12-5 Perturbation at an Atom in the Simple H ̈
Chapter 12 Time-Independent Rayleigh–Schr ¨odinger Perturbation Theory12-5 Perturbation at an Atom in the Simple H ¨
uckel
MO Method
Perturbation theory can be used to estimate the effect of a change in the value of the coulomb integral Hkk at carbon atom k. This is normally given the value of α, but it might be desirable to consider a different value due, for example, to replacement of an attached hydrogen by some other atom or group. We take the unperturbed H¨uckel MOs and orbital energies as our starting point and let the perturbed value of Hkk be a + δα.
Therefore, H = δα (at center k only)
(12-52)
The first-order correction for the energy of φi, the ith MO, is W (1) = φ
i
i |H |φi
(12-53)
Now φi is a linear combination of the AOs χ:
φi =
cjiχj
(12-54)
j
and so W (1) =
c∗
1
j i cli χj |H |χl
(12-55)
j
l
but H is zero except at atom k, where it is δα, and so the sum reduces to one term: W (1) = c∗
i
ki cki δα
(12-56)
Summing over all the MOs times the number of electrons in each MO gives the firstorder correction to the total energy:
W (1) = δα nic∗kicki = δαqk
(12-57)
i where qk is the π-electron density at atom k. Thus the energy change to first order is equal to the change in Hkk times the unperturbed electron density at that atom. This can also be seen to be an immediate consequence of the alternative energy expression
E = qlαl + 2
plmβlm
(12-58)
l
l
EXAMPLE 12-5 Consider the methylene cyclopropene molecule, C4H4, in the HMO approximation. (See Appendix 6 for data.) At which carbon will a perturbation involving α affect the total π energy the most? Calculate the energy to first order if the value of α at that atom increases to α + 0.1000β. Calculate the energy change, to first order, of each of the four MOs.
Section 12-5 Perturbation at an Atom in the Simple H ¨uckel MO MethodSOLUTION The total energy will be most affected if the perturbation occurs at the atom having the greatest π -electron density, that is, at atom 4. To first order, the total π energy changes by (0.1000β)(1.4881) = 0.1488β, giving an energy to first order of 4α + 4.9264β + 0.1488β = 4α + 5.1112β. Each MO energy drops by (using Eq. (12-56), from lowest to highest, 0.00794β, 0.06646β, 0.0000β, 0.02559β. The sum of the first two of these, times 2 (because of having two electrons each), gives 0.1488β, which equals the first-order change in total π energy found above.
Calculation of first-order corrections to the MOs proceeds in a straightforward man ner using Eq. (12-19) (see Problem 12-17). One of the results of interest from such a calculation is the change in π -electron density at atom l due to a change in the coulomb integral at atom k. The differential expression is δq1 = (∂ql/∂αk)δαk = πl,kδαk
(12-59)
The quantity of interest πl,k is called the atom-atom polarizability. We will now derive an expression for πl,k in terms of MO coefficients and energies.
We assume that our MOs fall into a completely occupied set φ1, . . . , φm and a completely empty set φm+1, . . . , φn. We take δα to be positive, which means that center k is made less attractive for electrons. According to Eq. (12-19) the j th MO is, to first order, n
φj|H|φi
φj + φ(1) = φ
φ
j
j + i
(12-60)
Ej − Ei i=1,i=j
where Ej is the orbital energy of φj . As indicated above, the integral vanishes except over AO k. Therefore,
n
c∗ c
kj ki δαk φ(1) =
φ
j
i
(12-61)
Ej − Ei i=1,i=j
Expanding φi in terms of AOs, we have
n
n
c
kj cki φ(1) = δα c
(12-62)
j
k
li χ
E
l
j − Ei i=1,i=j
l=1
We can write this as n
=
c(1)χ
(12-63)
lj
l
l=1
where
n
ckjckicli
c(1) = δα
(12-64)
lj
k
Ej − Ei i=1,i=j
The change in density at atom l equals the square of the perturbed wavefunction minus that of the unperturbed wavefunction and involves only the doubly occupied MOs 1 − m:
m
δql = 2 (clj + c(1))2 − c2
(12-65)
lj
lj
j =1
m
= 4 clj c(1)
(12-66)
lj
j =1
Chapter 12 Time-Independent Rayleigh–Schr ¨odinger Perturbation Theory where we have neglected (c(1))2 because the perturbation is assumed small. Substitut lj
ing Eq. (12-64) into (12-66) gives
m
n
cljckjck˙ιcli
δql = 4δαk
(12-67)
Ej − Ei j =1 i=l, i=j
This equation can be simplified further by recognizing that, for j and i