18. Oct 17 (Monday!!!!!!)

Monday Oct 17
Molecular Orbitals
Chapter 5; 149-158

HW: Ex 5.4,Ex 5.5, Ex 5.6, Ex 5.7, Ex 5.8, Problems 21, 22, 23
Help sessions 121LGRT    T (4-5:30) and Th (7:30-9)

Chapter_5_MT

SLIDES: Oct_17_Mon.ppt

Goals: Use LGOs to construct MOs for linear molecules

2 thoughts on “18. Oct 17 (Monday!!!!!!)

  1. crfields

    For exercise 5.7, I keep arriving at an MO diagram with only three bonding pairs and three lone (non-bonding) pairs (plus an empty sigma n-b orbital, slightly higher in energy than the Ï€n-b). This seems at odds with the Lewis structure, which has two bonds between each nitrogen and two lone pairs on each terminal nitrogen. I know that there’s hybridization between s and p electrons and that the answer to my question is probably related to that, but otherwise I’m not really sure how to arrive at the correct answer.

  2. mjknapp Post author

    Hi,
    It sounds like you are on the right track!
    The MO model leads to a total bond order of 3, which makes each N-N bond have a BO=1.5; and this seems to require 3 or 5 lone pairs (weird, I know!). This disagrees with the Lewis model, but that definitely happens when using different bonding models – the crucial test is connection with experiments (like bond energies, or bond lengths). A slightly different question is: how does a student develop intuition to ‘trust’ their MO diagrams? To achieve this takes time. But to help you along, I got a similar answer to yours!

    The book has an answer on page 687, which starts with the LGOs on p.154. The book predicts 4 bonding pairs, one antibonding pair, and 3 lone pair; the HOMO is a degenerate pi* (b2g and b3g orbitals, if you use D2h symmetry). I worked my own answer in D2h, and let the 2s and 2p subshells talk to each other, which led to 4 bp, 0 antibonding, 5 lp; my HOMO is actually a degenerate pair of half-occupied pi* (b2u* and b3u*). The orbital energies (lowest to highest) for my model: Ag, B1u, B2u/B3u, Ag(non); B1u(non); B2g(non)/B3g(non), B2u*/B3u*, B1u*, Ag*.

    My answer has some issues, because I know that azide is not a di-radical. So I probably should push my A1g(non) orbital higher in energy, and make that my HOMO. In any event, the total bonding is unchanged – just the predictions of chemistry (diradical vs. Lewis Base).

    I realize that this is somewhat unsatisfying, as finding the “RIGHT” answer is really challenging. I have to view construction MO diagrams as a hypothesis: I propose a bonding diagram, and a proposed HOMO/LUMO. Then I test that hypothesis by evaluating data (like bond length or reactivity patterns).

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