Mo Of Beh2, Hf2-, Co2 6v4e4n

5.2. MO Theory of Larger Molecules Suggested Textbook Reading: Chapter 6, Sections 6.6 – 6.7 and 6.9 - 6.10 •

For diatomics, deriving the MO's was a simple matter of "visually" combining the available AO on the two bond forming atoms.

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The geometry of the molecule is predetermined (i.e., linear).

For polyatomic molecules we face two problems: 1) What are the meaningful and allowed LCAO’s leading to the MO’s? 2) How do the shapes and energies of these MO’s relate to the structure and properties of the molecule? The concept of group orbitals: SALCs (See Resource Section 5 of your textbook) SALC: Symmetry Adapted Linear Combinations of Atomic Orbitals Concept: Instead of mixing all atomic orbitals at once to make molecular orbitals of a large molecule, we will BUILD SMALLER FRAGMENTS and mix those. •

In molecules of the type EXn (n = 2, …, 9) consider the orbitals on the central atom E and the set of terminal Xn atoms separately:

E

→ s, px, py, pz, dxy, dxz, dyz, dz2, dx2-y2

X

→ groups of symmetry equivalent orbitals e.g. n × px and linear combinations thereof such as px(a) + px(b) and px(a) - px(b)

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Determine how the sets of group orbitals interact with the AO of appropriate symmetry on the central atom (I.E., USE THE CHARACTER TABLES)

Example #1: Linear EH2 (e.g., BeH2) 1)

Which are the valence AO’s of Be (i.e., the ones we will use for our MO)? 2s, 2px, 2py, 2pz

2)

How do we make SALCs from the two terminal H atoms? Only 1s orbitals are available for H atoms, so we make one in-phase and one out-of-phase combination of these 1s orbitals. (2 AO’s gives 2 MO’s)

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3)

How do we mix the AO’s of the central Be atom with the SALCs of the terminal H atoms? This example is easy enough that we can do it visually. For more difficult examples, we will have to determine the point group, set a coordinate system, and assign symmetry labels to the AO’s and SALCs.

Example #2: Linear EX2 (e.g., FHF- and CO2, both point group D h.) ∞

Questions to ask: •

How can we group both s and p atomic orbitals of the terminal atoms (F or O) into symmetry equivalent sets (i.e., form the SALCs)?

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How do these SALCs interact with the s, px, py, pz orbitals of the central atom (H or C, respectively)?

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The orbitals in FHF•

Only one orbital (1s) on central hydrogen atom

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Only SALCs #1 and #3 have the right symmetry to give bonding or anti-bonding interactions; BUT the F2s are too low in energy to mix!

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H1s interacts with SALC #3 to give one bonding and one antibonding MO (2 mix to give 2)

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SALCs #2 and #4 - #8 are nonbonding with respect to H1s.

The resulting MO Scheme: Source: Miessler & Tarr, Inorganic Chemistry, PrenticeHall, 1998.

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The orbitals in CO2 with symmetry labels using the reduced symmetry of point group D2h The group orbitals (SALCs) formed by the two oxygen atoms:

The orbitals on the central carbon atom:

The orbital energies from (PES or quantum calculation) are: Orbital Carbon Oxygen

2s -19.5 eV -32.4 eV

2p -10.7 eV -15.9 eV

→ The oxygen 2s orbital is too low in energy to have interactions with the other orbitals. •

In order to arrive at the bonding and anti-bonding interaction the orbital(groups) of matching symmetry are combined, i.e. B2u(carbon) with B2u(oxygen), Ag(carbon) with Ag(oxygen) etc. 121

The MO scheme resulting from the symmetry allowed combinations:

Source: Miessler & Tarr, Inorganic Chemistry, Prentice-Hall, 1998. (amended)

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So far we only covered linear molecules (see also CHEM 206 for extension to cyclic molecules). IF YOU CAN’T , GO BACK AND LOOK AGAIN!

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How do we approach non-linear molecules?

5.3. The Group Theoretical Approach to MOs Goal: Have an algorithm that allows us to derive the physically meaningful MOs for any molecule. The great theorem: Any allowed and physically meaningful LCAO ( = group orbitals + central atom orbitals) resulting in an MO must form a basis for an irreducible representation of the point group of the molecule under consideration. (For a rigorous derivation of this theorem see: F.A. Cotton, "Chemical Applications of Group Theory") Based on this theorem we can formulate our algorithm: 1. Determine the point group of the molecule. 2. Assign a coordinate system to the molecule - by convention: z = principal rotation axis of the molecule 3. Group the terminal atoms into symmetry equivalent sets - these are the SALCs. 4. In each set, group the terminal AOs according to their bonding symmetry with respect to the central atom, typically there will be s, pσ, pπ, dσ, dπ. Note: The s and pσ SALCs will have identical symmetry pattern, but in many cases the s orbitals are too low in energy to interact and therefore in a 1st approximation can often be taken as non-bonding stable orbitals (see the CO2 example earlier). 5. Sketch the SALCs. 6. Sketch the AOs of the central atom(s). 7. Find the characters of the representation for each SALC in the point group of the molecule under consideration by evaluating the effect of each symmetry operation on the SALC.

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Rules: a) Any orbital that remains unchanged contributes 1. b) Any orbital that remains in its original position, but reverses its sign contributes -1. c) Any orbital that changes position contributes 0. → Leads to an (ir)reducible representation that describes how the SALC transforms under the symmetry operations of the point group. 8. Deconvolute the reducible representation into irreducible representations using

(see earlier in your notes)

9. Find the central atom’s AOs with the same irreducible representations (see basis functions in the character table). 10. Combine the SALCs and central atom AOs of equal symmetry and similar energy (educated guess based on quantum numbers- see approximation made above) to form MOs. 11. Check: # of resulting MOs must be equal to # of AOs used !!!

Example #3: Non-linear EH2 (e.g., water) H2O: •

Point group C2v

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SALCs consist of H 1s orbitals only

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Designate a coordinate system.

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The procedure:

Source: Miessler & Tarr, Inorganic Chemistry, Prentice-Hall, 1998.

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The resulting MO scheme:

Source: Miessler & Tarr, Inorganic Chemistry, Prentice-Hall, 1998. •

What is the main difference between this picture and that given by VB theory?

Related Textbook Exercises 2.20, 2.21

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