Thread: Why the bond angle of methane is 109 instead of 90?

I know that the sp3 hybrid orbitals form 109 angle, but why the angle is not 90. If the angle is 109 the repulsion is minimized or what? Show some comparations. Thank u.

Looks like homework...

Yeah, as far as I've learnt, the angle in methane is 109 (angle in a tetrahedral) to minimize the repulsion. In CH4, Carbon is the central atom with 4 sp3 lops pointing out to bond with Hydrogen, each lop contains 2e (after bonding with H), so they will repel each other. The tetrahedral is thus obtained, and it's explained using Valence bond theory (VB), and Lewis structure, though it's not really correct now (with MO theory). You actually can draw a tetrahedral, and calculate the angle.

But isn't the 2px and 2py orbitals make angle of 90 degrees? By ur theory all off the atoms will tend to have 109 angle to minimize the repulsion right?

Normally, the 2px and 2py is perpendicular... But in CH4, hybridization takes place, so three 2p(x,y,z) orbitals mix with 2s orbital generating 4 sp3 orbitals... It's what explained using VB theory.



Yep, I think 109 degree is to minimize the repulsion.[/img]

What about water?
H2O has an angle of 104.45° between the O2-molecules as well.
And a minimum of repulsion would seem to me 180° rather than 104.45°.
I know it's a dipole molecule, but I don't understand the connection.

sp3 hybridization is expected to form in water as well, so an angle of 109.5 is expected. However, because of the greater repulsion between the two lone pairs of Oxygen, the HOH angle thus becomes smaller.



(I draw it myself so it's not nice)

Water has the formula H2O, that doesn't mean it is a linear molecule. The 2s and three 2p (x,y,z) orbitals of O mixed together forming FOUR sp3 orbitals, two of them have one electron each (so that bond with 2 hydrogen atoms with 1e each). The other two, each has a lone pair of electrons. The repulsion between the two lops containing lone pair is greater than the two lops containing bonding electrons. Therefore, an angle of approximately 104 degree is observed instead of 109.5.

Valence bond theory, Lewis structure and VSEPR are applied again here. These two are quite convinient in explaining simple molecule structures so are still often used. However, it's not correct, since electrons do NOT localize in a small region as they predicted, but a large region over the whole molecule. That's when Molecular Orbital theory (MO Theory) comes.

Hope that can help .

It's the explanation of CH4 structure using Valence bond theory and VSEPR. Although it's no longer a correct explanation, it's still ok in almost all the cases. We now still use VSEPR to predict the structure and bond angles in a molecule. But to understand deeply, we must use MO theory.




This is the structure of water in MO theory:

Originally Posted by mastermind

What about water?
H2O has an angle of 104.45° between the O2-molecules as well.
And a minimum of repulsion would seem to me 180° rather than 104.45°.
I know it's a dipole molecule, but I don't understand the connection.

In water each proton's 1s orbital forms a sigma bond to a p orbital on the oxygen. Since the p orbitals are all 90 degrees from each other the idea bond angle would be 90 degrees, but repulsion between the protons pushes them apart to around 105 degrees.