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konichiwa2x
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Hi, can someone explain the hybridisation of Cl atom in Cl2O7 or XeOF4?
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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.Cesium said:SP5? There are only 3 p orbitals so it uses the next 2 d orbitals. Sp3d2.
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 .ShawnD said:It's been a while so I might be a bit wrong.
Your ultimate equation is:
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.
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.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.
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 .
I can try.konichiwa2x said:thats a doubt I alwrays had. When should you promote an electron and when should you not?? Can you please explain that?
Hybridisation in chemistry is a concept that explains the mixing of atomic orbitals to form new hybrid orbitals that have different shapes, energies, and orientations than the original atomic orbitals.
The hybridisation of a Cl atom in Cl2O7 is sp3d3, also known as seven-fold hybridisation. This means that the Cl atom has seven hybrid orbitals formed by mixing one s orbital, three p orbitals, and three d orbitals.
The hybridisation of an atom in a molecule can be determined by counting the number of electron groups around the atom. Each electron group (bonded or lone pair) corresponds to one hybrid orbital. The total number of hybrid orbitals will determine the type of hybridisation.
The hybridisation of a Xe atom in XeOF4 is sp3d2, also known as six-fold hybridisation. This means that the Xe atom has six hybrid orbitals formed by mixing one s orbital, three p orbitals, and two d orbitals.
Hybridisation affects the shape of a molecule by determining the arrangement of atoms in space. The type of hybridisation will determine the number and type of bonds and lone pairs, which in turn determines the molecular geometry. For example, sp3 hybridisation results in a tetrahedral molecular geometry, while sp3d2 hybridisation results in an octahedral molecular geometry.