# Hybridized orbital scheme for H2O

1. ### newton2008

3
this is the question from our project/ assignment:

The experimentally observed H-O-H bond angle of water is 104.5 degrees in water. Determine a hybridized orbital scheme that could account for the observed bong angle of 104.5 degrees. Redraw a fully labeled orbital overlap diagram using the hybridized orbitals.

I know that H2O has a tetrahedral shape, but how am i supposed to draw a orbital overlap diagram and hybridize water?

2. ### AbedeuS

134
Orbital outlay on O:

$$1s^2 2s^2 2p^6$$

Unhybradized bonding would have a sigma bond between the s orbital of hydrogen and a parallel p orbital, the next bond would also be between an s orbital of hydrogen and another, parallel yet available p orbital (lets say pz for one py for the other). This would result in the lone pairs being in the 2s and 2px orbitals, and give a funny othogonal geomety.

Hybradization of the orbitals to $$sp^3$$ would darastically decrease the repulsion the previous model produces. so the hydrogens would bond to the two avaliable sp3 orbitals.

As for the molecular orbital diagram, the lowest energy bonding interaction is a sigma interaction between hydrogen and the afore mentioned hybridization. So you will have two low energy bonding interactions, and two higher energy lone pairs that are not involved in bonding.

This is my guess anyway, when I do molecular orbital diagrams recently it is usually after doing a whole load of group theory, and we generally dont take lone pairs into consideration in lectures (yet).

3. ### pkleinod

111
Here is a thought about hybrid orbitals to get you started.

You could construct hybrid orbitals h1 and h2
from an s-orbital, s, and two p-type orbitals p1 and p1.
Choose p1 to be px and p2 to be

$$p_2 = p_x \cos{\theta} + p_y \sin{\theta}$$,

where $$\theta = 104.5^o$$ is the bond angle and the
px and py are orthonormal. The hybrid
orbitals are then

$$h_1 = N(s + a p_1)$$

$$h_2 = N(s + a p_2)$$.

The requirement that h1 and h2 be orthogonal

$$a^2 = -1/\cos{\theta}$$.

N is obtained by normalising. If desired, a third hybrid could be
constructed with a third p-type orbital, p3, to
accommodate a lone pair, leaving the other lone pair in the pz
orbital. Or a set of four orbitals could be constructed by including
pz in the linear combination defining the hybrids.

Perhaps this will give you a start in the right direction.

4. ### vivesdn

104
Just from a qualitative point of view: you have 4 valence orbitals and 6 electrons. With or without hybridation, you have to place 6+2 electrons (coming from the hydrogen) on these 4 orbitals (+ 2 antibonding orbitals that will be produced when combined with 2 s-type hydrogen atoms). So all of them will have pairs of electrons, two of them being bonding orbitals.

Now you can play with some drawings.
* You can try to bond one H with s-type orbital and the other with one p-type orbital. This makes nonsense as the combination will have differents energies for both bondings.
* You can first combine the s-type orbital with a p-type one. You'll loose the spherical symetry of s-type to get two orbitals following the axis of the p orbital. One of them will have more e-density on one side and the other, on the opposite side. This configuration will produce two sigma bonding with both H of the same energy, as they are equivalent. the orthogonal p-type orbitals populated each with an electron pair, will have 90º interactions with each other and with the bonding pairs.
* You can combine first the s-type with two p-type. You'll get three orbitals that now lay on a plane pointing at 120º intervals. This combination will still interact with one lone pair at 90º, and at 120º with the other. This configuration is not symmetric, as both lone pairs will have different energy.
* You can combine the s-type with the 3 p-type. This configuration is tetrahedral, as you'll get 4 equivalent orbitals pointing to four directions maximizing the angular distance among them. You'll have 2 bonding pairs with the Hs and two lone equivalent pairs. You'll get the greatest angular distance in this way.

As the two bonding pairs and the two lone pairs are not equivalent, you can expect a slight stretch from the fully symmetrical tetrahedral shape.