(i)

Bond order ∝ stability (i.e., bond strength)

Helium molecule (He 2 ) : He 2 : ( σ 1s) 2 ( σ *1s) 2

Bond order of He 2 is ½(2 – 2) = 0

The molecular orbital description of He 2 predicts two electrons in a bonding orbital and two electrons in an antibonding orbital, with a bond order of zero - in other words, no bond. The noble gas He has not significant tendency to form diatomic molecules and, like the other noble gases, exists in the form of free atoms.

Carbon molecule (C 2 ) : ( σ 1s) 2 ( σ *1s) 2 ( σ 2s) 2 ( σ *2s) 2 ( π 2p 2 x = π 2p 2 y ) or KK ( σ 2s) 2 ( σ *2s) 2 ( π 2p 2 x = π 2p 2 y )

Lithium molecule (Li 2 ) : ( σ 1s) 2 ( σ *1s) 2 ( σ 2s) 2

Peroxide (O 2 2– ) : ( σ 1s) 2 ( σ *1s) 2 ( σ 2s) 2 ( σ *2s) 2 ( σ 2p z ) 2 ( π 2p 2 x = π 2p 2 y ) ( π *2p x 2 = π *2p 2 y )

As all electrons are paired so C 2 , Li 2 and O 2 2– are diamagnetic.

Fluorine molecule (F 2 ) : ( σ 1s) 2 ( σ *1s) 2 ( σ 2s) 2 ( σ *2s) 2 ( σ 2p z ) 2 ( π 2p 2 x = π 2p 2 y ) ( π *2p x 2 = π *2p 2 y )

(ii)

(iii) Oxygen molecule (O 2 ) : ( σ 1s) 2 ( σ *1s) 2 ( σ 2s) 2 ( σ *2s) 2 ( σ 2p z ) 2 ( π 2p 2 x = π 2p 2 y ) ( π *2p x 1 = π *2p 1 y )

Bond order = 1/2(10 – 6) = 2.0,

O 2 + : ( σ 1s) 2 ( σ *1s) 2 ( σ 2s) 2 ( σ *2s) 2 ( σ 2p z ) 2 ( π 2p 2 x = π 2p 2 y ) ( π *2p x 1 = π *2p 0 y )

Bond order = 1/2(10 – 5) = 2.5.

Nitrogen molecule (N 2 ) : ( σ 1s) 2 ( σ *1s) 2 ( σ 2s) 2 ( σ *2s) 2 ( π 2p 2 x = π 2p 2 y ) (σ 2p z ) 2

The bond order of N 2 is 1/2(10 – 4) = 3.

N 2 + : ( σ 1s) 2 ( σ *1s) 2 ( σ 2s) 2 ( σ *2s) 2 ( π 2p 2 x = π 2p 2 y ) (σ 2p z ) 1

Bond order = 1/2(9 – 4) = 2.5.