Hybrid Orbitals & Molecular Orbital Theory:

Problem 7.82:  What is the difference in spatial distribution between electrons in a p bond and electrons in a s bond?

A s bond has its shared electrons along the bond axis between two nuclei.  On the other hand, electrons in a p bond are equally distributed above and below the bonded nuclei.  The following little picture shows this better.

The above picture gives the basics.  If you are in to gruesome details, the next picture (of ethylene) shows how p bonds really come into their glory:

Incidentally, p bonds do not have to be in the same plane.  This is shown dramatically for the molecule, allene, H₂C=C=CH₂.  (In my quantum course, this is given as a good example of D_2d symmetry... .)  Anyway, here is allene:

Note that, in this case, the central carbon has to have sp hybridization!

The above are simply a quotation of the rules.  The various shapes have already been shown with the VSEPR model.

Problem 7.86:  What hybridization do you expect for the central atom in a molecule that has the following geometry [shape]?

This is another example of just quoting some rules--but note that, here, we occasionally have more than one correct answer.
 

(a)	Tetrahedral	sp3
(b)	Octahedral	sp3d2
(c)	Bent	sp2 or sp3
(d)	Linear	sp or sp3d
(e)	Square pyramidal	sp3d2

Problem 7.88:  What hybridization would you expect for the indicated atom in each of the following?
 

(a)	H2C=O	sp2
(b)	BH4-	sp3
(c)	XeOF4	s3pd2
(d)	SO3	sp2

This is quite routine--if you understand electron-dot structures.  There only just so many hybrids possible!

Pretty much the same sort of stuff we have been doing all along here... .  The hybridizations here are actually simpler than the problem just above.

Maybe these are not that easy!  Draw the electron-dot structures to convince yourself that, when I said they were easier, I was not just whistlin' Dixie.

First, here is the image:

This diagram actually applies to this problem and to the next.  Note that we need different diagrams here since the energies of the s and p 2p orbitals are reversed for case (a) vs. case (b) above.  (With N2 it does seem strange in any event that the two p bonds would form before the s bond.  Do you think this is right?)

To handle this better and give simpler diagrams, I shall take the relevant energy level diagrams from the instructor's manual rather than fiddling with the above combined diagram.  After all,

Anyway, here are the electrons put into the appropriate orbital energy diagrams for all three oxygen species of this problem.

(These are shown in red since that is the traditional color of oxygen in molecular models.)  We can gauge the stability of these by calculating the bond order.  This is given simply by

(Unlike the authors, I tend to use real symbols!)  Doing the calculations, we see immediately that all three of these species are stable.  We see that the bond orders decrease from left to right, incidentally.  Here are the actual calculations.

As said before, all are stable.  In addition, all have one or more unpaired electrons.  The first of these, incidentally, takes about as much energy to form as a Xe⁺ ion.  This is what initially led to the discovery of xenon compounds.

This problem goes pretty much like the previous one.  First, we draw the various orbital occupation energy level diagrams:

(In this case, the traditional molecular model color for nitrogen is blue.)  We see that the two ions have unpaired electrons whereas N₂ has all electrons paired.  These are easily shown to be all stable as evidenced by the following bond orders:

A bond order of three, incidentally, is about as high as they get with representative elements.