Question

which of the following diatomic molecular species has only π bonds according to molecular orbital theory? a)Be2 b)O2 c)N2 d)C2. don't give only option, otherwise I will report​

Answer 1

C₂ is the diatomic molecular species which has only π bonds according to molecular orbital theory.

• In O₂ there is one sigma bond and one pi bond
• In N₂ there is one sigma bond and two pi bonds.

Molecular orbital diagram of N₂ is shown in the attached figure.

Electronic configuration of N₂ is 1s² 2s² and 2p³. In this configuration 1s and 2s is completely filled so molecular orbital diagram will be of 2p.

And we can clearly see from the attached figure in case of N₂ all the electrons are present in bonding molecular orbital and bond order of N₂ is 3.

So for Nitrogen it is clear that two electrons are present in sigma and 4 electrons are present in pi.

In C₂, electronic configuration of carbon is 1s² 2s² 2p². Here also 1s and 2s is completely filled so we consider only 2p orbitals.

From the attached figure of C₂ it is clear that all the 4 electrons are present in pi orbitals. And bond order of C₂ is also 2 which means both the bonds are formed due to pi orbitals.

So correct answer is C₂

Answer 2

Answer:

option d

Explanation:

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. Firstly, let us calculate the bond order of oxygen molecules. Number of electrons in O2 are 16.

Therefore, its electronic configuration will be σ1s2 σ∗1s2 σ2s2 σ∗2s2 σ2pz2 π2px2 π2py2 π∗2px1 π∗2px1

As we can see there are 10 bonding orbitals and 6 antibonding orbitals.

Therefore, B.O=12[10−6]=42=2.

Therefore, the bond order of O2 is 2. And it has 1 sigma and 1 pi-bond.

Then we have N2. It has 14 electrons. Therefore, electronic configuration will be-σ1s2 σ∗1s2 σ2s2 σ∗2s2 σ2pz2 π2px2 π2py2

There are 10 bonding and 4 anti-bonding orbits.

Therefore, B.O=12[10−4]=62=3. It has 2 pi bonds and 1 sigma bond.

Next, we have C C2 . It has 12 electrons. So, its electronic configuration will be σ1s2 σ∗1s2 σ2s2 σ∗2s2  π2px2 π2py2  . Here, we have 8 bonding and 4 anti-bonding orbits.

Therefore, B.O=12[8−4]=42=2.

Here, we have 2 bonds and both will be pi-bonds.

And lastly, we have Be2. It has 8 electrons. So, its electronic configuration is σ1s2 σ∗1s2 σ2s2 σ∗2s2 . 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 has only π bonds according to the molecular orbital theory.

Therefore, the correct answer is option d C2.