Using molecular orbital theory, determine the bond order and magnetic property of $O_2$.
Using molecular orbital theory, determine the bond order and magnetic property of $O_2$.
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 3mo ago
▲ 7
Total electrons in $O_2$ = 16. MO configuration:
$$(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p})^2(\pi_{2p})^4(\pi^*_{2p})^2$$
$$Bond Order = \frac{N_b - N_a}{2} = \frac{10 - 6}{2} = 2$$
The two electrons in $\pi^*_{2p}$ orbitals are unpaired (by Hund's rule):
$$O_2 is paramagnetic$$
Log in to post your own answer or join the discussion.
Discussion (4)
VC
This finally made it click for me — thank you!
T
This finally made it click for me — thank you!
M
I solved it a slightly different way and got the same answer, good sign.
RJ
How do we know the approximation is valid here?