Question

write molecular orbital configuration of oxygen molecule.calculate the bond order and predict its magnetic behavior​

Answer 1

Explanation:

oxygen molecule contain 2 unparied electrons .so,its magnetic property is paramagnet

Answer 2

The molecular orbital configuration of the oxygen molecule is as follows-

$\sigma_{1s}^2 \sigma_{1s}*^2 \sigma_{2s}^2 \sigma*_{2s}^2 \sigma_{2px}^2 \pi_{2py}^2 \pi_{2pz}^2 \pi_{2py}*^1 \pi_{2pz}*^1 \sigma*_{2px}$.

• The molecular orbital configuration of the oxygen molecule is as follows-
• $\sigma_{1s}^2 \sigma_{1s}*^2 \sigma_{2s}^2 \sigma*_{2s}^2 \sigma_{2px}^2 \pi_{2py}^2 \pi_{2pz}^2 \pi_{2py}*^1 \pi_{2pz}*^1 \sigma*_{2px}$.
• It has 10 electrons in the bonding molecular orbital and 6 electrons in the antibonding molecular orbital.
• Bond order=(bonding electrons- antibonding electrons)/2.
• B.O=(10-6)/2
• B.O=2
• The bond order of oxygen gas is 2.
• It has two unpaired electrons so, it is paramagnetic.