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 among the following will have a triple bond order ? .

Options

A. `CO`
B. `CN^(Θ)`
C. `NO^(o+)`
D. All of these

Answer

D

Explanation

`: C underset(larr)(=)underset(..)overset(..)O` (Bond order =3)
(b) `:^(Θ)C -= N` (Bond order =3)
( c) `underset(..)overset(o+)N underset(rarr)overset(=) O:` (Bond order =3)
Accrding to `MOT` the bond order is a follows
`CO,CN^(Θ)` and `NO^(o+)` are isoelectronic species with `e^(-')s`
each hence they have same bond order .
.
Alternatively `CO,CN^(ɵ)` and `NO^(o+)` are isoelectronic species with `14e^(-)'s` each hence they have same bond order.