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.

Explanation

a) Molecular Orbital Energy diagram of `N_(2)` :

Electronic configuration of molecular orbital of `N_(2)` :
`sigma1s^(2),sigma^(star)1s^(2), sigma2s^(2), sigma^(star)2s^(2), pi2p_(y)^(2)=pi2p_(z)^(2), sigma2p_(x)^(2)`
Calculation of bond order of `N_(2)` :
Number of bonding electrons `(N_(b))` = 10
Number of anti-bonding electrons `(N_(a))` = 4
Bond order `=1/2(N_(b)-N_(a))`
`=1/2(10-4)=3`
Nitrogen is diamagnetic since all the electrons in `N_(2)` molecule are paired.
b) Molecular Orbital Energy diagram of `O_(2)` :

Electronic configuration of molecular orbital of `O_(2)` :
`sigma1s^(2), sigma^(star)1s^(2), sigma2s^(2), sigma^(star)2s^(2), sigma2p_(x)^(2), pi2p_(y)^(2)=pi2p_(x)^(2), pi^(star)2p_(y)^(1)=pi^(star)2p_(z)^(1)`
Calculation of bond order of `O_(2) :`
Number of bonding electrons `(N_(b))=10`
Number of anti-bonding electrons `(N_(a))=6`
Bond order `=1/2(N_(b)-N_(a))=1/2(10-6)=2`
Oxygen is paramagnetic as there are two unpaired electrons in two antibonding `pi^(ast)2p_(y) " and " pi^(ast)2p_(z)` molecular orbitals. So oxygen is diamagnetic.