Ch 5 Lecture 1 Molecular Orbitals for Diatomics

I. IntroductionA. Molecular Orbital (MO) Theory uses Group Theory to describe bonding

1) Simple Bonding Theories don’t explain all molecular properties

2) MO Theory often better explains some properties (ex: magnetism)

B. How MO Theory Works

1) Atomic orbitals of like symmetry interact to form molecular orbitals

2) Electrons fill the MO’s and determine bonding, stability, and properties

3) A simple pictorial approach is often adequate to find MO’s

4) More rigorous Group Theory approaches are sometimes needed

II. Forming Molecular OrbitalsA. Linear Combinations of Atomic Orbitals (LCAO’s)

1) Atomic Orbitals (s, p, d, f…) are pictorial representations of solutions to the quantum mechanical equations describing electron locations in atoms

a) A wave function is a solution to the quantum mechanical equation

b) We use to represent a given wave function

c) The orbitals are actually pictures of 2 functions

2) Molecular Orbitals are formed by combining AO’s

a. = caa + cbb

b. = a molecular wave function

c. a = an atomic wave function from atom A

d. c = constant that can be + or –

B. Bonding in Diatomic Molecules

1) As 2 atoms approach each other, their atomic orbitals overlap in space

2) Electrons from both atoms occupy the same space in this overlap

3) MO’s form from the interaction of these electrons in the same space

4) Bonding MO’s occupy the space between the nuclei

a) In a bonding interaction, the overall electronic energy is lowered

b) Electronic energy is lowered when an electron can occupy more space

5) Conditions for bonding

a) Symmetry of AO’s must allow overlap of like signed areas (+/+)

b) Energies of the AO’s must be similar

c) A—B distance must be appropriate for overlap

C. Bonding in H2 (Ha—Hb)

1) Only the occupied AO’s are important in forming MO’s

2) For H only the 1s orbital is occupied

3) There are two possible LCAO’s for H2

a) N = normalizing factor to keep total probability = 1

b) ca = cb = 1 since the two H atoms are identical

babbaa

babbaa

1s1s2

11sc1scN*σΨ

1s1s2

11sc1scNσΨ

4 signifies that the MO is symmetric to rotation about internuclear axis

a. means the lower energy bonding MO

b. * means the higher energy antibonding MO

5) N Atomic Orbitals (2) produce N Molecular Orbitals (2)

6) Nonbonding orbitals are also possible, although none form in diatomics

a. A nonbonding MO has the same energy as its substituent AO’s

b. If symmetry doesn’t match, the MO is made up of only 1 atom’s AO’s

c. Sometimes the AO/MO energies are identical by coincidence

D. MO’s produced from p-orbitals

1) p-orbitals are more complex than s-orbitals due to +/- lobes and more complex symmetry

a) A + lobe only means the sign of the amplitude is positive, not charge

b) A – lobe only means the sign of the amplitude is negative, not charge

2) We must choose a common z-axis connecting the 2 nuclei

a) 1 MO and 1 * MO result from pz/pz overlap

b) 2 MO’s and 2 * MO’s result from px/px and py/py overlap

signifies sign change upon C2 rotation about the internuclear axis

3) Because of differing symmetries, no overlap is possible between s/p or between px/py/pz orbitals (equal overlap with +/- lobes cancels out)

4) Energies Match and Molecular Orbitals

a. Overlap of AO’s with the same energy leads to strong interaction (greatly stabilized bonding MO and greatly destabilized antibonding MO)

b. Overlap of AO’s with differing energies progressively weakens interaction

a. 1s/2s overlap can’t occur for a diatomic even though symmetry ok

b. Heteronuclear diatomics must use these types of MO’s

III. Homonuclear DiatomicsA. Why do we need MO’s

1) The N2 Lewis structure explains bonding and reactivity quite well

2) Problem: The O2 Lewis structure doesn’t explain why it is paramagnetic

3) Problem: The B2 Lewis structure doesn’t have an octet

4) MO descriptions are better

B. MO’s of the First Row Elements

1) Correct ordering only if identical orbitals interact

2) Relative energy depend on element

3) Electrons fill MO’s from bottom up

4) Bond Order

a) BO = ½[bonding-antibonding]

b) O2: BO = 2

5) MO symmetry labels

a) g = gerade = symmetric to i

b) u = ungerade = antisymmetric to i

N N

O O

B B

C. Orbital Mixing

1) Orbitals of similar energies can interact if symmetry allows

2) This leads to adjustments in the energy ordering of diatomic MO’s

a) Example: g(2s) and g(2p) can mix lowering g(2s) and raising g(2p)

b) Example: u(2s) and u(2p) can mix lowering u(2s) and raising u(2p)

3) Effects of mixing

a) Energy ordering changes as we move across the first row elements

b) Bonding/Antibonding nature of a given MO may change

D. MO Theory and Properties of Homonuclear Diatomics

1) Magnetism

a) Paramagnetic = strongly attracted by magnetic field due to upaired e-

b) Diamagnetic = weakly repelled by magnetic field by all paired e-

2) H2 [g2(1s)]

a) BO = 1 and BL = 74 pm

b) H2+ has BO = ½ and is less stable with BL = 106 pm

3) He2 [g2(1s)u*2(1s)] BO = 0, Noble gas that does not exist as a diatomic

4) Li2 [g2(2s)] BO = 1

5) Be2 [g2(2s)u*2(2s)] BO = 0, not stable

6) B2 [u1u

1(2p)]

a) Paramagnetism of B2 is not explained by Lewis structure

b) BO = 1

c) Without mixing, all e- paired = diamagnetic

d) Mixing makes degenerate u(2p) lower than g(2p)

e) Filling degenerate orbitals equally (Hund’s Rule) makes it paramagnetic

B B

7) C2 [u2u

2(2p)]

a. BO = 2, mixing suggest that both are -bonds with no -bond?

b. C2 is rare, but C22- (with a -bond as well BO = 3) is more stable

8) N2 [g2u

2u2(2p)]

a. BO = 3 agrees with short strong bond (BL = 109.8 pm, D = 945 kJ/mol)

b. After nitrogen the “normal” energy ordering initially predicted returns

i. Nuclear charge gets higher as we go across first row

ii. Mixing becomes less important because the energy difference between 2s and 2p orbitals increases (s closer to nucleus, p more diffuse)

iii. Large energy difference means mixing is less important

9) O2 [g2u

2u2g*1g*1 (2p)]

a. BO = 2, paramagnetic (not explained by Lewis structure)

b. O2 ions correlate BO and BL very well

O O

10) F2 [g2u

2u2g*2g*2 (2p)] BO = 1, diamagnetic

11) Ne2 BO = 0 Noble gas, diatomic not stable

E. Photoelectron Spectroscopy: Where Orbital Energies Come From

1) Radiation can dislodge an electron from a molecule

a) UV radiation removes outer e-

b) X-Ray radiation can remove inner e-

2) The KE of the expelled e- tells us the energy of the orbital it came from

a) Ionization energy is equivalent to the orbital energy

b) IE = h – KE

3) N2 spectrum

a) Lower E at top (outer orbital)

b) Fine structure is due to vibrationalEnergy levels within electronic levels

i. Many levels = bonding orbitalii. Few levels = less bonding

4) Correlating Spectrum with Orbitals

a. Sometimes spectrum has orbitals out of order, because energies shift as electrons are removed from molecule

b. Higher level calculations can correct for these “errors”

c. N2 levels u(2p) and g(2p) are very close; some debate as to order

5) O2 Spectrum: ordering is

correct as shown in the spectrum

F. Correlation Diagrams

1) Calculated effect of bringing atoms together to a merged nucleus

2) Example: 2 H atoms approaching and merging into 1 He

a) Won’t happen chemically

b) Useful to see what happens to orbitals

c) MO’s for a homonuclear diatomics are somewhere in the middle, depending on inherent energy of the element

Larger F2 farther to right, Smaller B2 farther to left (closer together)

3) As Nuclei Approach

a) Bonding MO’s decrease in energy

b) Antibonding MO’s increase in energy

c) Atomic Orbitals are connected to MO’s by symmetry

The 2 1s orbitals become 1g and 1u*, then 1s and 2pz

4) The Noncrossing Rule: orbitals of the same symmetry interact so that their energies never cross on a correlation diagram

a) Helpful in assigning correlations

b) If 2 orbital sets of the same symmetry cross, you must change the grouping to avoid it