Download EXAMPLE #2: Carbon Monoxide, CO Recall: The MO energy level

 CHEM 2060 Lecture 29: Heteronuclear Diatomics MO L29-1 EXAMPLE #2: Carbon Monoxide, CO
Recall: The MO energy level diagram for O2 is not the same as the MO energy
level diagram for the C2 gas phase fragment.
How do we create an MO energy level diagram for a heteronuclear diatomic
species in which both atoms have valence s and p orbitals?
Step #1: We need to figure out the relative energies of the valence orbitals.
• We know that O is more electronegative than C, therefore the orbital
“manifold” of O should be lower in energy than that of C.
• We know that the energy gap between the 2s and 2p orbitals increases across
the period…so a larger separation in O than in C.
• If we have IP values for the valence orbitals, we use these. If not, we use the
above knowledge to estimate the relative atomic orbital energy and plot them.
CHEM 2060 Lecture 29: Heteronuclear Diatomics MO L29-2 Step #2: We need to figure out which orbitals are of the correct symmetry to
mix with each other.
• This requires defining your Cartesian coordinates. Usually, the z-axis is the
bond axis.
• The s orbitals and pz orbitals of both atoms are the correct symmetry to form σ
interactions.
MIX 4 ATOMIC ORBITALS (σ) ⇒ GET 4 MOs (σ)
• The px orbitals of each atom are π symmetry along x.
MIX 2 ATOMIC ORBITALS (π in x) ⇒ GET 2 MOs (π in x)
• The py orbitals of each atom are π symmetry along y.
MIX 2 ATOMIC ORBITALS (π in y) ⇒ GET 2 MOs (π in y)
CHEM 2060 Lecture 29: Heteronuclear Diatomics MO L29-3 Carbon Monoxide
LUMO = lowest unoccupied
molecular orbital.
⇒ 1π* is largely C character
HOMO = highest occupied
molecular orbital.
⇒ 3σ is largely C character
(actually non-bonding…C lone pair)
Carbon monoxide is reactive at the
C atom, not the O atom!
CHEM 2060 Lecture 29: Heteronuclear Diatomics MO L29-4 CO poisoning
Hemoglobin is the iron-containing oxygen-transport metalloprotein in red blood
cells of all vertebrates (except some families of fish). Hemoglobin in the blood
carries O2 from the respiratory organs (lungs/gills) to the tissues where it
releases the O2 to burn nutrients and provide energy. It collects the waste CO2
from respiration and transports it back to the respiratory organs to be dispensed
from the organism.
Human hemoglobin has 4 proteins (2 α and
2 β) and 4 iron-containing heme groups.
The heme groups are actually simple
porphyrins with an Fe(II) ion in the middle.
Each hemoglobin can transport 4 molecules
of O2…one on each Fe(II) atom.
CHEM 2060 Lecture 29: Heteronuclear Diatomics MO L29-5 Carboxyhemoglobin (COHb) is a stable complex of carbon monoxide and
hemoglobin that forms in the red bloods cells when CO is inhaled (or produced
through normal metabolism).
Large quantities of CO inhibit the delivery of O2 to the body because:
1) CO binds approximately 230 times more strongly to the Fe(II) than O2
2) When CO is bonded to one of the Fe(II) ions, it affects the other three…it
actually increases the O2 affinity so much that the O2 is not released to the
tissues properly. (Bright red blood in CO poisoning victims!)
CO binds via the C atom
CHEM 2060 Lecture 29: Heteronuclear Diatomics MO L29-6 Larger Molecules
One of the more important distinctions between atomic orbitals of different
energy is the number of nodes ⇒ more nodes = higher energy
e.g., 1s orbital (no nodes) is lower in energy than 2s orbital (1 node)
e.g., 1s orbital (no nodes) is lower in energy than 2p orbital (1 node)
e.g., in H atom, 2s and 2p orbital have exactly the same energy
This is also true for molecular orbitals.
• All Molecular Orbital Nodes must be symmetrically disposed.
• Provides some “intuition” of orbital shapes.
CHEM 2060 Lecture 29: Heteronuclear Diatomics MO L29-7 “Linear” Molecules
1. Successively higher energy orbitals have MO nodes symmetrically placed
and increasing in number by one, in order from most stable to least stable.
2. An MO node through a nucleus means that the AO from that atom does not
contribute to that MO. (the coefficient is zero)
3. The MO energy increases with increasing number of nodes because the net
number of bonding interactions (between neighboring atoms) decreases as
the number of MO nodes increases.
First, let’s start with a row of 1s orbitals. Once we have done this you will see
how easy it is to draw MO’s from both s and p using symmetry.
(No math involved).
CHEM 2060 Lecture 29: Heteronuclear Diatomics MO L29-8 Rules for constructing linear MOs:
1. The lowest energy MO has no nodes…so when we draw it as a linear
combination of AOs, all the AOs have the same phase (all +; all white; all
vanilla)
2. In successively higher MOs, # of nodes increases by 1 and these are
symmetrically placed …so the next highest energy MO has a node exactly in
the middle (this may or may not intersect the nucleus of an atom, depending
on whether the chain has an odd or even number of atoms).
3. The phase changes at each node, so negative/black/chocolate on one side and
positive/white/vanilla on the other.
4. More nodes = higher energy.
5. The lowest energy MO is the most bonding. The highest energy MO is most
antibonding.
CHEM 2060 Lecture 29: Heteronuclear Diatomics MO L29-9 These diagrams can represent the
interaction between 1s orbitals
(spherical)...
…or really any of the atomic s orbitals…
…or they can represent the π bonding p
orbitals (looking down on top of a p
orbital).
Since we normally call the “long”
direction z, it would be the px and py
orbitals that we are using.