hybridization of phosphorus (Sapling Q.11)

[Monica Soliman 3F]

I confused about how is PO4^3- has four hybrid orbitals and not five? I feel like I m not understanding how to do it? can someone tell me the right way to do reach four hybrid orbitals and not five.

[Isabella Chou 1A]

Looking at the Lewis structure, you see that the P atom has four regions of electron density. This means that there should be four hybrid orbitals.

Even though there is a double bond, this still counts as one region of electron density. This might make a bit more sense if we think about the molecule being represented by a resonance hybrid, rather than there being 3 single bonds and 1 double bond. The resonance hybrid of this molecule suggests that each P-O bond has partial double bond character. I hope this helps!

[Shrey Pawar 2A]

The type of bonds don't need to be taken into account to find the hybridization, just the electron density. In this case there are 4 regions of electron density so as a result it is sp3. Hope this helps!

[Raashi Chaudhari 3B]

I am also confused about this question.
It asks for the hybridization of P, but each structure has 4 regions of electron density so they would all have the same answer. So, is it just a trend that if P has 4 regions of electron density it will always be sp^3

[Shrey Pawar 2A]

It wouldn't be a trend for just P, I believe it would work for all hybridization. The hybridization is based on the amount of electron density regions so because there are 4 regions around P in this case, it is sp3. If there were 3 regions it would be sp2 and so forth. This is my understanding and I could be wrong but hope this helps!

[Abby Lam 3F]

The double bond counts as one region of electron density.

[Madisen Brown -1C]

Though one of the oxygens is double-bonded to phosphorus, this only constitutes as one region of electron density

[Taber Ball 1F]

Hi! This is another part of Sapling Q.11. I understood how your answers applied to the other diagram, but this diagram threw me for a loop. If any of you have advice on how to better understand and complete these questions it would be much appreciated! Thank you!

[Mingzi Yang 1E]

Hi! This is another part of Sapling Q.11. I understood how your answers applied to the other diagram, but this diagram threw me for a loop. If any of you have advice on how to better understand and complete these questions it would be much appreciated! Thank you!

P have three bonding pairs and one lone pair. There are four regions of electron density, so the hybridization will be sp^3.

[arisawaters2D]

During hybridization, are the "s" valence electrons separated if they are in an s orbital to make hybrid orbitals? In an example in the lecture for CH4, it showed 4 separate orbital arrows being created from 1 full 2s orbital and two half full 2p orbitals for the sp3 hybridization. Do we always separate the 'arrows' in the s orbitals in hybridization?

[Adam Bustamante 1I]

PO4 ^3- has four regions of electron density. Even though it has a double bond, it still counts as just one region (single, double, triple bonds = 1 region). The hybridization for P would then be sp3.

[Yichen Fan 3A]

Like most people have mentioned, double bond still counts as one region of electron density in hybridization. Also don't forget lone pairs as region of electron density as well.

[Will Skinner]

There are four regions of electron density because there are four bonds. The double bond only counts as one region of electron density, so the hybridization orbitals would be sp3.

[Shalyn Kelly 3H]

The above replies helped my understanding as well. For this question, we just look at phosphorus and count the bonds (regardless of single, double, etc) and lone pairs surrounding it. That number is the regions of electron density and we can find the hybridization from there.

[Talia Dini - 3I]

PO4^3- has four hybrid orbitals and not five because the double bond on the phosphorus atom only counts as one region of electron density.

[Chance Herbert 3A]

There are only four regions of electron density which results in the hybridization of sp³. It is important to note that we only consider bonding electron pairs and lone pairs as regions of electron density. This molecule would have the VSEPR formula AX₄, meaning that there are four regions of electron density. Double or triple bonds are considered as a singular region of electron density. Hope this helps :)

[Hannah Lechtzin 1K]

Hi! This is another part of Sapling Q.11. I understood how your answers applied to the other diagram, but this diagram threw me for a loop. If any of you have advice on how to better understand and complete these questions it would be much appreciated! Thank you!

I always think about it in terms of bonded electrons and lone pairs. In this part of the question each P atom has 3 bonds and 1 lone pair, so 3 unpaired electrons and 1 pair of electrons. That would mean that the electrons would be spread across 4 orbitals. Three orbitals would have an unpaired electron and one orbital would have the lone pair of electrons. sp3 would be 4 orbitals because s contributes one orbital and p3 contributes 3, so sp3 has to be the right hybridization. I hope this helps!

[Shannon Moore 2L]

The phosphorus atom has a sp3 hybridization because it has three regions of electron density (the double bond only counts as one region).
I find it helpful to count the number of atoms is attached to the atom in question (in this case 4) and add the number of lone pairs on the atom in question (in this case O) and use this number to determine the hybridization.
4 atoms attached + 0 lone pairs = 4 ----> sp3
1 --> s
2 --> sp
3 --> sp2
4 --> sp3
5 --> sp3d1
6 --> sp3d2

[Tobie Jessup 2E]

Hybridization correlates with the number of regions of electron density. This problem is confusing since there is a double bond, but a double bond is only considered to be 1 region of electron density, so for this problem there is 4 regions and therefore the hybridization is sp^3.

[rhettfarmer-3H]

Hi! This is another part of Sapling Q.11. I understood how your answers applied to the other diagram, but this diagram threw me for a loop. If any of you have advice on how to better understand and complete these questions it would be much appreciated! Thank you!

for this one there are multiple structures in one element so we are counting of each P element. it is clear that there is 4 regions of electron density of each p which corrects to sp^3.

[Megan Lu 3D]

Hi! This is another part of Sapling Q.11. I understood how your answers applied to the other diagram, but this diagram threw me for a loop. If any of you have advice on how to better understand and complete these questions it would be much appreciated! Thank you!

Hi! Like others have detailed above, we only need to consider electron densities when finding hybridization, not the type of bond or lone pair. We can see that phosphorous has four regions of electron density (one lone pair and three bonds); thus, the hybridization would be sp3.

[Arnav Saud 2C]

The main reason why this occurs is because there are only 4 regions of electron density. The double bond does not count as an extra region, which is why there are only 4.