CHM 501

12. Calculate Z* for the highest energy electron in the following ions: K+, Ca+, Sc+, Ti+, V+. Does Z* account for the observed values of the second ionization potential of these elements? Why or why not? Some have argued that Ca+ should be considered a transition metal. Give an argument to support this contention.

Answer

Atom

neutral configuration

cation configuration

Slater Z*

Second Ionization Potential

K

[Ar]4s1

[Ar] = (1s)2(2s2p)8(3s3p)8

Z*=19 – [7(0.35) + 8(0.85) + 2(1.00)] = 7.75

31.625 eV

Ca

[Ar]4s2

[Ar]3d1 = (1s)2(2s2p)8(3s3p)8(3d)1

Z*=20 – [18(1.00)] = 2.0

11.871 eV

Sc

[Ar]4s23d1

[Ar]3d2 = (1s)2(2s2p)8(3s3p)8(3d)2

Z*=21 – [1(0.35) + 18(1.00)] = 2.65

12.80 eV

Ti

[Ar]4s23d2

[Ar]3d3 = (1s)2(2s2p)8(3s3p)8(3d)3

Z*=22 – [2(0.35) + 18(1.00)] = 3.30

13.58 eV

V

[Ar]4s23d3

[Ar]3d4 = (1s)2(2s2p)8(3s3p)8(3d)4

Z*=23 – [3(0.35) + 18(1.00)] = 3.95

14.65 eV


 

Z* nicely parallels the second ionization potential. It is hard to ionize K+ and this is indicated by the large Z*. All of the other elements ionize more readily and the second ionization potentials steadily increase along with Z*.

The electron configuration for Ca+ is d1, the same as any transition metal element. If the ion were 4s1, then the Slater Z* would be = 20 – [8(0.85) + 10(1.00)] = 3.20, which is out line with the observations. Z* = 3.2 suggests an ionization potential closer to Ti+.