# Hybridized orbital scheme for H2O

• newton2008
In summary, the question is asking for a hybridized orbital scheme that could account for the observed 104.5 degrees bong angle of water.
newton2008
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?

pleasee.. any help is appreciated

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 don't take lone pairs into consideration in lectures (yet).

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
leads to the condition

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

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.

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.

## 1. How does the hybridized orbital scheme for H2O explain its molecular shape?

The hybridized orbital scheme for H2O explains its molecular shape by utilizing the concept of hybridization, in which the atomic orbitals of the central oxygen atom undergo mixing to form hybrid orbitals. In the case of H2O, the 2s and 2p orbitals of oxygen mix to form two sp3 hybrid orbitals, which are oriented at an angle of 104.5 degrees from each other. This results in the characteristic bent shape of the H2O molecule.

## 2. What is the significance of the hybridized orbitals in H2O?

The hybridized orbitals in H2O play a crucial role in determining the molecule's geometry, bond angles, and bond strengths. They allow for a more accurate representation of the molecule's shape compared to the simple valence bond theory, which only considers the overlap of atomic orbitals. The hybridized orbitals also explain the observed tetrahedral bond angle of 104.5 degrees, which is not possible with the individual atomic orbitals.

## 3. How do the hybridized orbitals affect the polarity of H2O?

The hybridized orbitals in H2O contribute to its polarity by influencing the distribution of electron density in the molecule. The two lone pairs of electrons on the oxygen atom occupy two of the sp3 hybrid orbitals, while the other two are used to form sigma bonds with the hydrogen atoms. This results in a bent molecular geometry with a slight negative charge on the oxygen atom and a slight positive charge on the hydrogen atoms, making H2O a polar molecule.

## 4. Can the hybridized orbital scheme for H2O be applied to other molecules?

Yes, the hybridized orbital scheme can be applied to other molecules with similar geometries, such as NH3 (ammonia) and CH4 (methane). These molecules also have sp3 hybridized orbitals, resulting in a tetrahedral geometry with bond angles of 107 degrees and 109.5 degrees, respectively. However, the specific hybridization of each atom may vary depending on factors such as electronegativity and molecular geometry.

## 5. How does the hybridized orbital scheme for H2O differ from the molecular orbital theory?

The hybridized orbital scheme for H2O is a type of valence bond theory, which describes the bonding in terms of overlapping atomic orbitals. In contrast, the molecular orbital theory considers the entire molecule as a whole, taking into account the interactions between all of the atomic orbitals. While both theories can explain the molecular properties of H2O, they use different models and approaches to do so.

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