Give the Molecular Orbital Energy diagram of a) $N_{2}$ and b) $O_{2}$ . Calculate the respective bond order. Write the magnetic nature of $N_{2}$ and $O_{2}$ molecules.

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Solution

$B o n d o r d e r = \frac{(N o . o f e l e c t r o n s i n b o n d i n g o r b i t a l) - (N o . o f e l e c t r o n s i n a n t i - b o n d i n g o r b i t a l)}{2}$
Bond order of $N_{2} = \frac{10 - 4}{2} = \frac{6}{2} = 3 \Rightarrow N \equiv N$
Bond order of $O_{2} = \frac{10 - 6}{2} = \frac{4}{2} = 2 \Rightarrow O \equiv O$
If orbitals contains unpaired electrons, it is paramagnetic. Otherwise diamagnetic.
$N_{2}$ does not have unpaired electrons. So, it is diamagnetic.
$O_{2}$ have 2 unpaired electrons. So, it is paramagnetic.

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