Question

Two $\pi $ bonds and half $\sigma $ bonds are present in:
[A] $N_{2}^{+}$
[B] ${{N}_{2}}$
[C] $O_{2}^{+}$
[D] ${{O}_{2}}$

Answer 1

Hint: To answer this question, find out the bond order of each of the given molecules. We can find out the bond order from the molecular orbital theory. The molecule having 2.5 bond order will be the correct answer. We can find the bond order by subtracting the number of electrons in the antibonding orbital from the electrons in the bonding orbital and dividing it by 2.

Complete step by step answer:
We know that sigma and pi- are covalent bonds. We can find out the type of bonds by finding out the bond order.
 Bond order is the number of covalent bonds shared between a pair of atoms. It indicates the stability of a bond. Bond order is given by, $B.O=\dfrac{1}{2}\left[ \left( No.of\text{ }{{\text{e}}^{-}}\text{ in bonding molecular orbital} \right)-\left( No.of\text{ }{{\text{e}}^{-}}\text{ in antibonding molecular orbital} \right) \right]$

Firstly, let us calculate the bond order of oxygen molecules. Number of electrons in ${{O}_{2}}$ are 16.
Therefore, its electronic configuration will be $\sigma 1{{s}^{2}}\text{ }{{\sigma }^{*}}1{{s}^{2}}\text{ }\sigma 2{{s}^{2}}\text{ }{{\sigma }^{*}}2{{s}^{2}}\ \sigma 2{{p}_{z}}^{2}\text{ }\pi 2{{p}_{x}}^{2}\ \pi 2{{p}_{y}}^{2}\text{ }{{\pi }^{*}}2{{p}_{x}}^{1}\text{ }{{\pi }^{*}}2{{p}_{x}}^{1}$
As we can see there are 10 bonding orbitals and 6 antibonding orbitals.
Therefore, $B.O=\dfrac{1}{2}\left[ 10-6 \right]=\dfrac{4}{2}=2$
Therefore, the bond order of ${{O}_{2}}$ is 2. And it has 1 sigma and 1 pi-bond. Therefore, this is not the correct option.
Next, we have $O_{2}^{+}$. It has 15 electrons. Its electronic configuration will be $\sigma 1{{s}^{2}}\text{ }{{\sigma }^{*}}1{{s}^{2}}\text{ }\sigma 2{{s}^{2}}\text{ }{{\sigma }^{*}}2{{s}^{2}}\ \sigma 2{{p}_{z}}^{2}\text{ }\pi 2{{p}_{x}}^{2}\ \pi 2{{p}_{y}}^{2}\text{ }{{\pi }^{*}}2{{p}_{x}}^{1}\text{ }$
There are 10 bonding electrons and 5 antibonding electrons.
Therefore, $B.O=\dfrac{1}{2}\left[ 10-5 \right]=\dfrac{5}{2}=2.5$

Here, even though the bond order is 2.5 but we have 1 sigma bond and 1.5 pi bond. Therefore, this is also not the correct answer.
Next, we have ${{N}_{2}}$. It has 14 electrons. Therefore, electronic configuration will be $\sigma 1{{s}^{2}}\text{ }{{\sigma }^{*}}1{{s}^{2}}\text{ }\sigma 2{{s}^{2}}\text{ }{{\sigma }^{*}}2{{s}^{2}}\ \sigma 2{{p}_{z}}^{2}\text{ }\pi 2{{p}_{x}}^{2}\ \pi 2{{p}_{y}}^{2}$
There are 10 bonding and 4 anti-bonding orbits.
Therefore, $B.O=\dfrac{1}{2}\left[ 10-4 \right]=\dfrac{6}{2}=3$. It had 3 pi-bonds therefore this option is incorrect.

Lastly, we have $N_{2}^{+}$. Total number of electrons $N_{2}^{+}$ = 13.
Electronic configuration of $N_{2}^{+}$= $\sigma 1{{s}^{2}}\text{ }{{\sigma }^{*}}1{{s}^{2}}\text{ }\sigma 2{{s}^{2}}\text{ }{{\sigma }^{*}}2{{s}^{2}}\ \pi 2{{p}_{x}}^{2}\ \pi 2{{p}_{y}}^{2}\text{ }\sigma 2{{p}_{z}}^{1}$

As we can see there are 9 electrons in the bonding orbital and 4 electrons in antibonding orbital.
Therefore, $B.O=\dfrac{1}{2}\left[ 9-4 \right]=\dfrac{5}{2}=2.5$
As we can see here that 1s and 2s – electrons are already cancelling each other by having equal bonding and antibonding thus we are only left with 2p-orbital. As we can clearly see that there are 4 pi-bonding electrons and each pi-bonding requires 2 electrons thus, there are 2 pi-bonds and 1 is sigma bond. Therefore, this is the correct answer.
So, the correct answer is “Option A”.

Note: In order to form a covalent bond, the atoms need to be in a specific arrangement which will allow the overlapping between the orbitals. It is difficult to break a sigma bond because sigma bonds are stronger than pi- bonds. A sigma bond is formed by the overlapping of atomic orbitals along the axis and pi-bond is formed by overlapping of two lobes of the atomic orbitals.