Recall the formula for Coulomb's law:

$\overline{){\mathbf{E}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{k}}{\mathbf{}}\frac{{\mathbf{Q}}_{\mathbf{1}}\mathbf{}\mathbf{\times}\mathbf{}{\mathbf{Q}}_{\mathbf{2}}}{\mathbf{d}}}$

where

k = 231 aJ·pm

Q_{1} and Q_{2} = charge of the ion

d = distance between the two ions

Use Coulomb's Law to calculate the energy, in aJ, of a cadmium ion and a sulfide ion at their equilibrium ion-pair separation distance. An aJ is an attoJoule and equal to 1×10^{-18} J. Recall that Coulomb's Law is: E=Q_{1}Q_{2}/d where k is a proportionality constant equal to 231 aJ·pm, Q_{1} and Q_{2} are the charges of the two ions, and d is the distance between the two ions. Cd^{2+} has a radius of 95 pm and S^{2-} has a radius of 184 pm.

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