Computational Chemistry

5.5.4.3 An Example of Population Analysis: H–Heþ

As a simple illustration of the calculation of atom charges and bond orders, consider
H–Heþ. From our Hartree–Fock calculations on this molecule (Section 5.2.3.6.5)
we have

P¼

0 :2020 0: 5097

0 :5097 1: 2864



and S¼

1 :0000 0: 5017

0 :5017 1: 0000



ð 5 : 229 Þ

Therefore ðPSÞ¼

0 :2020 0: 2557

0 :2557 1: 2864



ð 5 : 230 Þ

From Eq. (5.228), (PS) gives us

n 1 ¼ 0 : 2020

n 2 ¼ 1 : 2864

n 1 = 2 ¼n 2 = 1 ¼ 20 ðÞ¼: 2557 0 : 5114

Charge on H,qHFor this we needNH, the sum of all theNron H (Eqs. (5.216)
and (5.215)). There is only one basis function on H,f 1 , so there is only one relevant
Nrfor H, and forf 1 there is only one overlap, withf 2 , so the summation involves
only one term,n1/2. Using Eq. (5.215):

Nr¼N 1 ¼nrþ

1

2

X

s 6 ¼r

nr=s¼n 1 þ

1

2

ðn 1 = 2 Þ¼ 0 : 2020 þ

1

2

ð 0 : 5114 Þ¼ 0 : 4577

The sum of all theNron H has only one term,N 1 , since there is only one basis
function on H. Using Eq. (5.216):

NA¼NH¼

X

r 2 H

Nr¼N 1 ¼ 0 : 4577

The charge on H,qH, is the algebraic sum of the gross electronic population and
the nuclear charge: (Eq. (5.217)):

qA¼qH¼ZH#NH¼ 1 # 0 : 4577 ¼ 0 : 5423

Charge on He,qHeFor this we needNHe, the sum of all theNron He (Eq. (5.216).
There is only one basis function on He,f 2 , so there is only one relevantNrfor He,
and forf 2 there is only one overlap, withf 1 , so the summation involves only one
term,n2/1(¼n2/1):

350 5 Ab initio Calculations