Question

Can you show me mathematically that the p orbital has 2 phases?
And how did we deduce this well known shape of p orbitals?

I have part of an explanation here with trig:

http://sketchtoy.com/60156646

What I don't understand in this is that, to me, the length indicated by cos²φ is supposed to be cosφ but they're considering cosφ to be the hypotenuse!

What do you think? Can you help? I'm really confused
Thanks in advance.

Answer 1

Is this for physical chemistry?

Answer 2

Yes! It's not really required as it seems that students take such things on faith and move on but I'm curious

Answer 3

Notice that this is really a function in polar coordinates saying \[r= \cos \phi\] You're thinking about what happens when you have a constant radius, but here the radius changes depending on what angle we're at.

But don't fear, what you really should think is: What is cos(phi)? It's the adjacent side divided by the hypotenuse, correct? \[\cos \phi = \frac{\cos^2 \phi}{\cos \phi}\]

I don't know if that clears it up, but you just need to remember that the radius depends on phi and is not a constant. Here's a little graph to show: http://www.wolframalpha.com/input/?i=r%3Dcos%20theta&t=crmtb01

Answer 4

Thank you! I understand better now but one thing still bothers me. How come it's okay to just say that the length is equal to cos²φ? You can't just pick any length as long as it cancels out and gives the right thing, right?

Answer 5

Well you're thinking of two different triangles as if they're the same.

All the graph of r=cos(theta) is saying is that you can start plugging in angles and at every angle the radius changes based on that.

So if we graph a bunch of points, maybe it'll be clearer

(theta, r)

a=(0,1)
b=(pi/4, sqrt(2)/2)
c=(pi/2, 0)
d=(3pi/4, -sqrt(2)/2)
e=(pi, -1)

Answer 6

yeah, i just accepted it. I didn't, nor do I have plans on becoming a physical chemist.