Question

Use MO diagrams to place B2+, B2, and B2- in order of (a) decreasing bond energy; (b) decreasing bond length.

Answer 1

We use the MO diagram for a homonuclear diatomic species (since C and N are neighbours, we treat them as the "same").

The first two electrons contribute to bonding. The next two are anti-bonding.

The next six contribute to bonding, and the following six are anti-bonding.

So, if we start with CN+, which has 4+5-1 (8) valence electrons, we note that the first two electrons contribute to bonding, while the next two cancel this out; the next four contribute to bonding, so the bond order is 4/2 = 2.

If we add one more electron to get CN, there are now 5 bonding electrons, giving bond order 5/2=2.5.

Adding one more to give CN- would give the bond order 6/2 = 3. (If we added more electrons, each one would lower the bond order.)

Given a series of molecules with identical skeletal structures, the one with the highest bond order has the highest bond energy:

CN+ < CN < CN-

Lewis structures will verify that CN- has a triple bond, but they do not work particularly well for CN+ and CN.

learn more about bond orders at

brainly.com/question/9713842

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