Question

What is meant by the term bond order? Calculate the bond order of ${N_2},{O_2},{O_2}^ + $ and ${O_2}^ - $

Answer 1

Hint:The bond order shows the number of chemical bonds present between a pair of atoms. Its formula is defined as half the difference between the number of electrons in bonding orbitals and antibonding orbitals. Higher is the bond length, higher is the stability of the molecule.

Formula used:
Bond order $ = \dfrac{1}{2}[a - b]$
Where a is the number of electrons in bonding molecular orbitals and
B is the number of electrons in antibonding molecular orbitals
Complete step by step answer:
The bond order of a covalent bond is the total number of covalently bonded electron pairs between two atoms in a molecule. Further, it can be found by drawing the Lewis structure of the molecule and counting the total number of electron pairs between the atoms. The single bonds have a bond order of1, double bonds have a bond order of 2 and triple bonds have a bond order of 3.
Now, let’s calculate the bond order of the given molecules.

1) ${N_2}$
Basically, the bond order of this molecule is 3 as it possesses triple bond.
So, according to the formula let’s calculate the bond order
 $
   = \dfrac{1}{2}(10 - 4) \\
   = \dfrac{1}{2}(6) \\
   = 3 \\
 $

 2)${O_2}$
According to the formula,
 $
  a = 8 \\
  b = 4 \\
 $
Bond order $ = \dfrac{1}{2}(8 - 4)$
=2

 3)${O_2}^ + $ - Now in this case there is one extra electron
So, according to the formula bond order will be:
 $ = \dfrac{1}{2}(10 - 5)$
=2.5

 4)${O_2}^ - $
According to the formula,
Bond order $ = \dfrac{1}{2}(10 - 7)$
=1.5

Note:If the bond order of a covalent bond is zero, then the two atoms in the question are not covalently bonded i.e. no bond exists. Moreover, the difference in the electronegativities of the atoms participating in the chemical bond also contributes to the bond energy.