Lecture 14 Chemistry 362 M. Darensbourg 2017 Spring term Molecular orbitals for diatomics
Molecular Orbital Theory of the Chemical Bond Simplest example - H 2 : two H atoms H A and H B Only two a.o.'s (1s A, 1s B ) to form linear combinations. General rule: n a.o.'s n m.o.'s So we can only construct 2 m.o.'s for H 2 - and these are: ψ b = 1s A 1s B and ψ a = 1s A - 1s B i.e. the sum (ψ b ) and the difference (ψ a ) of the constituent a.o.'s. Consider the electron distribution in each of these:
Consider in each case the INTERNUCLEAR REGION Probability of finding electron there is: ψ b > 1s A, 1s B > ψ a Electron in this region attracted to BOTH nuclei, therefore most favourable position. Hence, electron in ψ b will be at lower energy than in non-interacting a.o.'s, and electron in ψ a will be at higher energy still. Thus an electron in ψ b will hold the nuclei together, one in ψ a will push them apart. ψ b is a BONDING m.o., ψ a is an ANTI-BONDING m.o.
H A H B 1s atomic orbitals either reinforce or cancel each other
Thus we can draw ENERGY LEVEL DIAGRAM for m.o.'s of H 2 : ψ a 1s A 1s B ψ b H A H 2 H B By aufbau & Pauli principles - the 2 electrons go into ψ b - with paired spins.
M.O.Energy Level Diagram for A 2 (A = Li, Be) 2s σ*2s 2s Remember: 1s orbitals effectively non-bonding, A A 2 σ2s A Use Aufbau, Pauli, Hund - just as in filling atomic orbitals Li 2 Only two valence electrons, i.e. 2 0 σs σ*s. Bond order = 1. Diamagnetic Li 2 exists in gas phase over metallic lithium. "Be 2 " σ s 2 σ*s 2 Bond order = 0 - no net bonding energy, so molecule does not exist. Beryllium in gas phase is monatomic.
p orbitals: p z p z overlap => sigma orbitals, bonding and anti-bonding in same area as sigma derived from s-orbital overlap. p z /p z.. σ2p z p z p z overlap everywhere positive BONDING M.O... p z p z M.O. σ*2p z overlap everywhere negative ANTI-BONDING
p orbitals: p x p x overlap => orbitals with one unit of angular momentum; 1 angular Node; or π orbitals, bonding and anti-bonding; p y p y overlap identical giving π MO p x /p x or p y /p y.. π2p x or π2p y π*2p.. x or π*2p y
Nodes, (not Toads): Angular => Sigma and Pi and Delta Angular Defined by change of sign when rotated around The bond axis. If no change: Angular Momentum of electrons in that orbital is zero: a Sigma, σ MO If one change in sign, a Pi, π MO If 2 changes in sign, a delta, δ MO A complexity: Orbitals of same symmetry, either on same atom or on adjacent atoms can mix. Thus sigma MO s from s s overlap or from p z p z overlap can mix and affect each other if sufficiently close in energy. How do we know if they are?
Mixing of s and p orbitals No mixing of s and p orbitals
Consequence of no mixing on energy level diagram: Homonuclear Diatomics No mixing of sigma orbitals Derived from 2s and 2p z Core (closed shell) electrons; little effect.
Homonuclear Diatomics No mixing vs. mixing of sigma orbitals Derived from 2s and 2p z No mixing (O 2 and F 2 ) Mixing (B 2, C 2, N 2 ) Copyright 2014 Pearson Education, Inc.
σ*2p z M.O.Energy Level Diagram for A 2 (A = O ) 2p π*2p x,π*2p y π2p x,π2p y 2p 2s σ2p z σ*2s 2s σ2s A A 2 A O 2 Electronic configuration: σ s 2 σ*s 2 σpz 2 πpx 2 πpy 2 π*px 1 π*py 1 Note Hund's rule again! Bond order = (8-4)/2 = 2 (double bond) and PARAMAGNETIC. V.B. theory could not explain paramagnetism.
Electron Configurations: Assign electrons to energy levels but put in horizontal scheme: For no mixing: O 2, F 2, Ne 2 [KK] < σ 2s < σ* 2s <s 2pz <(π 2px, π 2py ) < (π* 2px, π* 2py ) < σ* 2pz For mixing: B 2, C 2, N 2 [KK] < σ 2s < σ* 2s < (π 2px, π 2py ) <σ 2pz < (π* 2px, π* 2py ) < σ* 2pz Assign electrons and determine bond order, magnetism, and Term Symbols! B.O. = (# bonding electrons - # anti-bonding electrons)/2 But what about heteronuclear diatomics: CN -, NO, CO?
HETERONUCLEAR DIATOMIC MOLECULES Simplest would be HHe. Differs from H 2 in two ways: (1) A.O. energies for H, He different. He - greater nuclear charge, electrons more tightly bound. (2) Now three electrons to feed into m.o.'s. Energy level diagram is now: ψ a Ψ b = c 1 ( 1s H) c 2 ( 1s He) where c 2 > c 1 Ψ a = c 1 ( 1s H) - c 2 ( 1s He) where c 1 > c 2 1s H average energy of a.o.'s ψ b 1s He H HHe He For heteronuclear diatomics, m.o.'s formed symmetrically above and below AVERAGE energy of constituent a.o.'s
For HHe, bond order = (2-1)/2 = 1/2 i.e. v. wk. "1/2" bond - not formed under normal conditions - v. unstable. Unpaired electron, PARAMAGNETIC. Note for "He 2 " - extra electron in antibonding m.o. - therefore bond order = 0. Molecule does not exist - no force to hold atoms together. He is monatomic gas. More Dramatic for HF: Must use H 1s F 2pz
s/p z gives bonding and anti-bonding pair. s/p z.... σspz σ*sp z s/p x or p y and p z /p x or p y all non-bonding (positive and negative overlaps cancel. No overlap at all for p x /p y. p z /p x s/p x or or non-bonding p z /p y s/p y non-bonding Before moving on to show the energy level diagram for A 2 molecules - we need to be clear about the labels for m.o. s
MO Energy Level Diagrams for HF and CO from Shriver, 2.21
Carbon Monoxide VSIE O 2s = 32.3 ev O 2p = 15.8 ev C 2s = 19.4 ev C 2pz = 10.6 ev
Carbon Monoxide Consequence: CO binds to metals, such as iron, Through Carbon rather than Oxygen. Can produce dative bond by donating Electron density from the 3σ to empty orbital On metal, and will accept electron density From the metal via π-backbonding.
Molecular Orbitals in Polyatomic Molecules In the vast majority of cases, bonds can be described as LOCALISED between pairs of nuclei - therefore can use VB approach. At this stage we only need to use molecular orbitals (m.o s) when DELOCALISATION of electrons occurs - i.e. when several Lewis structures have to be drawn. Even in these cases we can use the VB approach for the σ-framework, and construct m.o. s only for the π-bonds. Just look at TWO examples - one inorganic (CO 3 2- ), one organic (benzene, C 6 H 6 ).
2- Carbonate ion, CO 3. Regular trigonal planar, equal bonds, all angles 120 0. σ-framework: use sp 2 hybrids on both C and O's. 2 electrons in each bond lone pairs in remaining sp 2 hybrids on O's = 18 electrons O Total number of valence electrons = 4 (C) 18 (3 O's) 2 (negative charges) = 24 O C O Therefore have 6 electrons to put into π-type orbitals Which orbitals are available for π-bonding? On each atom, sp 2 hybrid formation uses p x and p y orbitals. Therefore one p z orbital on each atom is available for π-bonding - 4 orbitals in total.
z O C - - O - - Remember general rule: n a.o's n m.o.'s Therefore we will form 4 m.o.'s Details of how to calculate what these look like left until later.
By analogy with diatomic m.o.'s - most strongly bonding m.o. will be the one which increases electron density in all of the bonds: O C - - O - - The most antibonding m.o. will have NODES on all of the bonds: - O - C O - - nodes
The remaining two turn out to be non-bonding, giving the energy level for the π-bonding in CO 3 2- as follows: π The 6 electrons just fill the bonding and non-bonding m.o.'s. p z π 0 p z One π-bond shared out equally in a delocalised m.o. π Each bond therefore 1σ 1/3π C CO 3 2-3O All electrons paired - therefore DIAMAGNETIC
p orbitals: p x p x overlap => orbitals with one unit of angular momentum; 1 angular Node; or π orbitals, bonding and anti-bonding; p y p y overlap identical giving π MO p x /p x or p y /p y.. π2p x or π2p y p x p x overlap everywhere positive BONDING M.O. p x, p y orbitals are perpendicular to axis, but can still interact
π*2p.. x or π*2p y p x p x overlap everywhere negative ANTI-BONDING M.O. Also exactly analogous pair from p y. Need to consider all possibilities (could be needed for heteronuclear diatomics)