In `MO` energy diagram for heteronuclear diatomic molecule is similar However, the energies of the `AO` s of the atom having higher atomic number being lower, the diagram will be unsymmetrical, but that will not make a difference in the electron count The bond order is half the difference in the number of electrns of the bonding `(sigma and pi)` and anti-bonding `(sigma and pi)` `MO s` For a bond to have been formed the bond order the shorter is the bond distance and the greater is the bond dissociation energey But if the bond order is smae in the above two cases, then the bond distance will be greater and the bond dissocation energy smaller in the case which has more populated anti-bonding orbitals The presence of unpaired electron(s) in a molecualr orbital will make the system paramagnetic
Which of the following orders is correct in respect of bond dissociation energey ? .

Options

A. `N_(2)^(o+) gtN_(2)^(Θ)`
B. `O_(2)gtO_(2)^(Θ)`
C. `NO^(o+)gtNO`
D. All of these

Answer

D

Explanation

Higher the bond order higher is the bond dissociation energy
(a) Bond order of both `N_(2)^(o+)` is ``2.5 Bur in `N_(2)^(o+)` all `5e^(-')s` are presenct in bonding orbitals while in `N_(2)^(Θ)` one `e^(-)` is present in antibonding orbital Thsu `N_(2)^(o+)` is more stable and hence have high dissociation energy
(b) Bond orders of `O_(2)^(o+)` and `O_(2)^(Θ)` are 2.5 and 1.5 respectively hence bond dissociation energy of `O_(2)^(o+) gt O_(2)^(Θ)`
(c ) Bond orders of `NO^(o+)` and `NO` are 3 and 2.5 Hence bond dissociation energy of `NO^(o+) gt NO`
Hence the correct answer is (d) .