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Answer 1 · 2015-10-07T17:30:37

The energy required to separate the ion pair = $4.297xx10^(-19)"J"$

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

For 2 charges $q_1$ and $q_2$ separated by a distance $r$ the force of attraction is given by Coulomb's Law:

$F=(1)/(4piepsilon_0).(q_1q_2)/(r^2)$

The constant $(1)/(4piepsilon_0)$ can be written as $k$ and has the value $9xx10^(9)"m/F"$

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

$W=-k.(q_1q_2)/(r) " "color(red)((1))$

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

The ionic radii are:

$r^+=87"pm"$ for $Cu^(2+)$

$r^(-)=126"pm"$ for $O^(2-)$

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

$126+87=213"pm"=r$

The electronic charge = $-1.602xx10^(-19)"C"$

So the charge on the $Cu^(2+)$ ion =

$+2(1.602xx10^(-19))=+3.204xx10^(-19)"C"$

The charge on the $O^(2-)$ must therefore be:

$-3.204xx10^(-19)"C"$

Putting these values into $color(red)((1))rArr$

$W=-(9xx10^(9)xx(3.204xx10^(-19))xx(-3.204xx10^(-19)))/(215xx10^(-12))color(white)(x)J$

$W=4.297xx10^(-19)"J"$