In the simplest terms covalent bonds are formed when atomic orbitals on two nonmetal atoms overlap to share electrons. So consider the hydrogen halides;
HF
HCl
HBr
HI
The electron configuration for the halogen is ns2np5 and the electron configuration for H is 1s1. The covalent bond that is formed between the hydrogen atom and the halogen is formed from the overlap of the 1s orbital on hydrogen and the np orbital on the halogen. When we look at the orbital diagram for hydrogen and the halogen we see the 1s orbital on hydrogen contains one unpaired electron, and an np orbital on the halogen also contains one unpaired electron. When the 1s orbital on hydrogen overlaps with the np orbital on the halogen the covalent bond is formed. We can describe the bonding more specifically in the following way;
Similarly if we consider the elemental form of the halogens;
F2
Cl2
Br2
I2
we can describe the covalent bond formed between the atoms using simple atomic orbitals. In each case the np orbital with one unpaired electron on the halogen overlap with the same np orbital on the other halogen. The covalent bond is formed along the internuclear axis between the two atoms. We can describe the bond between the halogen atoms in the elements as;
Can we use atomic orbitals to describe the covalent bonds in other compounds?
Problems Start Appearing
If we try to use atomic orbitals to describe the covalent bonds in;
H2O
NH3
CH4
we run into some problems. Lets see what happens.
In the case of water, H2O (see the movie), the oxygen atom has two 2p orbitals each with an unpaired electron. There are two hydrogen atoms so we might try to describe each O–H in water using a 1s orbital on hydrogen and a 2p orbital on oxygen. Using 2p orbitals on oxygen would result in a water molecule with an H–O–H bond angle of 90˚. (Remember the three p orbitals in a p subshell are all 90˚ from each other.) However, experimentally the H–O–H bond angle is actually 105˚, considerably larger.
Similar conflicts arise for ammonia and methane.
The expected bond angle using atomic orbitals does not compare to the experimentally determined bond angle. In the case of ammonia (see the movie)we would predict 90˚ again using the 2p orbitals on nitrogen to overlap with the 1s orbital on hydrogen. However, the experimental bond angle in ammonia is 107˚.
In methane the problem is even more interesting. Carbon contains two 2p orbitals with unpaired electrons, so we might expect the compound between hydrogen and carbon would CH2, not the experimentally observed compound, methane, CH4!
How do we handle these three cases?
Current models describe the bonding by mixing atomic orbitals on the oxygen, nitrogen or the carbon. By mixing the 2s and three 2p orbitals on the central atom we produce a new set of hybridized orbitals on the central atom that overlap with the 1s atomic orbital on hydrogen.
Look at the movie showing how the four (a 2s and three 2p) atomic orbitals on carbon mix to form four hybrid orbitals. The hybrid orbitals are given a label of sp3, because an s and three p orbitals are mixed. Notice the bond angle for sp3 hybrid orbitals is 109.5˚, the same bond angle is experimentally measured in CH4. In the case of water and ammonia, the same sp3 hybrid orbital are used by the central atom, because an s and three p orbitals are mixed. Notice the hybrid orbitals are used on the oxygen and the nitrogen atom, respectively.
Why are not the same bond angles…109.5˚, observed?
The H–O–H and N–N–H bond angles in water and ammonia are slightly smaller than 109.5˚ because of the greater repulsion that occurs between lone-pair electrons and bonding pair electrons compared to bonding pair/bonding pair repulsions.
So any time a central atom has four groups of electrons the orbitals on the central atom that form covalent bonds with the terminal atoms will be sp3 hybrid orbitals.
What type of orbitals on boron would overlap in BF3? Can we use sp3 hybrid orbitals? Possibly, but consider the F–B–F bond angle, it is 120˚…hmmm…nothing like 109.5˚, or the slightly smaller bond angle observed in water and ammonia. Also when we look at the atomic orbitals containing the valence electrons on boron (See the movie), we have two electrons in the 2s orbital and one electron in the 2p orbital. If we promoted one of the 2s electrons into an empty 2p orbital and mixed the one 2s orbital and the two 2p atomic orbitals we form three sp2 hybrid orbitals. So the B–F bond is formed from an sp2 hybrid orbital on boron and a 2p atomic orbital on fluorine.
What type of orbitals on beryllium would overlap in BeCl2? Can we use sp3 hybrid orbitals? Can we use sp3 hybrid orbitals? Possibly, but consider the Cl–Be–Cl bond angle, it is 180˚…hmmm…nothing like 109.5˚ if sp3 hybrid orbitals are used, or nothing like 120˚ if sp2 hybrid orbitals are used. Also when we look at the atomic orbitals containing the valence electrons on beryllium (See the movie), we have two electrons in the 2s orbital and no electrons in a 2p orbital. If we promoted one of the 2s electrons into an empty 2p orbital and mixed the one 2s orbital and the one 2p atomic orbitals we form two sp hybrid orbitals. So the Be–Cl bond is formed from an sp hybrid orbital on beryllium and a 3p atomic orbital on chlorine.
Hybridization (CA)
#Domains of Electrons (CA)
Bond angles
sp3
4
109.5˚
sp2
3
120˚
sp
2
180˚
Lets now consider ethylene, C2H4. A Lewis structure shows two central carbon atoms, each with 3 groups of electrons. So the H–C–H and H–C–C bond angles are 120˚. The hybridization on the carbon is sp2. In carbon (See the movie) we promote one of the 2s electrons into the empty 2p atomic orbital, however, we’ll only mix the 2s and two of the 2p orbitals to form three sp2 hybrid orbitals and leaving one pure 2p atomic orbital with one electron. Note that each of the sp2 hybrid orbitals and the 2p atomic orbital have a single electron. Now the two carbon atoms form a sigma bond by overlap of an sp2 hybrid orbital on each carbon, there are two more sp2 hybrid orbitals on each carbon to bond with the electron in the 1s orbital on each of the four hydrogen atoms. That leaves a 2p atomic orbital on each carbon atom. Note the orientation of the 2p atomic orbitals. These two atomic orbitals overlap to form a second bond between the two carbon atoms, a pi bond. A bond formed from two sp2 hybrid orbitals is called a sigma bond. A sigma bond is formed from orbitals (hybrid or atomic orbitals) that are along the internuclear axis between the two atoms. A bond formed from overlapping 2p atomic orbitals that are parallel is called a pi-bond. A double bond consists of one sigma and one pi bond. A triple bond would consists of one sigma and two pi bonds.
Lets now consider ethyne, C2H2. A Lewis structure shows two central carbon atoms, each with 2 groups of electrons. So the H–C–H and H–C–C bond angles are 180˚. The hybridization on the carbon is sp. In carbon (See the movie) we promote one of the 2s electrons into the empty 2p however, we’ll only mix the 2s and one of the 2p orbitals to form two sp hybrid orbitals leaving two pure 2p atomic orbitals with one electron. Note that each of the sp hybrid orbitals and the 2p atomic orbitals have a single electron. Now when the two carbon atoms form a bond by overlap of an sp hybrid orbital on each carbon, there is one more sp hybrid orbital on each carbon to bond with the electron in the 1s orbital on each of the two hydrogen atoms. That leaves two 2p atomic orbital on each carbon atom. Note the orientation of the 2p atomic orbitals. Two of the 2p atomic orbitals overlap to form a second bond between the two carbon atoms. The other two 2p atomic orbitals overlap to form a third bond between the two carbon atoms. A bond formed from two sp hybrid orbitals is called a sigma bond. A sigma bond is formed from orbitals (hybrid or atomic orbitals) that are along the internuclear axis between the two atoms. A bond formed from overlapping 2p atomic orbitals that are parallel is called a pi-bond. A triple bond consists of one sigma and one pi bond.