Question

Question 4.40 What is meant by the term bond order? Calculate the bond order of: N2, O2, O2(+) and O2(–).

Class XI Chemical Bonding and Molecular Structure Page 131

Answer 1

Bond order :- Bond order is defined as one half the difference between the number of electrons present in bonding and antibonding orbitals .
$e.g., B.O =\frac{1}{2}[N_b-Na]$
A positive bond order means a stable molecule .but negative or zero bond order means an unstable molecule.
stability of a molecule is directly proportional to bond order .

Calculation of the bond order of N2, O2, O2+ and O2- :-

1. electronic configuration of N2 ( 14 electrons )
σ1s²,σ*1s², σ2s²,σ*2s², (π2px²≈π2Py²),σ2pz²
so, Na = 4, Nb = 10
now, B.O = 1/2[ 10 - 4 ] = 3

2. electronic configuration of O2(16 electrons)
σ1s²,σ*1s², σ2s²,σ*2s², (π2px²≈π2Py²),(π*2Px¹≈π*2Py¹)
Na = 6 , Nb = 10
now, B.O = 1/2 [ 10 - 6 ] = 2

3. electronic configuration of O2+(15 electrons)
σ1s²,σ*1s², σ2s²,σ*2s², (π2px²≈π2Py²),(π*2Px¹≈π*2Py)
Na = 5 , Nb = 10
now, B.O = 1/2[ 10 - 5 ] = 2.5

4. electronic configuration of O2-(17 electrons)
σ1s²,σ*1s², σ2s²,σ*2s², (π2px²≈π2Py²),(π*2Px²≈π*2Py¹)
Na = 7, Nb = 10
now, B.O = 1/2[10 - 7] =1.5

Answer 2

Answer:

Bond order is defined as the number of covalent bonds in a covalent molecule.It is equal to one half of the difference between the number of electrons in the bonding & antibonding molecular orbitals.

Explanation:

BOND ORDER OF N2

Number of bonding electrons = 10

Number of anti-bonding electrons = 4

Bond order of nitrogen molecule = 1/2 (10-4) = 3

Bond order of O2

Number of bonding electrons = 8

Number of anti-bonding electrons = 4

Bond order = 1/2 (8-4) =2

Hence, the bond order of oxygen molecule is 2.

Bond order of O2+

Bond order of O2+=1/2 (8-3)= 2.5

Thus, the bond order of O2+ is 2.5.

Bond order of O2-

Bond order of = O2- =1/2 (8-5) = 1.5

Thus, the bond order of ion is O2- = 1.5.