What is meant by the term Bond order? Calculate the bond orders in the following.
(a) `N_(2) " "` (b) `O_(2)" "` (c) `O_(2)^(+)" "` and (c) `O_(2)^(-)`

Explanation

Bond order is the number of bonds between two atoms. It is one half the difference between the number of electrons present in the bonding and the antibonding orbitals..
Bond order`=1/2[N_(b)-N_(a)]`
Where `N_(b)` is the number of electrons in bonding molecular orbitals while `N_(a)` is the number of electrons in antibonding moleculars orbitals. For calculating bond order electrons present in valence only are considered.
a) MO electronic configuaration of `N_(2)` molecule is
`(sigma_(z)1s)^(2)(sigma_(z)^(star)1s)^(2)(sigma_(z)2s^(2))(sigma_(z)^(star)2s^(2))(pi2p^(2)=pi2p^(2))(sigma_(z)2p)^(2)`
Bond order `=(10-4)/2=6/2=3`
b) MO electronic configuration of `O_(2)` molecule is
`(sigma1s)^(2)(sigma^(star)1s)^(2)(sigma2s)^(2)(sigma^(star)2s)^(2)(sigma2p_(z))^(2)(pi2p_(x)^(2)pi2p_(y)^(2))(pi^(ast)2p_(x)^(1)pi^(ast)2p_(1)^(y))`
Bond order `=1/2[N_(b)-N_(a)]=1/2(10-6)=2`
c) MO electronic configuration of `O_(2)^(+)` molecule is
`(sigma1s)^(2)(sigma^(star)1s)^(2)(sigma2s)^(2)(sigma^(star)2s)^(2)(sigma2p_(2))^(2)(pi2p_(x)^(2)pi2p_(y)^(2))(pi^(star)2p^(1))`
Bond order `=1/2[N_(b)-N_(a)]=1/2(10-5)=2.5`
d) MO electronic configuration of `O_(2)^(-)` molecule is
`(sigma1s)^(2)(sigma^(star)1s)^(2)(sigma2s)^(2)(sigma^(star)2s)^(2)(sigma2p_(2))^(2)(pi2p_(x)^(2)pi2p_(y)^(2))(pi^(ast)2p_(x)^(2)pi^(ast)2p_(y)^(1))`
Bond order `=1/2[N_(b)-N_(a)]=1/2(10-7)=1.5`