# Determining covalent bond strengths without hybridisation

Assuming pure $\ce{2s}$ and $\ce{2p}$ orbitals of carbon are used in forming $\ce{CH4}$ molecule, which of the following statements is false? (single choice question)

1. Three $\ce{C-H}$ bonds will be at right angles
2. One $\ce{C-H}$ bond will be weaker than the other three $\ce{C-H}$ bonds
3. The shape of the molecule will be tetrahedral
4. The angle of $\ce{C-H}$ bond formed by $\ce{s-s}$ overlapping will be uncertain with respect to other three bonds.

According to me, the second option is false because $\ce{1s-2s}$ overlapping is stronger than $\ce{1s-2p}$. I just randomly found this question and I don't know the correct answer.

As I already stated, asking this question is wrong on so many levels, but let's just have a look anyway.

We are assuming a set of $3$ rigid $\ce{2p}$ and $1$ rigid $\ce{2s}$ orbitals at carbon. With rigid I mean, hybridisationnote is forbidden - which is wrong, $\text{[very]}_n$ wrong.

As a consequence, a head on overlap between the hydrogen $\ce{s}$ and the carbon $\ce{p}$ results in three bonds, each perpendicular , i.e. $90^\circ$, against each other. This is the feature of the degenerate $\ce{p}$ orbitals. (Answer 1 would be correct.)

As a conclusion from that, the molecule cannot adapt a tetrahedral geometry, because with this geometric arrangement, it is impossible to form a $S_4$ symmetry axis, which is the main feature of the $T_\ce{d}$ or $T$ point groups. Answer 3 would be wrong.

The bond between the carbon $\ce{s}$ and hydrogen $\ce{s}$ also needs to be weaker than the others, since these orbitals are totally symmetric. That means there is a good portion of the carbon $\ce{s}$ orbital already blocked by the other hydrogens. On the other hand, the orbital overlap of the carbon $\ce{p}$ and the hydrogen $\ce{s}$ is quite optimal, since the main contribution of $\ce{p}$ orbitals lie outside of the carbon $\ce{s}$ orbital. (Answer 2 would be correct.)

This $\ce{s-s}$ bond would also be quite wobbly, because neither of the orbitals is directed. Therefore the position of the last hydrogen can easily change without effecting the orbital overlap, and in first approximation the bond strength. (Answer 4 would be correct.)

As a last comment, the molecule in this conformation is not stable, probably not even as an highly excited state. The kinds of thought games actually ruin the more important thing of understanding how orbitals behave. Orbitals are not rigid. They are quite flexible in the end.

Note:
Hybridisation in the generic sense means the linear combination of atomic orbitals of one atom, that contribute to a molecular orbital.

• This question may have been designed to demonstrate why we need hybridization/molecular orbital theory to explain observed geometries, symmetries, and bond strengths. I find that many times beginning students have trouble making the conceptual leap from atomic orbitals to molecular orbitals, and that walking them through examples to show why atomic orbitals can't explain bonding on their own helps them to appreciate the need for molecular orbitals. Good answer, btw! – thomij Jul 10 '14 at 12:13
• @thomij Thanks, while I think what you say is correct, I still believe that asking these questions is wrong, because they confuse more than they help. Showing shenanigans like this is alright, but deducing things from the impossible should never been done ;) But maybe it was just ripped out of context in this case. – Martin - マーチン Jul 10 '14 at 12:19
• Thankyou Martin!! I'm really sorry for the incorrect format. As a newbie to chemstackexchange I wasn't aware of it. Your explanation is really awesome! – user7087 Jul 10 '14 at 16:09
• @thomij I agree with MArtin. While I appreciate the good will behind such questions, my experience is,too, that it often confuses more people than helps. For example molecular orbitals and such localized hybrid orbitals are (rather) different as soon as we depart from methane. – Greg Jul 10 '14 at 16:31
• @Martin, Greg - This might be a bad example of the method (especially when taken out of context), and it might not be appropriate for chemistry.stackexchange, but it is effective when used properly. It is a common approach in teaching, science, and debate. It works because it takes advantage of natural human learning processes. If you would like to learn more, check out Socratic method, thought experiments, and reductio ad absurdum. – thomij Jul 11 '14 at 18:21

I found this question in a book, and the answer given is option 3. The justification provided is as follows.

In the unhybridized state of carbon, $\mathrm{2p}$ orbitals are $90^{\circ}$ to one another and each one will overlap with $\mathrm{1s}$ orbital of three hydrogen atoms, thus three $\ce{C-H}$ bonds are formed which are $90^{\circ}$ to one another. For the fourth hydrogen atom, its $\mathrm{1s}$ orbital may overlap with non-directional $\mathrm{2s}$ orbital of the carbon and this σ-bond will be stronger than $\unicode[Times]{x3c3}(\ce{C-H})$ bonds formed by $\mathrm{2p - 1s}$ overlap. In such situation $\ce{CH_4}$ molecule can never have tetrahedral geometry.

• Except that $\ce{CH4}$ is most definitely tetrahedral. – bon Nov 21 '15 at 11:29
• @bon Come here and read the question and Martin’s answer to it ;) – Jan Nov 21 '15 at 17:06