Question

Compare Relative Stability of following O2,O2+,O2-​

Answer 1

Answer:

Electronic configurations:

O2 = ? 1 s2  ?*1 s2  ?2 s2 ?*2 s2  ?2 p 2 z (?2p 2 x  ?2 p 2 y )  (?*2 p 1 x = ?*2 p1 y )  

O2+ = ? 1 s2  ?*1 s2  ?2 s2 ?*2 s2  ?2 p 2 z (?2p 2 x  ?2 p 2 y )  ?*2 p 1 x

O2 - = ? 1 s2  ?*1 s2  ?2 s2 ?*2 s2  ?2 p 2 z (?2p 2 x  ?2 p 2 y )  (?*2 p 1 x = ?*2 p1 y ) ?*2 p1 z

O2 2-  = ? 1 s2  ?*1 s2  ?2 s2 ?*2 s2  ?2 p 2 z (?2p 2 x  ?2 p 2 y )  (?*2 p 1 x = ?*2 p1 y ) ?*2 p2 z  

The electronic configuration of the O2 + ion containing 15 electrons can be written as:

? 1 s2  ?*1 s2  ?2 s2 ?*2 s2  ?2 p 2 z (?2p 2 x  ?2 p 2 y )  (?*2 p 1 x = ?*2 p y )  ?*2 p z

The bond order can be found as:

B.O = (Nb-Na)/2

Nb = Number of electrons in the bonding orbitals = 10

Na = Number of electron in the anti-bonding orbitals = 5

B.O = (10-5)/2 = 5/2 = 2.5

The electronic configuration of the O2 - ion containing 17 electrons can be written as:

? 1 s2 < ?*1 s2 < ?2 s2 < ?*2 s2 < ?2 p 2 z < (?2p 2 x = ?2 p 2 y ) < (?*2p 2 x = ?*2 p 1 y ) < ?*2 p z

The bond order can be calculated as:

B.O = (10-7)/2 = 3/2 = 1.5

Similarly, Bonbd order of O2 is = B.O = (10-6)/2 = 4/2 = 2

Bonbd order of O2 2- is = B.O = (10-8)/2 = 2/2 = 1

The higher the value of bond order, higher is the stability of the bond, so on the basis of above information, we can say that, O2 + ion is more stable than O2,  O2 - and O22- ions.