Sample 1Medium
A molecular orbital energy-level diagram for a homonuclear diatomic molecule of a second-period element shows σ1s, σ1s, σ2s and σ2s each with two electrons, σ2pz with two, π2px and π2py with two each, and π2px and π2py with one electron each, the two spins parallel. Which statement about this molecule is correct?
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The answer is D.
The diagram holds 16 electrons, so the molecule is O2. Bond order = ½(10 − 6) = 2, and the two unpaired electrons in the π* orbitals make it paramagnetic.
- A.
N2 has 14 electrons and no electrons in the π* orbitals; this diagram holds 16.
- B.
The two unpaired electrons in π2px and π2py make the molecule paramagnetic.
- C.
O2⁻ has 17 electrons, with three in the π* orbitals; this diagram has two.
NCERT: Class 11 Chemistry, Chapter 4