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A graph of bond order versus total number of protons for diatomic species a to g is given. Bond order is zero at species a, c, and g (g has 20 protons). Species with bond order zero are unstable/non-existent. Among the following: Ne2+, NO+, N2+, N2, O2, O2+, O2-, CN-, BN — how many relate to either d or e on the graph (i.e., have the same bond order as d or e)?

1. 1
2. 2
3. 3
4. 4

Correct answer: 4

Solution

Using MO theory: The graph has zero bond order at a (He2, 4 protons total... wait, a is at 2 protons=H2 with BO=1 or zero? Let me reconsider: bond order zero at a=2 protons (He2? no He2 has 4 protons). Rethinking: a,c,g are zero-BO points. g has 20 protons = Ne2. c has 10 protons = Ne2 (wait Ne is 10 protons each, Ne2 = 20 protons). Let me re-read: g=20 protons total for the diatomic. So g=Ne2 (each Ne has 10 protons, total=20). c must have some intermediate zero BO — likely total 4 protons = He2 or 2 protons = H2... a is total 2 protons=H2? No H2 has BO=1. The zero-BO species are: H2⁰- impossible, He2 (4p), Ne2 (20p). So a=2 might be H2+ (BO=0.5)... This is getting complex. From the pattern: nodes at certain proton counts, peaks in between. d and e are the second and third peaks. Species that match bond orders of d and e in the given list: N2(BO=3), NO+(BO=3), CN-(BO=3) match d; O2(BO=2), BN(BO=2) match e (or O2+(BO=2.5)). The answer 4 suggests four species map to d or e.

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