Trick to find bond order

Today , I am going to share a trick to find the bond order of any elements/molecules/ions whose number of electrons range from 10 to 18.

BOND ORDER is defined as "TOTAL NUMBER OF ELECTRONS IN BONDING MO - TOTAL NUMBER OF ELECTRONS IN ANTI-BONDING MO /2". For the application of this formula , one need to draw "MOLECULAR ORBITAL DIAGRAM". But this might take some time.

Now , I am going to share you a trick to find "BOND ORDER".

The formula that I will use is    3- (0.5*n)
Firstly , we need to find out the value of  'n',
then substitute the value of  'n' in the formula.
'n' is nothing but "TOTAL NUMBER OF ELECTRONS PRESENT IN ANY GIVEN ELEMENT/ION/MOLECULE - 14"

LET US TAKE SOME  EXAMPLES:)

1)
FIND OUT THE BOND ORDER OF O2.

FORMULA IS 3-(0.5*n)
n = TOTAL NUMBER OF ELECTRONS PRESENT IN O2   -  14
n = 16 - 14
n=2
BY SUBSTITUTING THE VALUE OF n IN THE FORMULA,
=3 - (0.5 * 2)
=3 - 1
= 2
HENCE THE BOND ORDER OF O2 IS 2.

2)
FIND OUT THE BOND ORDER OF N2,

FORMLULA IS 3-(0.5*n)
n = TOTAL NUMBER OF ELECTRONS PRESENT IN N2  -  14
n = 14 - 14
n = 0
BY SUBSTITUTING THE VALUE OF n IN THE FORMULA,
= 3 - (0.5 * 0)
= 3 - 0
= 3
HENCE THE BOND ORDER OF N2 IS 3.

3)
FIND OUT THE BOND ORDER OF CO

FORMULA IS 3-(0.5*n)
n = TOTAL NUMBER OF ELECTRONS PRESENT IN CO  -  14
n = (6 + 8) - 14
n = 14 - 14
n = 0
BY SUBSTITUTING THE VALUE OF n IN THE FORMULA ,
= 3 - (0.5 * 0)
= 3 - 0
= 3
HENCE THE BOND ORDER OF CO IS 3.

4)
FIND OUT THE BOND ORDER OF O2-

FORMULA IS 3 - (0.5 * n)
n = TOTAL NUMBER OF ELECTRONS PRESENT IN O2-  -  14
n = (2 * 8) +1 - 14   { we are adding 1 because of the anionic charge of -1}
n = 16 + 1-14
n = 17-14
n = 3
BY SUBSTITUTING THE VALUE OF n IN THE FORMULA,
=  3 - (0.5 * 3)
= 3 - 1.5
= 1.5
HENCE THE BOND ORDER OF O2- IS 1.5


PRACTICE EXERCISE:
FIND OUT THE BOND ORDER OF THE FOLLOWING;
a) o2 2-
b)NO
c) O2 +
d) O2 2+

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