Question

Calculate the bond order of N2,O2,O2^+and O2^-

Answer 1

Bond order= No of Bonds between the compound

Bond orders are calculate using Molecular orbital theory,which is quite time consuming and confusing. Just remember the method I  Give you :)


10=1
11=1.5
12=2
13=2.5
14=3               ⇔       Just remember this,all other reduce by 0.5
15=2.5
16=2
17=1.5
18=1

Now what are all these numbers?
LHS is the number of electrons in the atom or compound
RHS = bond order.

Now lets get back to the question:
N₂=N+N = 7e⁻+7e⁻ = 14e-  which has bond order =3 ,from the above data
O₂=8e⁻+8e⁻ = 16e⁻  which has bond order = 2
O₂⁺=8e⁻+8e⁻ - 1e⁻ = 15e⁻ which has bond order= 2.5
O₂⁻ = 8e⁻ +8e⁻ +1e⁻ =17 e⁻ which has bond order=1.5

This method is for easily getting answer.

The formal ways is by filling the no of electron in the MOT table given below,and using the formula

No of Bond order= (No of bonding electrons- No of anti bonding electrons)/2

You get the same result. :)

Answer 2

Answer:

Explanation: bond order of N2 =Nb-Na/2=8-2/2=3. O2=8-4/=2. O2^+ =8-3/2=5/2. O2^- =8-5/2=3/2.

Bond lenght:O2^->O2>O2^+