Computational Chemistry

(Steven Felgate) #1

[ 56 a] for ignoring internuclear repulsion and using a simple sum of one-electron
energies was that when the relative energies of isomers are calculated, by subtract-
ing two values ofEtotal, the electron repulsion and nuclear repulsion terms approxi-
mately cancel, i.e. that changes in energy that accompany changes in geometry are
due mainly to alterations of the MO energy levels. Actually, it seems that the (quite
limited) success of the EHM in predicting molecular geometry is due to the fact that
Etotalis approximatelyproportionalto the sum of the occupied MO energies; thus
although the EHM energy difference is not equal to the difference in total energies,
it is (or tends to be) approximately proportional to this difference [ 59 ]. In any case,
the real strength of the EHM lies in the ability of this fast and widely applicable
method to assist chemical intuition, ifprovided witha reasonable molecular
geometry.


4.4.2 An Illustration of the EHM: the Protonated Helium Molecule..


Protonation of a helium atom gives He–H+, the helium hydride cation, the simplest
heteronuclear molecule [ 60 ]. Conceptually, of course, this can also be formed by
the union of a helium dication and a hydride ion, or a helium cation and a hydrogen
atom:


He:þHþ! He:Hþ
or He^2 þþ:H"! He:Hþ
or HeþþH! He:Hþ

Its lower symmetry makes this molecule better than H 2 for illustrating molecular
quantum mechanical calculations (most molecules have little or no symmetry).
Following the prescription in points 1–7:



  1. Input structure
    We choose a plausible bond length: 0.800 A ̊(the H–H bond length is 0.742 A ̊
    and the H–X bond length is ca. 1.0 A ̊, where X is a “first-row” element
    (in quantum chemistry, first-row means Li to F, not H and He). The Cartesian
    coordinates could be written H 1 (0,0,0), He 2 (0,0, 0.800).

  2. Overlap integrals and overlap matrix
    The minimal valence basis set here consists of the hydrogen 1sorbital (f 1 ) and
    the helium 1sorbital (f 2 ). The needed integrals areS 11 ¼S 22 andS 12 ¼S 21 ,
    whereSij¼


Ð

fifjdv. The Slater functions forf 1 andf 2 are [ 61 ]

f 1 ðH1sÞ¼

z^3 H
p

! 1 = 2

e"zHjr"RHj (4.108)

160 4 Introduction to Quantum Mechanics in Computational Chemistry

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