- A0, 1, 0, 1
- B2, 1, 0, 1
- C0, 1, 2, 1
- D2, 1, 2, 1
View written solutionFree
Correct answer: C
- Write the MO ordering
For second-period homonuclear diatomic molecules:
- For :
- For :
We now find the highest occupied molecular orbital (HOMO) in each species and count its unpaired electron(s).
Each N has 7 electrons, so has electrons total.
MO configuration:
The HOMO is , which contains 2 paired electrons.
So, number of unpaired electrons in HOMO of is:
Removing one electron from removes it from the HOMO, i.e. from .
So configuration becomes: for the HOMO.
Hence the HOMO has one unpaired electron.
So, number of unpaired electrons in HOMO of is:
Each O has 8 electrons, so has electrons total.
For oxygen, MO order is:
Configuration:
The HOMO consists of the degenerate orbitals and , each with one electron.
Thus, in the highest occupied molecular orbital level, there are: unpaired electrons.
Removing one electron from removes it from the HOMO level.
So configuration becomes: spread over two degenerate orbitals. One orbital has a pair and the other has one single electron.
Thus, number of unpaired electrons in HOMO of is:
- Final sequence
For
the number of unpaired electrons in the HOMO are:
This matches Option C.
- Comparison with stored answer
Stored correct answer: C
Derived answer: C
Hence, the derived answer agrees with the stored correct answer.
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