Question

Question 4.36 Compare the relative stability of the following species and indicate their magnetic properties;

O2, O2(+), O2(–) (superoxide), O2(2–) (peroxide)

Class XI Chemical Bonding and Molecular Structure Page 131

Answer 1

Concept :- bond order is directly proportional to stability.

magnetic nature depends paired and unpaired electrons.
If molecule has one or more than unpaired electrons it means molecule is paramagnetic nature.
And if molecule has no unpaired electron{ e.g., all are paired electrons } then, molecule is diamagnetic nature.


electronic configuration of O2(16 electrons)
σ1s²,σ*1s², σ2s²,σ*2s², (π2px²≈π2Py²),(π*2Px¹≈π*2Py¹) 
Na = 6 , Nb = 10 
now, B.O = 1/2 [ 10 - 6 ] = 2 
It has two unpaired electrons.so, O2 molecule is paramagnetic.


electronic configuration of O2+(15 electrons)
σ1s²,σ*1s², σ2s²,σ*2s², (π2px²≈π2Py²),(π*2Px¹≈π*2Py)
Na = 5 , Nb = 10 
now, B.O = 1/2[ 10 - 5 ] = 2.5
It has one unpaired electron so, it is paramagnetic
 

electronic configuration of O2-(17 electrons)
σ1s²,σ*1s², σ2s²,σ*2s², (π2px²≈π2Py²),(π*2Px²≈π*2Py¹)
Na = 7, Nb = 10
now, B.O = 1/2[10 - 7] =1.5 
It has one unpaired electron so, it is paramagnetic .


electronic configuration of O2^2-(17 electrons)
σ1s²,σ*1s², σ2s²,σ*2s², (π2px²≈π2Py²),(π*2Px²≈π*2Py²)
now, B.O = 1/2[10 -8] = 1
It has no unpaired electron.so, it is diamagnetic

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

Hence, increasing order of stability is
O2+ > O2 > O2- > O2^2-