Question

Which of the following diatomic molecular species has only $\pi $ bonds according to the molecular orbital theory?
[A] ${{O}_{2}}$
[B] ${{N}_{2}}$
[C] ${{C}_{2}}$
[D] $B{{e}_{2}}$

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. 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-$B.O=\dfrac{1}{2}\left[ \left( No.of\text{ }{{\text{e}}^{-}}\text{ in bonding molecular orbital} \right)-\left( No.of\text{ }{{\text{e}}^{-}}\text{ in antibonding molecular orbital} \right) \right]$

Complete step by step solution:
We know that molecular orbital theory is used to explain the bonding between molecules that cannot be explained using the valence bond theory.
To answer this, let us proceed option wise.
Firstly we have ${{O}_{2}}$. 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.
Then 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 has 2 pi bonds and 1 sigma bond.
Next we have [C] ${{C}_{2}}$ . It has 12 electrons. So, its electronic configuration will be $\sigma 1{{s}^{2}}\text{ }{{\sigma }^{*}}1{{s}^{2}}\text{ }\sigma 2{{s}^{2}}\text{ }{{\sigma }^{*}}2{{s}^{2}}\ \text{ }\pi 2{{p}_{x}}^{2}\ \pi 2{{p}_{y}}^{2}\text{ }$ . Here, we have 8 bonding and 4 anti-bonding orbits.
Therefore, $B.O=\dfrac{1}{2}\left[ 8-4 \right]=\dfrac{4}{2}=2$ .
Here, we have 2 bonds and both will be pi-bonds.
And lastly, we have $B{{e}_{2}}$. It has 8 electrons. So, its electronic configuration is $\sigma 1{{s}^{2}}\text{ }{{\sigma }^{*}}1{{s}^{2}}\text{ }\sigma 2{{s}^{2}}\text{ }{{\sigma }^{*}}2{{s}^{2}}$ . It has 4 electrons in bonding as well as anti-bonding orbital. So, its bond order will be 0.
We can see from the above discussion that ${{C}_{2}}$ has only $\pi $ bonds according to the molecular orbital theory.

Therefore, the correct answer is option [C] ${{C}_{2}}$.

Note: We know that sigma and pi- are covalent bonds. 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.