1} bundles all the charge constants. Rather than looking those up, read AA straight off the measured force:

    A=Fr02=(1.49×10−8)(2.49×10−10)2=9.24×10−28 J⋅m.A=F r_0^{2}=\left(1.49\times 10^{-8}\right)\left(2.49\times 10^{-10}\right)^{2}=9.24\times 10^{-28}\ \mathrm{J\cdot m}.

  3. Write the Born model for the net energy. Attraction and short-range repulsion combine as

    Enet=−Ar+Brn,E_{\text{net}}=-\frac{A}{r}+\frac{B}{r^{n}},

    where the repulsive term rises steeply (here n=8n=8) and stops the ions collapsing into each other.

  4. Use the equilibrium condition to eliminate BB. At r=r0r=r_0 the energy is a minimum, so dEdr=0\dfrac{dE}{dr}=0:

    Ar02−nBr0 n+1=0 ⟹ B=A r0 n−1n.\frac{A}{r_0^{2}}-\frac{nB}{r_0^{\,n+1}}=0\ \Longrightarrow\ B=\frac{A\,r_0^{\,n-1}}{n}.

    Substituting back collapses the two-term model into a single clean expression:

    Enet=−Ar0+Anr0=−Ar0(1−1n).E_{\text{net}}=-\frac{A}{r_0}+\frac{A}{n r_0}=-\frac{A}{r_0}\left(1-\frac{1}{n}\right).

    This is why BB never has to be computed: the repulsion always cancels the fraction 1/n1/n of the attraction at equilibrium.

  5. Put in the numbers. With n=8n=8 the bracket is 78\tfrac78:

    Ar0=9.24×10−282.49×10−10=3.71×10−18 J,Enet=−78(3.71×10−18)≈−3.25×10−18 J.\frac{A}{r_0}=\frac{9.24\times 10^{-28}}{2.49\times 10^{-10}}=3.71\times 10^{-18}\ \mathrm{J},\qquad E_{\text{net}}=-\frac{7}{8}\left(3.71\times 10^{-18}\right)\approx -3.25\times 10^{-18}\ \mathrm{J}.

    The negative sign confirms a bound, stable ion pair.

  6. Cross-check the given force against Coulomb's law. With z1=+2z_1=+2, z2=−2z_2=-2, k=8.988×109k=8.988\times 10^{9} and e=1.602×10−19 Ce=1.602\times 10^{-19}\ \mathrm{C}:

    F=k (2e)(2e)r02=1.488×10−8 N,F=\frac{k\,(2e)(2e)}{r_0^{2}}=1.488\times 10^{-8}\ \mathrm{N},

    which matches the quoted 1.49×10−8 N1.49\times 10^{-8}\ \mathrm{N}, so the data are self-consistent.

Answer

Enet=−Ar0(1−1n)≈−3.25×10−18 JE_{\text{net}}=-\frac{A}{r_0}\left(1-\frac{1}{n}\right)\approx -3.25\times 10^{-18}\ \mathrm{J}

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