3. The energy needed for the following process is 1.96 x 10 4 kJ/mol: Li( g ) Li

Question

3. The energy needed for the following process is 1.96 x 104 kJ/mol:

Li(g) Li3+(g) + 3e- ; I1 + I2 + I3 = 1st, 2nd, 3rd ionization energies. The first

ionization energy of lithium is 520 kJ/mol, Li(g) Li+(g) + e-, I1 , 1st ionization energy.

Calculate the second ionization energy of lithium, Li+(g) Li2+(g) + e- , I2 , 2nd ionization

energy.

Energy needed to remove three electrons = I1 + I2 + I3

Z = 3; n= 1

I3 = E = E - E3

I3 = -(2.18 x 10-18 J)(3)2 () + (2.18 x 10-18 J)(3)2 () ; See equation in #8.93, p 284

I3 = +1.96 x 10-17 J

I3 = +1.96 x 10-17 J (6.022 x 1023 ions/1 mol) = 1.18 x 107 J/mol 1.18 x 104 kJ/mol

Energy needed to remove three electrons =    I1 + I2 + I3

   1.96 x 104 kJ/mol = 520 kJ/mol + I2 + 1.18 x 104 kJ/mol

   I2 = 7.28 x 103 kJ/mol (7300 kJ/mol; Table 8.2)

                                           3 of 3

Using this information and the procedure used to solve problem #8.93, p 284, determine the effective nuclear charges, Zeff, during the 1st, 2nd, and 3rd ionizations of lithium (i.e. when the first electron is removed, when the second is removed and when the third is removed).

Explanation / Answer

Zeff = Z – S where S is a screening constant; S the number of core electrons

Zeff The nuclear charge the electron “experiences” is different than the number of protons in the nucleus. Zeff takes into account the shielding from other electrons (which lowers Zeff) and the penetration of the electron (which raises Zeff).

The screening constant can be evaluated using the following

Using this and the electronic configuration of Li we can calculate

Zeff for Li+ : That is during 1st ionization

During 2nd ionization

During 3rd ionization

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