## Filling the hybrid orbitals?

$sp, sp^{2}, sp^{3}, dsp^{3}, d^{2}sp^{3}$

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### Filling the hybrid orbitals?

When we are drawing an Aufbau diagram for a hybridized orbital, should we fill those orbitals completely, with two spin-paired electrons in every orbital before putting electrons in the 2p orbital? Or should we place one electron in each hybrid orbital and p orbital (parallel spins throughout) before pairing electrons? In other words, which approach involves a lower energy difference?

Girija_3E
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### Re: Filling the hybrid orbitals?

I am not sure if this is what you're asking, but when I draw hybridization orbitals, I normally start by drawing the electrons in their appropriate (s or p) orbitals at their ground state (first fill the s orbital, then fill the p orbital with electrons that have a parallel spin before pairing them, etc.). Next, I move the paired electron from the s orbital to an additional, p-orbital, so that now each orbital has 1 electron, all with parallel spins.

For example: C2H4
In its ground state:
2s _ (two electrons, opposite spins)
2p _ _ (1 electron in each orbital, parallel spins, 2 electrons total)

When it's hybridized:
2s _ (1 electron)
2p _ _ _ (1 electron in each orbital, parallel spins, 3 electrons total)
• In this case, one of the electrons in the s-orbital moved to the p-orbital, and all electrons have a parallel spin, so carbon has a 2sp^2 hybridization!
• Additionally, now instead of having 2 unpaired electrons, the carbon has 4 unpaired electrons.

Elizabeth Bamishaye 2I
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### Re: Filling the hybrid orbitals?

Electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels, so you would complete the orbital until moving on to the next.

804899546
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### Re: Filling the hybrid orbitals?

For a molecule that has 2sp^2 hybridization like in C2H4 shown above, what is the difference between that and sp^2 hybridization? What does the first coefficient actually represent?

Shannon Wasley 2J
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### Re: Filling the hybrid orbitals?

Each carbon in C2H2 is sp^2 hybridized. There are two carbons in the equation, which means the hybridization is 2sp^2

McKenna disc 1C
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### Re: Filling the hybrid orbitals?

Elizabeth Bamishaye 1E wrote:Electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels, so you would complete the orbital until moving on to the next.

Yes, so spin-pair in the s-orbital and then parallel spins in the p-orbital before spin-pairing in the p orbital. (Right?)

Elizabeth Bamishaye 2I
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### Re: Filling the hybrid orbitals?

McKenna disc 1C wrote:
Elizabeth Bamishaye 1E wrote:Electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels, so you would complete the orbital until moving on to the next.

Yes, so spin-pair in the s-orbital and then parallel spins in the p-orbital before spin-pairing in the p orbital. (Right?)

Yes, that's correct!

Akshay Anand
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### Re: Filling the hybrid orbitals?

Why wouldn't the C2H4 molecule have sp^3 hybridization?

AtreyiMitra2L
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### Re: Filling the hybrid orbitals?

Akshay Anand wrote:Why wouldn't the C2H4 molecule have sp^3 hybridization?

There are 2 areas of electron density surrounding C. C has a triple bond with C and a single bond with hydrogen. Therefore, it is sp.

AtreyiMitra2L
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### Re: Filling the hybrid orbitals?

804899546 wrote:For a molecule that has 2sp^2 hybridization like in C2H4 shown above, what is the difference between that and sp^2 hybridization? What does the first coefficient actually represent?

We usually include the coefficient when it asks of for the composition of each atom. Then, we are describing each bond with the coefficients. When it asks for the hybridization of each atom, you don't include the coefficients.

AtreyiMitra2L
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### Re: Filling the hybrid orbitals?

Girija_3E wrote:I am not sure if this is what you're asking, but when I draw hybridization orbitals, I normally start by drawing the electrons in their appropriate (s or p) orbitals at their ground state (first fill the s orbital, then fill the p orbital with electrons that have a parallel spin before pairing them, etc.). Next, I move the paired electron from the s orbital to an additional, p-orbital, so that now each orbital has 1 electron, all with parallel spins.

For example: C2H4
In its ground state:
2s _ (two electrons, opposite spins)
2p _ _ (1 electron in each orbital, parallel spins, 2 electrons total)

When it's hybridized:
2s _ (1 electron)
2p _ _ _ (1 electron in each orbital, parallel spins, 3 electrons total)
• In this case, one of the electrons in the s-orbital moved to the p-orbital, and all electrons have a parallel spin, so carbon has a 2sp^2 hybridization!
• Additionally, now instead of having 2 unpaired electrons, the carbon has 4 unpaired electrons.

but should we be drawing both when its hybridized and non hybridized

Ilan Shavolian 1K
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### Re: Filling the hybrid orbitals?

McKenna disc 1C wrote:
Elizabeth Bamishaye 1E wrote:Electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels, so you would complete the orbital until moving on to the next.

Yes, so spin-pair in the s-orbital and then parallel spins in the p-orbital before spin-pairing in the p orbital. (Right?)

this only applies when talking about non-hybridized atoms. I believe when we start to talk about hybridized atoms, its only one electron in each region, all containing parallel spin. (someone plz correct me if i'm wrong)

Justin Lai 1C
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### Re: Filling the hybrid orbitals?

Doesn't sp^2 hybridization have three regions of electron density?