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anonymous
 one year ago
Hybridization
anonymous
 one year ago
Hybridization

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444279503247:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Which of the following clusters of orbitals could form a shape similar to that shown here (Figure 3) in the valence shell of an isolated atom or one about to enter into bonding with other atoms?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0These are the answers...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0five sp^3d and three sp^2 and one p orbital... Okay so I don't see how it could be five sp^3d orbitals

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Because an sp^3d orbitals looks like this dw:1444279684152:dw

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444279833146:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so if we had five of those laying on top of each other I don't see it looking the first image

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0those are unhybdirized D orbitals so also take a look at this: These are hybridized d orbitals dw:1444279990684:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so these are hybridizeddw:1444279982930:dw

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0Hybridization, orbitals this is YOU @Empty

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444280491106:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0answers are the following five sp^3d and three sp^2 and one p orbital...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0five sp^3d is separate from that of three sp^2 and one p orbital

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0@rafflesnaffle what element is this for?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no elements is given otherwise I wouldn't be asking for help lol

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I think this picture shows fairly well of how you can put the d orbital in there:

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0So we can get a better picture of this here's your link dw:1444280864261:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0nvm didn't do what I wanted it to do

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how is it five of those though?

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Well if you want to understand mathematically how they're combined, you have to realize that these orbitals are derived from quantum mechanics, solving the Schrodinger equation for a single electron atom. When they add, they add constructively or destructively and have interference. It's pretty hard to explain without actually just seeing the math but dw:1444281026465:dw So I'm trying to like draw them all on the same atom there, and there's a part you can't see is that when the lobes pinch off at a point, that node is where the sign of the wave function changes sign, so one lobe is positive and another is negative. I don't think I'm doing a very good job of explaining this so give me a second to think of a simpler way to introduce this.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can we start from scratch? Maybe that would help?

Empty
 one year ago
Best ResponseYou've already chosen the best response.3How much math and how much chemistry do you know and how much do you want to know? Because it gets pretty messy really fast to actually explain this I'm afraid!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444281352039:dw I am just showing you what I know. I've taken calc 14, differential equations and linear algebra, all physics, and all engineering courses. I have a strong back ground in math. This is my second chemistry course. it's been about a year since I have taken a chem course. This is gen chem 2. I took gen chem 1 about a year ago.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in gen chem 1 we only touched this subject. in gen chem 2 we are reviewing the material, but I feel like we are going into much more detail than when i learned it in chem 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are you still writing or is this just you username stuck?

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Oh awesome then you're good, since that's basically what you need to get to quantum mechanics. So all of these orbitals come from solving the time independent Schrodinger equation: \[\hat H \psi = E\psi\] What's important is \(\psi(x,y,z)\) is a solution to this differential equation above (which is written in an eigenvalue form don't worry about this too much since I know you can't see the derivatives in it, just letting you know it exists) The thing is that solutions to it are linear, and often times Fourier series represent arbitrary solutions to it, I won't go into too much detail here but I can go pretty indepth but all of this gets hairy so let's stay on track, the thing I want to talk about is the solutions \(\psi\) are called wave functions and can be chosen to take on positive and negative values. This is directly related to bonding and hybridization of orbitals. Orbitals extend throughout all space and tell you the probability of finding a particle there so if you integrate a wave function in this way over all space, you will get 1, which corresponds to 100% chance, for now I'll only look at one dimension since 3 dimensions isn't much more to get into. \[1 = \int_{ \infty}^\infty \psi^2 dx\] I am fedexing you through a lot, but the point is that now if you want to find out what an orbital is, you evaluate this integral: \[\int_{ \infty}^\infty x \psi^2 dx\] This will tell you the expectation value (most likely place to find an electron) of a point on the xaxis. So if we bring two wave functions together, an s and a p orbital might have wave functions that look like this: dw:1444282175920:dw Rember I said that the solutions are linear? This is hybridization right here, so we add them and get constructive and destructive interference: dw:1444282316571:dw So if we look at the squares of the wave functions, that's really what the orbitals look like, so this sort of interference stuff is sorta tucked away when you just see these nice pictures: dw:1444282369175:dw

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I didn't realize I wrote so much, so maybe start at the bottom with the picture to try to motivate yourself, I tried to make it like ODE/PDE Linear algebra lingoish so that you could at least relate and feel like you're getting a pretty accurate portrayal of the quantum mechanics but we've barely scratched the surface and I know I've left quite a few things unexplained... So don't hesitate to ask haha. I'll try not to avalanche you with any more stuff unless you ask for it haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That was a good explanation and it was a little more than I was asking for, but hey at least I learned something new.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am going to sleep on this. Will you be around tomorrow?

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Yeah I think overall the point is the pictures you look at are the squares of stuff, but the hybridization and bonding takes place 'before you square them' so seeing how adding things that appear to be positive and getting less of them in a picture is ultimately where your confusion is coming from I think on this. Yeah I'll be around totally.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay sounds good. I will return to this post tomorrow possibly with further questions. night

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sweetburger I've got a question.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Which of the following clusters of orbitals could form a shape similar to that shown here (Figure 3) in the valence shell of an isolated atom or one about to enter into bonding with other atoms?dw:1444320664039:dw
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