2
$\begingroup$

I was attempting the problems in Chapter 1 of Orbital Interactions in Chemistry, and found a slight problem.

ftp://ftp.wiley.com/public/sci_tech_med/orbital_interactions_2e/Chapter%201%20-%20Answers.pdf

For Q1 (d), I obtained the same secular determinant as the question. However, my answer was slightly different as solving the determinant yielded completely different answers for me.

For the question, H12 (resonance integral) was given as -14.18 eV, S12 (overlap) was given as 0.596, and the energy of the 1s hydrogen atomic orbital was given as -13.60eV. However, plugging these numbers into the secular determinant yielded me orbital energies of -18.26eV, -13.6eV and 41.07eV (the answer given was -33.657eV, -13.6eV, 6.457eV).

Here's the question, for reference. Is my answer incorrect, and if so, how is it incorrect? Thanks!

Here's the question.

enter image description here

Any help would be appreciated.

$\endgroup$
2
  • 2
    $\begingroup$ I don't see why it's too broad (even before the newer edit). 1d and 1e are very closely related questions, and the answer to them is practically the same, as I've already illustrated in my answer. The only thing that might be useful is if the question were typed out (instead of attached as a screenshot), but that has nothing to do with it being too broad or not. $\endgroup$ – orthocresol Mar 19 '19 at 12:08
  • $\begingroup$ I agree with orthocresol. The question itself is somewhat hard to read and would benefit from an entire reshape. Wiley is (unfortunately) also not very well known for stable links, so including only the relevant information from that pdf file would be advisable. All in all, I am leaning towards reopening, but I would at the moment not want to overrule a community decision. $\endgroup$ – Martin - マーチン Mar 19 '19 at 14:33
7
$\begingroup$

I ran the calculations through the MATLAB command window and obtained the same answers as you:

>> H = [-13.60 -14.18 0; -14.18 -13.60 -14.18; 0 -14.18 -13.60]
>> S = [1 0.596 0; 0.596 1 0.596; 0 0.596 1]
>> energies = eig(H,S)

energies = 

  -18.2615
  -13.6000
   41.0717

On the other hand, if you assume that $S_{12} = 0$ i.e. $S = I_3$ (3x3 identity matrix), then you get the same result as the authors:

>> energies = eig(H)       % or eig(H, eye(3))

energies = 

  -33.6535
  -13.6000
    6.4535

I find it slightly telling that the authors explicitly comment that the bonding orbital is stabilised by exactly as much as the antibonding orbital is destabilised. If you solve the secular equations without explicitly substituting in numbers, you should find that this can only be true if the overlap matrix is set to the identity matrix.

The same is true of part (e); you only get the authors' answer by setting $S = I_3$.

I would consider this an erratum in the provided answers.


To find the eigenvalues, use instead [energies, MOs] = eig(H,S). You can try this block of code online. Note that this website has Octave instead of MATLAB, so energies and MOs need to be reversed:

H = [-13.60 -14.18 0; -14.18 -13.60 -14.18; 0 -14.18 -13.60];
S = [1 0.596 0; 0.596 1 0.596; 0 0.596 1];
[MOs, energies] = eig(H,S);
disp(energies);
disp(MOs);

energies is a 3x3 matrix with the eigenvalues as the diagonal elements. MOs is a 3x3 matrix, where each column vector is the MO corresponding to the relevant eigenvalue (i.e. the left-most column is the MO corresponding to the first eigenvalue in energies).

$\endgroup$
2
  • $\begingroup$ could I inquire what the correct wave functions for (d) would then be? I got the Bonding MO as 0.3684𝜒1 + 0.5207𝜒2 + 0.3684𝜒3. $\endgroup$ – ANZGC FlyingFalcon Mar 19 '19 at 14:07
  • 1
    $\begingroup$ @ANZGCFlyingFalcon, if you learn a little bit of coding, it's easy to verify it for yourself instead of having to rely on me or anybody else. This is around five lines of code. For now, though, please see the addendum to the original answer. $\endgroup$ – orthocresol Mar 19 '19 at 16:36

Not the answer you're looking for? Browse other questions tagged or ask your own question.