# Explain the hybridisation of Cl atom in Cl2O7 or XeOF4?

1. Sep 13, 2006

### konichiwa2x

Hi, can someone explain the hybridisation of Cl atom in Cl2O7 or XeOF4?

Last edited: Sep 13, 2006
2. Sep 16, 2006

### ShawnD

It's been a while so I might be a bit wrong.

hybridization = [# of connected atoms] + [# of lone pairs]

1) Lone pairs of electrons?
First you look at oxidation numbers to see how many bonds something should have. In Cl2O7, the O7 is -14, which leaves each Cl at +7 (14 / 2 = 7). Remember that for each bond, only one of the electrons 'belongs' to that particular atom. This leaves us in a good position because chlorine should have 7 valence electrons, and we've found that each Cl has 7 bonds; that means no lone pairs.

2) How many atoms are connected?
Draw it out, what does it look like? I would guess Cl2O7 is two ClO3 groups joined with an oxygen. That means each Cl has 4 oxygens connected to it.

3) Fill in your ultimate equation from god
4 atoms connected + 0 lone pairs = 4 thingies.
Add your S and P to get 4. SP3 (1 + 3 = 4). Cl2O7 is SP3 hybridized.

XeOF4 is trickier. O is -2, F4 is -4, so that makes Xe +6. Xe should have 8. The 2 missing electrons are a lone pair.
5 atoms are connected
[5 atoms] + [1 lone pair] = 6 thingies
edit: Orbitals contain 2 electrons, so S cannot be bigger than 1, P cannot be bigger than 3, and D is limited something higher; can't remember. As Cesium says below, having 6 thingies makes it SP3D2.

It seems like I missed something. Can somebody verify this? (thank you Cesium)

Last edited: Sep 16, 2006
3. Sep 16, 2006

### Cesium

SP5? There are only 3 p orbitals so it uses the next 2 d orbitals. Sp3d2.

4. Sep 16, 2006

### ShawnD

You're right. Each orbital has 2, so 6p electrons is 3 orbitals. I knew that sounded wrong because I don't think I've ever seen "SP5" anywhere. Hopefully he hasn't read the original post, or he may be stuck remembering some wrong information.

5. Sep 17, 2006

### Gokul43201

Staff Emeritus
Merely having 4 hybrid orbitals doesn't mean they must necessarily be sp3 (instead of say, d2sp). In this particular case, if you go through the rigamarole of "promotion and hybridization" starting from the atomic orbitals of the Cl atom, you will find that it is indeed sp3 .

Again, this is not the only way to make 6 hybrid orbitals, but figuring between sp3d2 (high-spin) and d2sp3 (low-spin) is much trickier (depends on the ligand field) and is usually skipped in an introductory course.

6. Sep 18, 2006

### konichiwa2x

thanks a lot shawn! that was really helpful

thats a doubt I alwrays had. When should you promote an electron and when should you not?? Can you please explain that?

7. Sep 18, 2006

### Gokul43201

Staff Emeritus
I can try.

The reason for promotion is that, at the end of the process, you want to effect a reduction in energy. (If promotion and hybridization results in an increase in energy, it would not be favorable). First let's get some of the theory out of the way.

1. Promotion is an inherently energy increasing process - you are "promoting" a low energy electron to a higher energy orbital. The reason this is still favored is that it increases the number of available bonding orbitals, and hence provides scope for a larger reduction in energy from more hybridisation. The net effect must be a reduction in energy, so you can only promote electrons by a certain extent. Promotion from ns to np is usually pretty inexpensive, as is promotion from (n-1)d to ns and np. In some rare cases, it is feasible to promote from np to nd, but this is to be noted as a larger increase in energy.

2. You only promote electrons from fully (doubly) occupied orbitals. For one thing, you do not increase the number of available orbitals by promoting an electron up from a singly occupied orbital, so that's a dealbreaker. Also, an electron in a doubly occupied orbital has a little extra energy from the Coulomb interaction with its orbitalmate. So the removal of this interaction energy helps to make the promotion more energy inexpensive.

3. You promote electrons to higher orbitals to achieve the right number of hybrid orbitals with the correct number of vacancies in them to accomodate the required number of atoms. I'm almost certain, that sounds all too vague, so let's try a 3-part example.

C is 2s^2 2p^2. 4 hybrid orbitals needed in CH4 are made by promoting a 2s electron to the empty 2p orbital and hybridising the 2s with the three 2p orbitals.

If you wanted to make C2H4, you only need 3 hybrid orbitals on each C-atom (this follows from Shawn's post). You could argue that one could make the sp2 hybrid orbitals without having any need to promote an electron. But the problem then would be that these 3 hybrid orbitals will have 4 electrons in them, making it possible to bond only with 2 atoms (since 3 orbitals can hold no more than 6 electrons).

N is 2s^2 2p^3. For making NH3, we must use 4 valence orbitals to bond with 3 external atoms. This is achieved by hybridising the 2s with the three 2p, to make four sp3 hybrids with 3 vacancies. No promotion is required and we have the right number of half-filled orbitals.

So, there are two constraints which eventually determine whether or not to promote electrons:

(i) the total number of hybrid orbitals needed (=#lone pairs + #attached atoms), and

(ii) the number of half-filled orbitals needed (=#attached atoms)

8. Oct 9, 2006

### ssttaacceeyy

what if there are lone pairs?? i have a couple of questions where i have to show the electron promotion and there are lone pairs in all of them exceptr for one.. and the one without the lone pairs is the only one that works out for me...