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Question: Which of the following diatomic molecular species has only \(\pi \) bonds according to the molecular...

Which of the following diatomic molecular species has only π\pi bonds according to the molecular orbital theory?
[A] O2{{O}_{2}}
[B] N2{{N}_{2}}
[C] C2{{C}_{2}}
[D] Be2B{{e}_{2}}

Explanation

Solution

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=12[(No.of e− in bonding molecular orbital)−(No.of e− in antibonding molecular orbital)]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 O2{{O}_{2}}. Firstly, let us calculate the bond order of oxygen molecules. Number of electrons in O2{{O}_{2}} are 16.
Therefore, its electronic configuration will be σ1s2 σ∗1s2 σ2s2 σ∗2s2 σ2pz2 π2px2 π2py2 π∗2px1 π∗2px1\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=12[10−6]=42=2B.O=\dfrac{1}{2}\left[ 10-6 \right]=\dfrac{4}{2}=2
Therefore, the bond order of O2{{O}_{2}} is 2. And it has 1 sigma and 1 pi-bond.
Then we have N2{{N}_{2}}. It has 14 electrons. Therefore, electronic configuration will be-σ1s2 σ∗1s2 σ2s2 σ∗2s2 σ2pz2 π2px2 π2py2\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=12[10−4]=62=3B.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] C2{{C}_{2}} . It has 12 electrons. So, its electronic configuration will be σ1s2 σ∗1s2 σ2s2 σ∗2s2  π2px2 π2py2 \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=12[8−4]=42=2B.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 Be2B{{e}_{2}}. It has 8 electrons. So, its electronic configuration is σ1s2 σ∗1s2 σ2s2 σ∗2s2\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 C2{{C}_{2}} has only π\pi bonds according to the molecular orbital theory.

Therefore, the correct answer is option [C] C2{{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.