Question

Use molecular orbital theory to explain bond order , relative stability of O 2 and O 2 ^ +​

Answer 1

Explanation:

Bond order is directly proportional to stability.

Magnetic nature depends paired and unpaired electrons.

If molecule has unpaired electrons it means molecule is paramagnetic nature.

And if the molecule has no unpaired electron{ e.g., all are paired electrons } then, a molecule is diamagnetic nature.

Electronic configuration of O

2

(16 electrons):

σ1s²,σ∗1s²,σ2s²,σ∗2s²,(π2px²≈π2Py²),(π∗2Px¹≈π∗2Py¹)

Na(Anti bonding molecular orbitals) = 6 , Nb(Bonding molecular orbitals) = 10

Bond order=

2

1

[bonding molecular orbotals - anti bonding

molecular orbotals]

now, B.O =

2

1

[10−6]=2

It has two unpaired electrons.so, O

2

molecule is paramagnetic.

Electronic configuration of O

2

2−

(17 electrons):

σ1s²,σ∗1s²,σ2s²,σ∗2s²,(π2px²≈π2Py²),(π∗2Px²≈π∗2Py²)

now, B.O =

2

1

[10−8]=1

It has no unpaired electron.so, it is diamagnetic

Now, the bond order is directly proportional to stability so, higher the bond order will be a higher stable molecule or ion.

Hence, increasing order of stability is

O

2

>O

2

2−