Perturbation Theory H 0 is the Hamiltonian of for a known system for which we have the solutions: the energies, e 0, and the wavefunctions, f 0. H 0 f.

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Presentation transcript:

Perturbation Theory H 0 is the Hamiltonian of for a known system for which we have the solutions: the energies, e 0, and the wavefunctions, f 0. H 0 f 0 = e 0 f 0 We now change the system slightly (change a C into a N, create a bond between two atoms). The Hamiltonian will be changed slightly. For the changed system H = H 0 + H 1 H 1 is the change to the Hamiltonian. We want to find out what happens to the molecular orbital energies and to the MOs.

How are energies and wave functions affected by a change? Energy to e i 0 : Zero order (no change, no correction): e i 0 First Order correction: Wave functions corrections to f 0 i Zero order (no correction): f 0 i First order correction:

Example: Creating allyl system out of ethylene plus methyl radical Pi system only: “Unperturbed” system: ethylene + methyl radical “Perturbed” system: allyl system

Example: Creating allyl system out of ethylene plus methyl radical Pi system only: “Unperturbed” system: ethylene + methyl radical “Perturbed” system: allyl system

Working out of the predicted values for the “perturbed” system. MO 1 only. Energy e 1 Orbital 1 Note the stabilizing interaction. Bonding!

Mixes in bonding Mixes in anti-bonding

Projection Operator Algorithm to create an object forming a basis of an irreducible rep from an arbitrary function. Where the projection operator results from using the symmetry operators, R, multiplied by characters of the irreducible reps. j indicates the desired symmetry. l j is the dimension of the irreducible rep (number in the E column). Starting with the 1s A create a function of A 1 sym ¼(E1s A + C 2 1s A +  v 1s A +  v ’1s A ) = ¼ (1s A + 1s B + 1s A + 1s B ) = ( ½(1s A + 1s B )

Consider the bonding in NF 3, C 3v. We can see what Irreducible Reps can be obtained from the atomic orbitals. Then get the Symm Adapted Linear Combinations by using Projection Op.  A A B C D  B  C  D  A = A 2 + E  B =  C =  D = A 1 + E 1 23 } These p orbitals are set- up as sigma and pi with respect to the internuclear axis. Now we know which irreducible reps we can get out of each kind of atomic orbital. Have to get the correct linear combination of the AOs. The E irred. rep is the bad one.

Now construct SALCs first A 2 then E.  A = A 2 + E P A2 (p 1 ) = 1/6 (p 1 + p 2 + p 3 + (-1)(-p) 1 + (-1)(-1p 3 ) + (-1)(-p 2 ) No AO on N is A 2 Now construct an SALC of A 2 sym See what the different symmetry operations do to the one of the types of p orbitals.

Now get the SALC of E symmetry. Note that E is doubly degenerate. P E (p 1 ) = (2p 1 - p 2 - p 3 ) = E 1 Apply projection operator to p 1 But since it is two dimensional, E, there should be another SALC. Apply P E to p 2. P E (p 2 ) = (2p 2 - p 3 - p 1 ) = E’ But E 1 and E’ should be orthogonal. We want sum of products of coefficients to be zero. Create a linear combination of E 2 = E’ + k E 1. = (-1 +k*2) p 1 + (2 + k(-1)) p 2 + (-1 + k(-1))p 3 Have to choose k such that they are orthogonal. 0 = (2(-1 + k*2) -1 (2 + k(-1)) -1 (-1 + k(-1)) k = ½ E 2 = (3/2 p 2 - 3/2 p 3 ) = p 2 – p 3

For now we show the interaction of the N p orbitals (p x and p y ) of E symmetry with the SALC of the F p orbitals of E symmetry. We will return to C 3v molecules later on….

The geometries of electron domains Molecular shapes: When we discussed VSEPR theory Can this be described in terms of MO’s?

Hybrid orbitals s + p = 2 sp hybrids (linear) s + 2p = 3 sp 2 hybrids trigonal planar s + 3p = 4 sp 3 hybrids tetrahedral s + 3p + d = 5 dsp 3 hybrids trigonal bipyramidal s + 3p + 2d = 6 d 2 sp 3 hybrids octahedral

Molecular orbitals of heteronuclear diatomic molecules

The general principle of molecular orbital theory Interactions of orbitals (or groups of orbitals) occur when the interacting orbitals overlap. the energy of the orbitals must be similar the interatomic distance must be short enough but not too short A bonding interaction takes place when: regions of the same sign overlap An antibonding interaction takes place when: regions of opposite sign overlap

Combinations of two s orbitals in a homonuclear molecule (e.g. H 2 ) Antibonding Bonding In this case, the energies of the A.O.’s are identical

More generally:  c a  (1s a )  c b  (1s b )] n A.O.’sn M.O.’s The same principle is applied to heteronuclear diatomic molecules But the atomic energy levels are lower for the heavier atom

Orbital potential energies (see also Table 5-1 in p. 134 of textbook) Average energies for all electrons in the same level, e.g., 3p (use to estimate which orbitals may interact)

The molecular orbitals of carbon monoxide  c c  (C)  c o  (O)] 2s2p C O E(eV) Each MO receives unequal contributions from C and O (c c ≠ c o )

Group theory is used in building molecular orbitals

A1A1 A1A1 A1A1 A1A1 B1 B2B1 B2 B1 B2B1 B2 “C-like MO” “O-like MO” Frontier orbitals “C-like MO’s” “O-like MO’s” mixing Larger homo lobe on C Bond order 3

A related example: HF s (A 1 )2p(A 1, B 1, B 2 ) H (1s) F (2s) No s-s int. (  E > 13 eV) Non-bonding (no E match) Non-bonding (no symmetry match)

Extreme cases: ionic compounds (LiF) Li transfers e - to F Forming Li + and F - A1A1 A1A1

Molecular orbitals for larger molecules 1. Determine point group of molecule (if linear, use D 2h and C 2v instead of D ∞h or C ∞v ) 2. Assign x, y, z coordinates (z axis is higher rotation axis; if non-linear y axis in outer atoms point to central atom) 3. Find the characters of the representation for the combination of 2s orbitals on the outer atoms, then for p x, p y, p z. (as for vibrations, orbitals that change position = 0, orbitals that do not change =1; and orbitals that remain in the same position but change sign = -1) 4. Find the reducible representations (they correspond to the symmetry of group orbitals, also called Symmetry Adapted Linear Combinations SALC’s of the orbitals). 5. Find AO’s in central atom with the same symmetry 6. Combine AO’s from central atom with those group orbitals of same symmetry and similar E

F-H-F - D ∞h, use D 2h 1st consider combinations of 2s and 2p orbitals from F atoms 8 GROUP ORBITALS DEFINED

Group orbitals can now be treated as atomic orbitals and combined with appropriate AO’s from H 1s(H) is A g so it matches two group orbitals 1 and 3 Both interactions are symmetry allowed, how about energies?

-13.6 eV eV eV Good E match Strong interaction Poor E match weak interaction

Bonding e Non-bonding e Lewis structure F-H-F - implies 4 e around H ! MO analysis defines 3c-2e bond (2e delocalized over 3 atoms)

CO 2 D ∞h, use D 2h (O O) group orbitals the same as for F F But C has more AO’s to be considered than H !

CO 2 D ∞h, use D 2h No match Carbon orbitals

A g -A g interactions B 1u -B 1u interactions All four are symmetry allowed

Group orbitals 1(A g ) and 2 (B 1u ) Group orbitals 3(A g ) -4 (B 1u ) 2s(C) (A g ) and Groups 1 & 3 (A g ) 2s(C) Group 1(s) < 2s(C) Group 3(p z ) Primary A g interaction 2p z (C) (B 1u ) and Groups 2 & 4 (B 1u ) 2p z (C) Group 2(s) << 2p z (C) Group 4(p z ) Primary B 1u interaction

Primary A g interaction Primary B 1u interaction

Bonding  Bonding  Non-bonding  Non-bonding  4 bonds All occupied MO’s are 3c-2e

LUMO HOMO The frontier orbitals of CO 2

Molecular orbitals for larger molecules: H 2 O 1. Determine point group of molecule: C 2v 2. Assign x, y, z coordinates (z axis is higher rotation axis; if non-linear y axis in outer atoms point to central atom - not necessary for H since s orbitals are non-directional) 3. Find the characters of the representation for the combination of 2s orbitals on the outer atoms, then for p x, p y, p z. (as for vibrations, orbitals that change position = 0, orbitals that do not change =1; and orbitals that remain in the same position but change sign = -1) 4. Find the irreducible representations (they correspond to the symmetry of group orbitals, also called Symmetry Adapted Linear Combinations SALC’s of the orbitals). 5. Find AO’s in central atom with the same symmetry 6. Combine AO’s from central atom with those group orbitals of same symmetry and similar E

For H H group orbitals  v ’ two orbitals interchanged  E two orbitals unchanged C 2 two orbitals interchanged  v two orbitals unchanged    

No match

pzpz bonding slightly bonding antibonding pxpx bonding antibonding pypy non-bonding

 3 10 Find reducible representation for 3H’s Irreducible representations:  Molecular orbitals for NH 3

pzpz bonding Slightly bonding anti-bonding bonding anti-bonding LUMO HOMO

Projection Operator Algorithm of creating an object forming a basis for an irreducible rep from an arbitrary function. Where the projection operator sums the results of using the symmetry operations multiplied by characters of the irreducible rep. j indicates the desired symmetry. l j is the dimension of the irreducible rep. h the order order of the group. Starting with the 1s A create a function of A 1 sym ¼(E1s A + C 2 1s A +  v 1s A +  v ’1s A ) = ¼ (1s A + 1s B + 1s B + 1s A )

Consider the bonding in NF 3  A A B C D  B  C  D  A = A 2 + E  B =  C =  D = A 1 + E 1 23 }

Now construct SALC  A = A 2 + E P A2 (p 1 ) = 1/6 (p 1 + p 2 + p 3 + (-1)(-p) 1 + (-1)(-1p 3 ) + (-1)(-p 2 ) No AO on N is A 2 Now construct an SALC of A 2 sym See what the different symmetry operations do to the one of the types of p orbitals.

E: P A2 (p 1 ) = (2p 1 - p 2 - p 3 ) = E 1 Apply projection operator to p 1 But since it is two dimensional, E, there should be another SALC P A2 (p 2 ) = (2p 2 - p 3 - p 1 ) = E’ But E 1 and E’ should be orthogonal. We want sum of products of coefficients to be zero. Create a linear combination of E 2 = E’ + k E 1. = (-1 +k*2) p 1 + (2 + k(-1)) p 2 + (-1 + k(-1))p 3 Have to choose k such that they are orthogonal. 0 = (2(-1 + k*2) -1 (2 + k(-1)) -1 (-1 + k(-1)) k = ½ E 2 = (3/2 p 2 - 3/2 p 3 ) = p 2 – p 3

The geometries of electron domains Molecular shapes: When we discussed VSEPR theory Can this be described in terms of MO’s?

Hybrid orbitals s + p = 2 sp hybrids (linear) s + 2p = 3 sp 2 hybrids trigonal planar s + 3p = 4 sp 3 hybrids tetrahedral s + 3p + d = 5 dsp 3 hybrids trigonal bipyramidal s + 3p + 2d = 6 d 2 sp 3 hybrids octahedral