Concise Physical Chemistry

c17 JWBS043-Rogers September 13, 2010 11:28 Printer Name: Yet to Come

284 THE VARIATIONAL METHOD: ATOMS

Z:= 2. 000 a := 2. 000 b := 2. 000

ε(a,b) :=

[

a^2

2

−Z·a+

a·b(a^2 + 3 ·a·b+b^2 )

(a+b)^3

]

a:=root

(

d

da

ε(a,b),a

)

a= 1. 601 ε(a,b)=− 0. 812

b:= 1. 601

a:=root

(

d

da

ε(a,b),a

)

a= 1. 712 ε(a,b)=− 0. 925

b:= 1. 712

a:=root

(

d

da

ε(a,b),a

)

a= 1. 681 ε(a,b)=− 0. 889

FILE 17.3 Mathcad©C calculation of the ionization potential of helium. An approximate

screening constant of 0.3 givesa=(z−s)∼= 1. 7.

the third iteration, approaching the experimental value. Notice that the calculated IP 1

on the second iteration is too large. Sometimes the solution of the iterative procedure

is approached asymptotically, and sometimes the approach is oscillatory.

Example 17.2 The Slater Orbital of Oxygen

Find the Slater orbital of oxygen.

Solution 17.2 First observe that the nuclear charge isZ=8. The 1s^2 part of the

screening constant on the valence electron is 2(0.85)=1.70 and the part of the

screening constant for the electrons in the 2s^2 and 2p^3 shell iss=5(.35)=1.75.

The interior electrons are more effective (0.85) in screening the valence electron than

the electrons that share the second valence shell (0.35). We do not count the valence

electron because it cannot screen itself. The screening electrons are always one less

than the number of electrons in the neutral atom, in this case 2+ 2 + 3 =7. The

total screening constant is 1.70+1.75=3.45. This gives an STO for oxygen:

s=2(0.85)+5(0.35)= 1. 7 + 1. 75 = 3. 45

Z(shielded)=

8 − 3. 45

2

φ(r)=re−(z−s)/na=re−(^8 −^3.^45 )/^2 a=re−^2.^28 /a

Other atomic STOs can be found by applying the same routine.