CHEM 001B Lecture Notes - Lecture 5: Molecular Orbital Theory, Molecular Orbital Diagram, Antibonding Molecular Orbital

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harlequintortoise375

9 Mar 2018

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Mo guidelines: the total number of mos = the number of aos orbitals mixed, orbitals with similar energy/shape mix more effectively, orbitals of different n result in less effective mixing, a mo can accommodate two electrons. Optical isomers = chirality: chirality property of a molecule that is not superimposable on its mirror image. How to identify a chiral center: a chiral carbon is a carbon atom that is attached to four different types of atoms or groups of atoms. Bond order and stability: bond order = (number of bonding e- - number of antibonding e-, electro(cid:374)s fill mo diagra(cid:373)s accordi(cid:374)g to hu(cid:374)d"s rule. Mo for no: zeff alters mo diagram; atomic orbitals for o are lower in energy. Metallic bonds: for metals like copper, their atomic orbitals combine and form molecular bonds, as more mos form, their energies get closer together until a co(cid:374)ti(cid:374)uous e(cid:374)erg(cid:455) .

Related Questions

Somehow the answer to this is (4) but I dont see how that is possible so please EXPLAIN.

Which of the following statements relating to molecular orbital theory (MO) is incorrect?

(1) Combination of two atomic orbitals produces two molecular orbitals

(2) Combination of two 2p orbitals may result in either sigma or pi MOs

(3) A species with a bond order of zero will not be stable.

(4) Molecular orbitals are hybridizations of atomic orbitals.

(5) A diatomic molecule with an odd number of electrons will be paramagnetic

emeraldtortoise550

(a) Using only the valence atomic orbitals of a hydrogen atom and a fluorine atom, and following the model of Figure 9.46, how many MOs would you expect for the HF molecule? (b) How many of the MOs from part (a) would be occupied by electrons? (c) It turns out that the difference in energies between the valence atomic orbitals of H and F are sufficiently different that we can neglect the interaction of the 1s orbital of hydrogen with the 2s orbital of fluorine. The 1s orbital of hydrogen will mix only with one 2p orbital of fluorine. Draw pictures showing the proper orientation of all three 2p orbitals on F interacting with a 1s orbital on H. Which of the 2p orbitals can actually make a bond with a 1s orbital, assuming that the atoms lie on the z-axis? (d) In the most accepted picture of HF, all the other atomic orbitals on fluorine move over at the same energy into the molecular orbital energy-level diagram for HF. These are called “nonbonding orbitals.” Sketch the energy-level diagram for HF using this information and calculate the bond order. (Nonbonding electrons do not contribute to bond order.) (e) Look at the Lewis structure for HF. Where are the nonbonding electrons?

chocolatehedgehog590

The molecular orbitals associated with the Ï-bonding MOs insquare cyclobutadiene (C4H4) are, like benzene, form by overlap ofa single p-orbital on each carbon atom that is not involved in thesigma-bonding network of the molecule. These p-orbitals pointperpendicularly with respect to the plane defined by the carbonatoms. There is no central atom, so the only interactions thatmatter are between the p-orbitals among themselves. These MOs haveexactly the same phase relationship as the SALCs you just made inquestion 1. i) Use your results from question 1 to draw the fourÏ-orbitals of C4H4. ii) Use the idea that the energy of a MOincreases with number of new nodes made during bonding (i.e. nodesnot already present in initial AO functions) to sketch a molecularorbital energy level diagram for the Ï-molecular orbitals of C4H4.Remember that, by definition, degenerate orbitals are identicalexcept for trivial rotation in space, and have identical energy.iii) Fill the MOs with the electrons from the atomic p-orbitalsthat you used to form the Ï-MOs of C4H4. Use your diagram todetermine the overall bond order by considering each doublyoccupied bonding MO to contribute 1, each nonbonding orbital 0 andeach antibonding orbital -1. The result would be divided by 4 C-Cbonds to get the average contribution to each C-C bond fromÏ-bonding, which then should be added to a Ï-bonding contributionof 1.0. iv) Repeat step (iii) for the dication C4H4 2+ to determinethe net bond order for this species. Why do you think you got thisvalue? v) Predict which of C4H4 and C4H4 2+ could be observed byESR spectroscopy. vi) Which of C4H4 2+ and C4H4 is more likely tobe susceptible to electrophilic attack.

CHEM 001B Lecture Notes - Lecture 5: Molecular Orbital Theory, Molecular Orbital Diagram, Antibonding Molecular Orbital

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