Hybrid Orbitals -Normalization Question

I’m working on a chemistry assignment and can’t understand how our sample problems get to the answers they get to. We have to normalize an equation, and I’m having trouble understanding how to do it. Or rather, I don’t understand and assumption made, and my textbook is no help and the prof maintains we’ve covered this in class, but I have no recollection of any explanation.

The sample problem is something like this:

F(sp) = c1(2s) + c2(2p) where c1 and c2 are constants and 2s and 2p are the wave functions for the 2s and 2p orbitals.

We normalise the equation by sqaring it, then taking the integral of it equal to one. I would think that to square this, we would get (c1(2s) + c2(2p))[sup]2[/sup], which would give us something like ax[sup]2[/sup] + bx + c = 1, like we did in high school.

But the assumption in the samples is that each term is normalized individually, which, although it makes it a hell of a lot easier to solve for c1 and c2, makes no sense to me based on what I know of mathematics. I’m not very good at math, and am having a lot of trouble with this course, especially the calculus since I haven’t had to do any at all in 2 1/2 years.

Can someone explain to me (in as easy terms as possible) why 2s and 2p orbitals are normalised individually and not treated as one equation? Is this a mathematical rule, or something much more to do with the theory behind hybridization? Is this question to tough to answer over the net?

Do I need to explain anything further?

Thanks anyways!

mnemosyne returns to her homework, counting down the days til the end of the course in the back of her head…

[fixed coding]

[Edited by bibliophage on 11-19-2001 at 09:54 PM]

PREVIEW PREVIEW PREVIEW!! ESPECIALLY with subs!!

Ok, the thing to remember is the special properties of wavefunctions.

Psi[sub]m[/sub]*Psi[sub]n[/sub]=0 where m!=n (now I’ll be durned if I can remember the name for this property, so I’ll have to go look it up…aha–orthogonal)

So, (c[sub]1[/sub]2s+c[sub]2[/sub]2p)[sup]2[/sup]
=c[sub]1[/sub][sup]2[/sup]2s[sup]2[/sup]+2c[sub]1[/sub]c[sub]2[/sub]2s2p+c[sub]2[/sub][sup]2[/sup]2p[sup]2[/sup]
=c[sub]1[/sub][sup]2[/sup]2s[sup]2[/sup]+0+c[sub]2[/sub][sup]2[/sup]2p[sup]2[/sup]

I assume you know where to take it from there.

As for why it’s done…you will agree with me that it’s easier? Also, keep in mind that for anything past H, the exact orbitals are not known–we can only get (admittedly very good) approximations using assumptions of electron-electron repulsion and whatnot. So this lets you normalize an MO without actually knowing what the 2p and 2s orbitals are.

I see on preview that Myrr21 and I have the same thing to say. Eh, I’ll say it anyway.

Okay, I’ve only done some really fundamental quantum physics in my time, so I can only give you a best guess. Here goes…

Wave functions are as a rule normalized, so (2s) and (2p) are already normalized, since they are wave functions in themselves. So that means that:

Int((2s)[sup]2[/sup]) = 1 and Int((2p)[sup]2[/sup]) = 1

They are also orthogonal, which means:

Int((2s)(2p)) = 0

Now we want to normalize the wave function F, which is a linear combination of (2s) and (2p).

Int(F[sup]2[/sup]) = 1
= Int([c[sub]1/sub + c[sub]2/sub][sup]2[/sup])
= Int(c[sub]1[/sub][sup]2/sup[sup]2[/sup] + 2c[sub]1[/sub]c[sub]2/sub(2p) c[sub]2[/sub][sup]2/sup[sup]2[/sup])
= c[sub]1[/sub][sup]2[/sup]Int((2s)[sup]2[/sup]) + 2c[sub]1[/sub]c[sub]2[/sub]Int((2s)(2p)) + c[sub]2[/sub][sup]2[/sup]Int((2p)[sup]2[/sup])
= c[sub]1[/sub][sup]2[/sup] + c[sub]2[/sub][sup]2[/sup] = 1

Upon which Myrr21 realizes that he left out the (very important) integral bit…and is talking about himself in the third person.

So just put an integral in front of everything I wrote.

Thanks to both of you! My prof finally explained it today AFTER we handed in the assignment, but I think I still got it right. I just dropped a couple of terms and got something kinda like the sample answers…anyhow, it worked :-). I’m tempted to print out your explanations and give them to the prof saying “see?!? THIS is how you show us what to do! Going from the first line to the last, with no explanation, is CRUEL!!”.
Thanks again! and don’t be surprised to see me here again asking more questions (exams are a’comin’!!!) :slight_smile: