How to write the molecular orbital configuration of
O2^- and O2^2-
Also state the Bond Order, Magnetic nature and stability of the molecule
Answers
Answer:
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Explanation:
As it can be seen from the MOT of O
2, The electrons in the highest occupied molecular orbital are unpaired therefore it is paramagnetic in nature. Also, the bond order can be calculated as [N b −N a ]/2=[10−6]/2=2. Therefore there is a double bond present as O=O.
Answer:
(1) The electronic configuration of O
2
molecule is
KK(σ2s)
2
(σ
∗
2s)
2
(σ2p
z
)
2
(π2p
x
)
2
(π2p
y
)
2
(π
∗
2p
x
)
1
(π
∗
2p
y
)
1
Its bond order is
2
8−4
=2.
It contains 2 unpaired electrons and is paramagentic.
(2) The electronic configuration of O
2
+
ion is
KK(σ2s)
2
(σ
∗
2s)
2
(σ2p
z
)
2
(π2p
x
)
2
(π2p
y
)
2
(π
∗
2p
x
)
1
Its bond order is
2
8−3
=2.5.
It contains 1 unpaired electron and is paramagnetic.
(3) The electronic configuration of O
2
−
ion is
KK(σ2s)
2
(σ
∗
2s)
2
(σ2p
z
)
2
(π2p
x
)
2
(π2p
y
)
2
(π
∗
2p
x
)
2
(π
∗
2p
y
)
1
Its bond order is
2
8−5
=1.5.
It contains 1 unpaired electron and is paramagnetic.
(4) The electronic configuration of O
2
2−
ion is
KK(σ2s)
2
(σ
∗
2s)
2
(σ2p
z
)
2
(π2p
x
)
2
(π2p
y
)
2
(π
∗
2p
x
)
2
(π
∗
2p
y
)
2
Its bond order is
2
8−6
=1.
It contains all paired electrons and is diamagnetic.
Higher is the bond order, stronger is the bond, and higher is the stability.
The decreasing order of the stability is O
2
+
>O
2
>O
2
−
>O
2
2−
.
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