Hybridization of Atomic Orbitals
We can use Lewis dot structures to determine bonding patterns in molecules. We can then use VSEPR to predict molecular shapes, based on the valence electron pairs of the Lewis structures. Once we know a molecular shape, we can start to look at the physical properties of compounds. For example, we should now be able to predict which molecules will be polar. Polarity exists when there is a separation of charge within a molecule. This will arise from polar bonds within the molecule, due to differences in electronegativity values between bonded atoms. For example, HF is a polar compound. Fluorine is much more electronegative than hydrogen and the shared pair of bonding electrons will spend more time near the F nucleus, than near the H nucleus.
The direction of a dipole moment (charge imbalance) is usually indicated by the presence of an arrow, as shown below for HF.
This indicates that the H will carry a partial positive charge (d+) and the F will carry a partial negative charge (d -). All diatomic molecules containing atoms of different electronegativities will be polar molecules. This will affect their physical properties (melting and boiling points, solubilities, etc.).
In larger molecules (more than two atoms), the polarity of the overall compound will be determined by the presence of polar bonds and the molecular shape.
For example, we can compare carbon dioxide, CO2 to sulfur dioxide, SO2. Their Lewis structures are shown below.
Carbon and sulfur have the same electronegativity, much less than that of oxygen. So, in both compounds the bonds will be equally polar. However, they have very different physical properties, CO2 boils at -78oC, and SO2 boils at +22.8oC, a 100o difference. This must depend on more than just the presence of the two polar bond in each molecule. What makes the difference is the molecular shape.
CO2 will be a linear molecule, because there are only two electron pairs on the central carbon atom. It will have the shape shown below.
As indicated by the arrows, there are two very polar bonds in this molecule. However, because of the molecular shape of CO2, they are pointing in opposite directions, and will cancel out. CO2 is a non-polar compound, due to its molecular shape.
Shown below is the shape of an SO2 molecule. Its molecular geometry will be trigonal because of the three valence electron pairs on sulfur, two bonding pairs and a lone pair. This will give it a bent molecular shape. In this molecule, the dipoles are not pointing in opposite directions, and will not cancel out. They will, in fact add, and give a net dipole moment. SO2 is a polar compound, which explains (as we will soon see) its elevated boiling point.
(Quiz For a similar example, draw the Lewis structures of BF3 (-99) and NH3 (-33)
Another indication of the importance of molecular shape can be seen by comparing the physical properties of CHCl3 and CCl4. CHCl3 dissolves in water, and CCl4 does not. Why?
They will both have tetrahedral geometries, with 4 valence pairs of electrons on each C. The C-Cl bonds will all be polar. Their shapes are shown below.
In the CHCl3 molecule, the three polar C-Cl bonds add (vector addition) to give a net dipole moment to the molecule. In CCl4 , the four polar C-Cl bonds will cancel out, making this a non-polar molecule. Water is a polar solvent, which only interacts with other polar species, "likes dissolve likes".
We can use VSEPR to predict molecular geometries and
the physical properties of molecules. What this model does not
explain is the nature of chemical bonding, the chemical properties of
molecules. Lewis theory proposed that chemical bonds form when
atoms share electron pairs.
Now, let's consider methane, CH4. The Lewis structure is:
VSEPR predicts a tetrahedral shape, based on the four atoms bonded to the central atom.
Hydrogen atoms have the electron configuration 1s1.
Carbon atoms have the electron configuration [He] 2s2 2p2.
Each of the hydrogen atoms can contribute 1 electron to a C-H bond. However, there are only two p orbitals in the carbon atoms that have unpaired electrons, and four equivalent C-H bonds need to form. To create four equivalent bonding orbitals in carbon, the atomic wave functions, Y, for the 2s and three 2p orbitals are mixed mathematically, to give new molecular wave functions, and molecular orbitals. Mixing one s orbital with three p orbitals will produce four hybrid orbitals, called sp3 orbitals. The shape and orientation of these new molecular orbitals are shown below:
The molecular, sp3 orbitals are arranged in a tetrahedron, with bond angles of 109.5o. Each of the 1s orbitals of H will overlap with one of these hybrid orbitals to give the predicted tetrahedral geometry and shape of methane, CH4.
Hybridization also changes the energy levels of the orbitals. The 2s orbital of carbon is lower in energy than the 2p orbitals, since it is more penetrating.
After hybridization, all four hybrid orbitals have the same energy, lower than p orbitals, but higher than s orbitals. The four valence electrons on carbon can be added to the energy diagram ( ). Each of the hydrogens has one valence electron in its 1s orbital ( ). These will pair up with the carbon electrons to form four s (sigma) bonds. These are called sigma bonds (Greek for s) because they are formed from hybridized orbitals, which result from s orbitals.
The overlap of the hydrogen s orbitals and the carbon sp3 orbitals puts the electron density directly between the nuclei. This is a property of s-bonds.
We can now look at the bonding in ammonia, NH3. The Lewis structure is shown below.
VSEPR predicts tetrahedral geometry (one lone pair and three bonding pairs of electrons) and trigonal pyramidal shape. We will again need four hybrid orbitals, obtained by mixing one s and three p atomic orbitals in nitrogen. Nitrogen has five valence electrons ( ).
Three hydrogen atoms with one unpaired electron apiece ( ) will overlap their 1s orbitals with the three available sp3 orbitals on the nitrogen. This leads to the formation of three s bonds and a lone pair of electrons occupying the fourth hybrid molecular orbital.
Next, consider SF4. The Lewis structure is shown below.
VSEPR predicts trigonal bipyramidal geometry (one lone pair and 4 bonding pairs) and see saw shape. In order to have five hybrid orbitals, we need to mix five atomic orbitals from sulfur. The s and p orbitals give a total of only 4 hybridized orbitals (one s + 3p) so we will now add one d orbital to the mix. Since sulfur is in Period 3 (n = 3), it will have five d orbitals. Mixing one of the d orbitals with the other four atomic orbitals (s and three p) will give a hybridization called dsp3, which provides five equivalent molecular orbitals.
The fluorine atoms are sp3 hybridized (3 lone pairs and one bonding pair), and the overlap of each sp3 orbital on fluorine with a dsp3 orbital on sulfur will form a s bond.
For compounds, like SF6, which require six equivalent molecular orbitals, mix six atomic orbitals, s + p + p + p + d + d. These would have d2sp3 hybridization and would form six s bonds.
Next we can look at formaldehyde, CH2O. Its Lewis structure is shown below:
VSEPR predicts that this compound will have trigonal planar geometry and shape, since there are three atoms bonded to the central carbon atom, and no lone pairs. This means that there must be three equivalent molecular orbitals. Following the method we used in the previous examples, we will mix 3 atomic orbitals to form 3 hybridized molecular orbitals.
We will call the hybridized orbitals sp2 orbitals, since we mixed the s with two of the three p atomic orbitals. The sp2 orbitals have the predicted trigonal planar geometry. Notice that one of the p orbitals was not changed in the hybridization, and has a higher energy level than the hybridized orbitals. The unhybridized p orbital is perpendicular to the plane of the sp2 orbitals, as shown below.
We can now fill in the four valence electrons from carbon.
We can now add electrons from H and O to form the four bonds. Three of the bonds will be equivalent in energy but the 4th bond is different. It is not formed by the hybridized orbitals (s bond) but will be formed by the overlap of unhybridized p orbitals. This will be called a p bond (pi is Greek for p).
The Lewis structure for formaldehyde shows that the oxygen has two lone pairs and a bond to the central carbon. Again, this requires 3 equivalent bonding orbitals, sp2 hybridization.
The electron configuration in the hybridized orbitals shows that to of the orbitals will be occupied by lone pairs of electrons and the third sp2 orbital will form a s bond with carbon. The unhybridized p orbitals on C and O will form a p bond.
The bonds in the diagrams are color coded. Red bonds are s bonds, with the shared electrons held directly between the atoms. The blue bond is a p bond, formed by the side-to-side overlap of the unhybridized p orbitals on C and O. These electrons are not held directly between the bonded atoms. Instead, they are held in a loose cloud of electron density above and below the axis of the bond. Because of this poor overlap, p bonds are weaker than s bonds, and have a higher potential energy, making them unstable. Double bonds consist of one s bond and one p bond.
VSEPR predicts that formaldehyde is a trigonal planar compound, with 120o bond angles. Molecules with sp2 hybridization have 120o bond angles. The structure of formaldehyde is shown below.
It is a polar compound because of the electronegativity of oxygen and the molecular shape of formaldehyde.
The Lewis structure of carbon dioxide is shown below, again with color coded bonds.
Each atom has an octet of electrons. Carbon is making 2 s and 2 p bonds to the oxygen atoms. The 2 s bonds indicate that there are 2 equivalent molecular orbitals formed. To form 2 hybrid molecular orbitals, we need to mix 2 atomic orbitals, an s orbital and a p orbital. The resulting hybrid orbitals are called sp hybrids. The angle between them is 180o making CO2 a linear molecule as predicted by VESPR. The two unhybridized p orbitals on carbon form p bonds to the oxygen atoms.
The energy diagram for carbon in CO2 is shown below.
What is the hybridization of oxygen in CO2. Each oxygen has two lone pairs and forms one s bond and one p bond. This means that there must be three hybridized orbitals and one unhybridized p orbital to make the p bond. This is sp2 hybridization.
When considering molecules with more than an octet of electrons around the central atom, we will need to involve the d orbitals. An example of this is PCl5.
VSEPR predicts a trigonal bipyramidal geometry since there are five groups around the central atom. In order to have five molecular orbitals, we will need to mix five atomic orbitals, one s + three p + one d. This is called dsp3 hybridization. The shapes and orientation of these orbitals are shown below, next to the structure of PCl5
Finally, we come to the molecules with six orbitals around the central atom. An example is SF6, whose Lewis structure is shown below.
We need six molecular orbitals so we mix six atomic orbitals, one s + three p + two d to give d2sp3 hybridization and octahedral geometry.