## 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]4s^{1} |
[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]4s^{2} |
[Ar]3d^{1} = (1s)^{2}(2s2p)^{8}(3s3p)^{8}(3d)^{1} |
Z*=20 – [18(1.00)] = **2.0** |
11.871 eV |

Sc |
[Ar]4s^{2}3d^{1} |
[Ar]3d^{2} = (1s)^{2}(2s2p)^{8}(3s3p)^{8}(3d)^{2} |
Z*=21 – [1(0.35) + 18(1.00)] = **2.65** |
12.80 eV |

Ti |
[Ar]4s^{2}3d^{2} |
[Ar]3d^{3} = (1s)^{2}(2s2p)^{8}(3s3p)^{8}(3d)^{3} |
Z*=22 – [2(0.35) + 18(1.00)] = **3.30** |
13.58 eV |

V |
[Ar]4s^{2}3d^{3} |
[Ar]3d^{4} = (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 d^{1}, the same as any transition metal element. If the ion were 4s^{1}, 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^{+}.