Question

Bond order of $N_{2}^{+},N_{2}^{-}$ and ${{N}_{2}}$will be:
(a) 2.5,2.5 and 3 respectively
(b) 2,2.5 and 3 respectively
(c) 3, 2.5 and 3 respectively
(d) 2.5,2.5 and 2.5 respectively

Answer 1

Hint:First, calculate the number of bonding and antibonding electrons in each species given.
The difference between bonding electrons and antibonding electrons divided by 2 will give the final answer i.e. the bond order of the molecule.

Complete step-by-step answer:In the question, three species of molecules are given and are asked to calculate the bond order of the three molecules.
We are very much familiar with the molecular orbital theory in which the atomic orbitals combine to additive fashion or in subtractive fashion to form bonding and antibonding molecular orbitals.
The electrons present in the molecule are distributed among these bonding and antibonding orbitals.
And these bonding and antibonding orbitals are arranged in the increasing order of their energy. The bonding orbitals are expressed as $\sigma \,$and $\pi $ whereas antibonding orbitals are expressed as ${{\sigma }^{*}}$and ${{\pi }^{*}}$
The order of the molecular orbitals are as follows:$\sigma 1s<{{\sigma }^{*}}1s<\sigma 2s<{{\sigma }^{*}}2s<\sigma 2{{p}_{z}}<\pi 2{{p}_{x}}=\pi 2{{p}_{y}}<{{\sigma }^{*}}2{{p}_{z}}<{{\pi }^{*}}2{{p}_{x}}={{\pi }^{*}}2{{p}_{y}}$
Now let us calculate the total number of electrons in each molecular species.
In$N_{2}^{+}$, N has 7 electrons and two N will have a total of 14 electrons, and since a positive sign is present, deduct one electron from the total number.
Hence there are 13 electrons in total, and now distribute the electrons in the orbitals according to the increasing energy of orbitals.
The configuration will be:
$\sigma 1{{s}^{2}}<{{\sigma }^{*}}1{{s}^{2}}<\sigma 2{{s}^{2}}<{{\sigma }^{*}}2{{s}^{2}}<\sigma 2{{p}_{z}}^{2}<\pi 2{{p}_{x}}^{2}=\pi 2{{p}_{y}}^{1}$
Here there are 9 bonding electrons and 4 antibonding electrons.
Therefore the bond order is calculated as, $Bond\,order\,\,=\,\dfrac{Bonding\,{{e}^{-}}s-Antibonding\,{{e}^{-}}s}{2}$
$Bond\,order\,=\,\dfrac{9-4}{2}=2.5$
In$N_{2}^{-}$, since there are two atoms and there are 7 electrons in one N atom. The total electrons in the nitrogen molecule will be 14 electrons.
Since there is a negative charge on the molecule add one electron, hence the total number of electrons present in the molecular species will be 15 electrons.
The configuration will be:$\sigma 1{{s}^{2}}<{{\sigma }^{*}}1{{s}^{2}}<\sigma 2{{s}^{2}}<{{\sigma }^{*}}2{{s}^{2}}<\sigma 2{{p}_{z}}^{2}<\pi 2{{p}_{x}}^{2}=\pi 2{{p}_{y}}^{2}<{{\pi }^{*}}2{{p}_{x}}^{1}={{\pi }^{*}}2{{p}_{y}}^{0}$
$Bond\,order\,\,=\,\dfrac{Bonding\,{{e}^{-}}s-Antibonding\,{{e}^{-}}s}{2}$
$Bond\,order\,=\,\dfrac{10-5}{2}=2.5$
In${{N}_{2}}$, there are two atoms and a total of 14 electrons are present in the nitrogen molecule.
The configuration will be:$\sigma 1{{s}^{2}}<{{\sigma }^{*}}1{{s}^{2}}<\sigma 2{{s}^{2}}<{{\sigma }^{*}}2{{s}^{2}}<\sigma 2{{p}_{z}}^{2}<\pi 2{{p}_{x}}^{2}=\pi 2{{p}_{y}}^{2}$
$Bond\,order\,\,=\,\dfrac{Bonding\,{{e}^{-}}s-Antibonding\,{{e}^{-}}s}{2}$
$Bond\,order\,=\,\dfrac{10-4}{2}=3$

Therefore the correct answer for the given question is option (a).

Note:We draw the molecular orbitals to check whether the species of consideration is diamagnetic or paramagnetic in nature.
If all the electrons are paired up and produce zero spin then those species will have a diamagnetic character, whereas if a lone pair of electrons are present, then it has a spin and behaves as a paramagnetic substance.