# Question #fc060

Oct 7, 2015

The energy required to separate the ion pair = $4.297 \times {10}^{- 19} \text{J}$

I have assumed a copper(II) ion and an electrostatic model.

#### Explanation:

For 2 charges ${q}_{1}$ and ${q}_{2}$ separated by a distance $r$ the force of attraction is given by Coulomb's Law:

$F = \frac{1}{4 \pi {\epsilon}_{0}} . \frac{{q}_{1} {q}_{2}}{{r}^{2}}$

The constant $\frac{1}{4 \pi {\epsilon}_{0}}$ can be written as $k$ and has the value $9 \times {10}^{9} \text{m/F}$

The work done in separating the 2 charges from $r$ to infinity is given by:

$W = - k . \frac{{q}_{1} {q}_{2}}{r} \text{ } \textcolor{red}{\left(1\right)}$

I will assume a wholly electrostatic model where we have discrete ions in contact.

${r}^{+} = 87 \text{pm}$ for $C {u}^{2 +}$

${r}^{-} = 126 \text{pm}$ for ${O}^{2 -}$

This means the total distance between the centre of the 2 ions is:

$126 + 87 = 213 \text{pm} = r$

The electronic charge = $- 1.602 \times {10}^{- 19} \text{C}$

So the charge on the $C {u}^{2 +}$ ion =

$+ 2 \left(1.602 \times {10}^{- 19}\right) = + 3.204 \times {10}^{- 19} \text{C}$

The charge on the ${O}^{2 -}$ must therefore be:

$- 3.204 \times {10}^{- 19} \text{C}$

Putting these values into $\textcolor{red}{\left(1\right)} \Rightarrow$

$W = - \frac{9 \times {10}^{9} \times \left(3.204 \times {10}^{- 19}\right) \times \left(- 3.204 \times {10}^{- 19}\right)}{215 \times {10}^{- 12}} \textcolor{w h i t e}{x} J$

$W = 4.297 \times {10}^{- 19} \text{J}$