Sample Problem: Using MO Theory to Explain Bond Properties

Sample Problem:
Using MO Theory to Explain Bond Properties
Problem: Consider the following data for these homonuclear diatomic
species:
N2
N2 +
O2
O2+
Bond energy (kJ/mol)
945
841
498
623
Bond length (pm)
110
112
121
112
No. of valence electrons 10
9
12
11
Removing an electron from N2 decreases the bond energy of the
resulting ion, whereas removing an electron from O2 increases the
bond energy of the resulting ion. Explain these facts using M.O
diagrams.
Plan: We first draw the MO energy levels for the four species,
recalling that they differ for N2 and O2. Then we determine the
bond orders and compare them with the data: bond order is related
directly to bond energy and inversely to bond length.
Sample Problem - Continued
Solution: The MO energy levels are:
N2
N2 +
O2
O2+
σ2p*
σ2p*
π2p*
π2p*
σ2p
π2p
π2p
σ2p
σ2s*
σ2s*
σ2s
σ2s
BO: 1/2(8-2) = 3 1/2[7-2] = 2.5
1/2(8-4) = 2
1/2(8-3) = 2.5
Bonding in Heteronuclear vs. Heteronuclear
Diatomic Molecules
(Zumdahl Section 14.4)
Review from Chapter 13:
Covalent
Ionic
Ionic
Figure13.11
Covalent
Figure 13.12
Homonuclear:
H2
Electronegativity
Nonpolar covalent bond
(450kJ/mol bond)
Figure 14.26
Heteronuclear:
HF
Electronegativity
Polar covalent bond
(565kJ/mol bond)
Figure 14.45
Electrons are not equally shared
in heteronuclear bonds
HF
Electronegativity
Figure 14.45
Because F (χ = 4.0) is
more electronegative than
H (χ = 2.2), the electrons
move closer to F.
(Table 13.2 for electronegativities)
This gives rise to a polar
bond: H F
M.O.s of a Polar Covalent Bond: HF
σ Antibonding (σ*)
Mostly H(1s)
H
F
H
F
σ Bonding
Mostly F(2p)
When the electronegativities of the 2 atoms are
more similar, the bonding becomes less polar.
Example: NO
2p
2s
N
2s
NO
O
Electronegativity
2p
χ(N) = 3.0
χ(O) = 3.4
. ..
..N=O..
Sample Problem: Use the M.O. model to predict the changes in
magnetism and bond order upon oxidation of NO to make NO+
NO+
NO
oxidation
Bond Order = (8-3)/2 = 2.5
Paramagnetic
Bond Order = (8-2)/2 = 3
Diamagnetic
Combining the Localized Electron and Molecular
Orbital Models (into a convenient working model)
(Zumdahl Section 14.5)
Figure 14.47
Only the π bonding changes between these
resonance structures- The M.O. model
describes this π bonding more effectively
Atomic Orbitals
Molecular Orbitals
+ 3 others at
higher energies
Figure 14.51
Another example:
Benzene
σ bonding:
π bonding:
p atomic orbitals
π molecular orbital
Molecular Spectroscopy (Zumdahl 14.7)
Figure 14.52
Ozone – The Important Molecule
z
= O3
y
18 valence electrons
9 molecular orbitals
x
empty
πxy
πx
πz∗
weak
absorption
hν
UV strong
absorption
at 230 nm
πz
Electronic Spectroscopy
Electronic Transitions Give Molecules Color!
Ultraviolet
Blue
Green
Yellow
Red
Wavenumber
These experiments are used, for example, to determine the
M.O. diagram experimentally
Vibrational Spectroscopy
For fun- see animated vibrations @ http://www.cem.msu.edu/~parrill/AIRS/
CH stretch
HCH bend
CN stretch
Wavenumber ∝ 1/(Wavelength)
2000cm-1 = 1/(0.0005cm) = 1/(5000nm)
These experiments are used, for
example, to determine the
identities of molecules in solution
Molecular Vibrations
z
z
N free atoms: 3N independent motions
y
x
y
z
= 3N degrees of freedom
x
y
x
Molecule: N bound atoms
z
3 translational
3 rotational (non-linear molecule)
y
3N – 3 – 3 = 3N – 6 degrees of freedom
x
for vibrations
Ozone:
N=3
3 x 3 – 6 = 3 vibrational degrees of freedom
Acetonitrile – An Important Solvent
N=6
CH3CN
3 x 6 – 3 = 15 vibrational degrees of freedom
Maximum of 15 vibrational absorptions in the infrared (IR) spectrum
5 bonds = 5 bond-stretching vibrational modes
15 - 5 = 10 bending (deformation) vibrational modes
Rotational Spectroscopy