Answer: Electron affinity of F equals
275.8kJ/mol
Explanation: Electron affinity is the energy change when an atom gains an electron.
Let's first calculate the energy required -E(r)to dissociate KF into ions not neutral atom which is given.
E(r) = {z1*z2*e²}/{4π*permitivity of space*r}
z1 is -1 for flourine
z2 is +1 for potassium
e is magnitude of charge 1.602*EXP{-9}C
r is ionic bond length of KF(is a constant for KF 0.217nm)
permitivity of free space 8.854*EXP{-12}.
Now let's solve
E(r)= {(-1)*(1)*(1.602*EXP{-9})²} /
{4*3.142*8.854*EXP(-12)*0.217*EXP(-9)
E(r) = - 1.063*EXP{-18}J
But the energy is released out that is exothermic so we find - E(r)
Which is +1.603*EXP{-18}J
Let's now convert this into kJ/mol
By multiplying by Avogadro constant 6.022*EXP(23) for the mole and diving by 1000 for the kilo
So we have,
1.603*EXP(-18) *6.022*EXP(23)/1000
-E(r) = 640.2kJ/mol.
Now let's obtain our electron affinity for F
We use this equation
Energy of dissociation (nuetral atom)= electron affinity of F +(-E(r)) + ionization energy of K.
498kJ/mol
=e affinity of F + 640.2kJ/mol
+(-418kJ/mol)
(Notice the negative sign in ionization energy for K. since it ionize by losing an electron)
Making electron affinity of F subject of formula we have
Electron affinity (F)=498+418-640.2
=275.8kJ/mol.
