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+.