Question #fc060

1 Answer
Oct 7, 2015

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