# Predict the number and nature of bonds using Molecular Orbital Theory

I found the following question in a high school chemistry problem book:

According to Molecular Orbital Theory which of the following is true:

(A) In $$C_2$$ molecule both the bonds are $$\pi$$ bonds.

(B) In $$C_2^{2-}$$ ion there is one $$\sigma$$ and one $$\pi$$ bond.

Both of these are true. But how can one know this using Molecular Orbital Theory?

P.S.: I don't know anything about Quantum Mechanics as I am a high school student. My textbooks only provide a result-oriented explanation. I would appreciate if the answers keep this in mind.

• Is answer (B) one or two $\pi$ bonds? – Blaise Apr 10 '19 at 14:12
• @Blaise (A) is two. $C_2$ violates the one $\sigma$, one $\pi$ rule. (B) is one $\sigma$, one $\pi$. – user76377 Apr 10 '19 at 15:02

I presume you can draw the homonuclear diatomic MO diagram for these two species.

As you probably already know, the s orbitals of the carbons form bonding and antibonding orbitals, both of which are filled. (The net effect of this is the “lone pairs” on each carbon.)

The p orbitals for 2 pi bonds and a sigma bond. Due to s-p mixing, the pi bonds are lower in energy and are filled first, and these 2 bonds (with their 4 electrons) are the 2 bonds we see in C2.

Add another 2 electrons to the system and you realise they fill the sigma bonding orbital, which means that the C2 2- anion has 2 pi bonds and a sigma bond (for a total bond order of 3).

Hope this helped.

According to the same book (Resonance Chemistry for JEE Main and Advanced), bond order of $$1$$, $$2$$ and $$3$$ correspond to single, double and triple bond.

This is a rule of thumb and most common molecules, such as $$O_2$$, $$N_2$$, etc., seem to follow it.