Chemistry 11 Enriched
p. 341-348
Unit VII
Problem Set 7.6 - Hybridization
We have seen how the quantum picture leads to different shapes for s, p, d and f subshells.
The question is how do these orbitals lead to bonding? For example, a bond may form from some s and some p
electrons, but since the p electrons are in orbitals at 90 degrees to each other how does this work for bonds that
are at 109 degrees?
The answer to the question lies in the mashing together of these s and p orbitals in order to come up with new
orbitals.
These new orbitals are called hybrid orbitals.
For each s or p put into the mix, one hybrid must be made.
Consider CH4. The Lewis diagram looks like:
H
H
C
H
H
If we look closely at the outside shell of C we see that there are 2 2s electrons and 2 2p electrons. The s
electrons are in a spherical arrangement and the p electrons are in lobes as pictured above. This does not
connect to the 109.5 o we expect for the bonds that form. Since these 2s and 2p electrons are the ones used in
bonding, we must resolve this issue. In order to explain it, we take these 4 orbitals (1 2s orbital and 3 2p
orbitals – the 2px the 2py and the 2pz) and make 4 hybrid orbitals that will be called all be called sp3. The name
comes from the fact that the new orbitals are a mashing (hybridizing) of 1 2s orbital and 3 2p orbitals.
In short, whenever a molecule is tetrahedral with 4 bonds, it must be sp3 hybridized. The 4 bonds mean there
must have been 4 parts put in (the s and the 3 p’s). In fact for any central atom with four repelling items –
bonds or non bonding pairs, the hybridization will be sp3. So the oxygen on water, for example, is sp3, since
there are two bonds and two non bonding pairs on the oxygen.
Electron Configuration for
C Before Bonding
Electron Configuration for
C After Bonding
2p
PE
PE
sp3
2s
All of the unpaired sp3 hybrid orbitals now hold one electron. These 4 equal orbitals repel each other so that
they are all 109o apart, in a tetrahedral shape. Each one is shared with the one electron on a H atom to create the
C-H bonds.
If a molecule is trigonal planer (three bonds) it must have come from an s orbital and 2 p orbitals and so is sp2
hybridized. Any atom with three repelling items will be sp2 hybridized.
Chemistry 11 Enriched
p. 341-348
Unit VII
Consider the H2CO molecule:
H
C=O
H
Electron Configuration for
C Before Bonding
Electron Configuration for
C After Bonding
2p
PE
2p
sp2 hybrid
PE
2s
The three hybrid orbitals occupy the three equatorial positions that are 120o apart, and the unchanged p orbital
still has its two lobes above and below the three hybrid orbitals – in the axial positions.
half filled
hybrid orbitals
unhybridized p orbitals
Each of the three hybrid orbitals can hold 2 electrons each and the p orbital system (both lobes above and
below) can also hold two electrons.
In a similar way, consider the oxygen molecule in the H2CO molecule which also is sp2 hybridized:
Electron Configuration for
O Before Bonding
Electron Configuration for
O After Bonding
2p
PE
2p
PE
2s
sp2 hybrid
Chemistry 11 Enriched
p. 341-348
Unit VII
Unhybridized p orbitals
partly filled hybrid orbital
Filled hybrid orbitals
So, for the C and O together, the first bond is a sp2 – sp2 hybrid orbital overlap:
The second bond occurs due to unhybridized p orbital overlap
the electron density of the p system exists above and below the C and O nuclei, but TOGETHER, these two
overlaps create one bond.
Two H atoms bonded
to the C sp2 hybrid
orbitals
Non-bonding hybrid
orbitals on O
Pi bond from
unhybridized p orbital
overlap
Any hybrid orbital overalp is called a sigma bond.
Any non-hybridized p orbital overalp is called a pi bond.
In any multi bond system, there is always one sigma bond and one (or more) pi bonds.
If a molecule is linear (two bonds) it must have
come from an s orbital and a p orbital and is sp
hybridized.
This diagram is the bonding in ethyne, which has a
triple bond between C atoms. The first is an sp – sp
overlap, the second and third are unhybridized p
orbital overlaps (pi bonds)
Chemistry 11 Enriched
p. 341-348
Unit VII
Chemistry 11 Enriched
p. 341-348
Unit VII
The following chart summarizes all hybridizations.
Summary of bond angles, shape names and Hybridization
items repelling
bonds
lone pairs
bond angle ( o )
name
Hybridization of
central
atom
example
2
2
0
180
linear
sp
H-C=C-H
3
0
trigonal planer
BF3
3
120
sp2
2
1
bent
N=O
H
4
0
109.5
tetrahedral
3
1
~107
trigonal pyramid
2
2
~104
bent
CH4
sp3
NH3
4
H2O
Chemistry 11 Enriched
p. 341-348
items repelling
Unit VII
bonds
lone pairs
bond angle ( o )
name
5
0
90 and 120
trigonal
bipyramidal
PF5
4
1
90 and 120
see saw
SCl4
Hybridization of
central
atom
dsp3
5
6
example
3
2
90
T-Shaped
IF3
2
3
180
Linear
XeF2
6
0
90
Octahedral
SF6
5
1
90
Square
Pyramidal
4
2
90
Square Planer
d2sp3
ClF5
XeF4
Complete p. 348 Practice Problems #1,2 and p. 360 #3,4,5a
Question:
Going back to problem in the VSEPR Theory problem set, list the hybridization for those structures which
have a central atom.
Answers:
a) sp3
e) sp3
i) dsp3
b) sp2
f) sp3 (oxygen) sp2 (nitrogen)
j) d2sp3
c) sp2
g) dsp3
k) d2sp3
d) sp2 (carbon) and sp3 (oxygen)
h) d2sp3
l) dsp3