calculate the bond order of o2- and o2²- their magnetic character using molecular orbital diagrams??
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Explanation:
Oxygen molecule O
2
: The electronic configuration of oxygen (z=8) in the ground state is 1s
2
2s
2
2p
4
. Each oxygen atom has 8 electrons, hence in O
2
molecule there are 16 electrons.
Therefore, the electronic configuration of O
2
is as follows :
O
2
:KK(σ
2s
)
2
(σ
2s
∙
)
2
(π
2p
x
)
2
=(π
2p
y
)
2
(π
2p
x
)
1
=(π
2p
y
∙
)
1
The molecular orbital energy level diagram of oxygen molecule is given as follows :
Bond order
2
N
b
−N
a
=
2
8−4
=2
Thus, oxygen molecule has two bonds. i.e., one is bond and one p bond. The last two electrons in p
2px
∙
and p
2py
∙
orbitals will remain unpaired. Therefore, oxygen molecule has paramagnetic character due to the presence of two unpaired electrons.
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