Download pz pz σ bond

Chemistry 102 Summary July 2nd
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So far in our discussion of molecules we have focused on the
arrangement of valence electrons.
Today we are going to concentrate on the atomic orbitals that hold
these valence electrons and more specifically what happens to
these atomic orbitals when bonds form.
Example: H2
Question: Why does bonding represent a lower energy state?
Answer:
Example: F2 1s22s22p5
pz
-
pz
! bond
Electron density is on the internuclear axis (designated as z).
End-end or head to head overlap of orbitals, ex. pz+pz or 1s +1s as in
H2.
Sigma (σ) bonds can be rotated since orbitals still overlap when
atoms rotate.
Example: O2 1s22s22p4
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We have two bonds to account for.
Obvious choice of orbitals is the two 2p atomic orbitals with
unpaired electrons.
Assume the z axis is the internuclear axis, then the 2pz orbitals
overlap to form the sigma bond.
The second bond is formed from side to side overlap of the 2px
orbitals.
z-axis
px
-
px
! bond
Electron density in a π (pi) bond is above and below the
internuclear axis.
Pi bonds are formed through side to side overlap of parallel p
atomic orbitals (px + px or py + py).
Pi bonds cannot be rotated without breaking the bond.
All multiple bonds contain at least one pi bond.
A single bond consists of one sigma bond.
A double bond consists of one sigma bond and one pi bond (px+px
or py+py).
A triple bond consists of one sigma bond and two pi bonds (px+px
and py+py).
Example: N2
Bonding in more complicated molecules:
Example: CH4
-Simple overlap of atomic orbitals can’t explain the 109.5 degree bond
angles of CH4.
- if s,p and d orbitals overlapped to form all bonds then all bond angles
would be ninety and 180 degrees and this is not the case!
Hybrid Orbitals – used to explain the geometry seen in polyatomic
molecules. This is a human construct which mixes atomic orbitals to form
hybrid orbitals.
-
For the central carbon atom in CH4 we combine the carbon 2 s
orbital and three carbon 2 p orbitals to result in four sp3 hybrid
orbitals.
-
These sp3 hybrid orbitals are at bond angles of 109.5 degrees and
overlap with the hydrogen 1s orbitals to form four equivalent C-H
bonds in CH4..
Figures 9.3 and 9.6.
After overlap the electrons in the bonds are on average localized
between the two nuclei.
Examples:
(i)
NH3
(ii)
H2O
(iii)
C2H4
Question: What do all these molecules have in common?
Answer:
Example: C2H4
Question: What is the geometry around each carbon?
Answer:
-
Molecules which exhibit trigonal planar geometry exhibit sp2
hybridization.
In sp2 hybridization, the 2s orbital is combined with two of the three
2 p atomic orbitals to form 3 equivalent sp2 hybrid oribitals, one p
orbital is left unhybridized.
Example: C2H2
Question: What is the geometry around each carbon?
Answer:
-
In linear geometry we mix one s orbital with one p orbital to form
two sp hybrid orbitals and leave two p orbitals unhybridized.
Examples:
(i)
CO2
(ii)
O2
(iii)
N2
Requisite Skills
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Know how to predict hybridization around central atoms.
Know difference between sigma and pi bonds.
Be able to count the number of sigma and pi bonds.
Be able to predict which orbitals overlap to form sigma and pi
bonds.
Example: Fill in the following table and also list the sigma and pi bonds
and the orbitals that overlap to form them.
O
H
C1
C2
C3
H
N
H
Atom
Geometry
Bond Angles
C1
C2
C3
N
O
Example: Is allene (C3H4) a planar molecule?
Hybrid
Orbitals
Example: PCl5
Question: What is the geometry about phosphorous?
Answer:
-
In trigonal bipyramidal geometry dsp3 hybrids are employed – they
are formed through mixing one s orbital + one d orbital + three p
orbitals to give five dsp3 hybrid orbtials.
Example: SF6
Question: What is the geometry about sulfur?
Answer:
-
In octahedral geometry d2sp3 hybrid orbitals are employed – they
are formed through mixing one s orbital + two d orbitals + three p
orbitals to give six d2sp3 hybrid oribtials.
Delocalization
Example: NO2-
Experiment tells us that NO bonds are equal in length and strength.
We explain equal bond lengths in molecules that exhibit resonance
by saying the pi electrons are delocalized over the entire surface of
the molecule/ion.
One unhybridized p orbital from nitrogen and one from each of the
two oxygen atoms all overlap together to give a pi bond
delocalized above and below the entire plane of the NO2- ion.
There is an extra stability associated with the greater number of
resonance structures that can be drawn for a molecule because
electron-electron repulsions are minimized.
Example: Benzene (C6H6)