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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.

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

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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.

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