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Carbon has two electrons in its p orbital which should be able to form bonds, are there any examples in which this occurs instead of carbon hybridizing before bonding?

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  • $\begingroup$ Carbon (much like other period 2 elements) wants pretty badly to have a full octet, which this arrangement would not provide. $\endgroup$ – Ivan Neretin Nov 24 '15 at 17:12
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    $\begingroup$ The premise is incorrect. Carbon’s hybridisation is a mathematical model to help us understand its binding modes. All carbon compounds, when calculated ab initio start off from unhybridised carbon — and you can calculate a methane molecule without requiring pre-hybridisation. The nice thing is: If you then re-localise your bonds, you get something like an $\mathrm{sp^3}$ hybrid molecule on carbon. $\endgroup$ – Jan Nov 24 '15 at 17:12
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Hybridisation is a mathematical concept. It is one way of deriving chemical bonds from elements but not the only way. A different one (that computational chemistry typically uses) would be to just plant atoms at certain positions in space, and then mix their unhybridised atomic orbitals with each other to see what comes out. One can then back-calculate this to localise the molecule-spanning orbitals to single bonds and thereby deduce a hybridisation that would have worked to build up the molecule. Take the following example of methane (taken from Professor Klüfers’ internet scriptum for basic and inorganic chemistry at the university of Munich):

On the left you have four hydrogen s-orbitals, on the right you have the orbitals of carbon. As you see, there is no hybridisation assumed at all and the ground state is taken to be a fully populated $\mathrm{2s}$ orbital and three $\mathrm{2p}$ orbitals populated with two electrons amoung them for a total of $\mathrm{2s^2~2p^2}$. The hybrid orbitals are gained from symmetry considerations (in a tetrahedron, the p-orbitals transform as $\mathrm{t_1}$) and hydrogen orbitals were selected as was seen fit. The electrons are added a posteriori. Spin conservation would have produced a twice-excited state with very high total energy, so immediate spin pairing is assumed.

To localise the bonds, linear combine all four orbitals. If you take every orbital with factor $+1$, you arrive at the $\ce{C-H}$ bond to the upper right hydrogen. By shifting around a factor of $-1$ among the three $\mathrm{t_1}$ molecular orbitals during linear combination, you can address every other $\ce{C-H}$ bond. The resulting bonding orbital looks like n $\mathrm{sp^3}$ orbital mixed with a hydrogen $\mathrm{s}$ orbital.

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Hybridization isn't actually an operation carried out on the orbitals. It is an operation carried out on the mathematical functions that define the orbitals (refer chapter 1 of A Guidebook to Organic Reaction Mechanisms by Peter Sykes). As @Ivan Neretin said, Carbon, like all other period 2 elements, wants to attain octet badly. It does this by promoting two 2s electrons to 2p orbitals. The energy required for this is much less than that released during formation of two extra bonds (for eg.: formation of 2 C-H bonds in methane would release 97kcal/mol).

Now, carbon atom could form 3 directed bonds, which are mutually perpendicular, using three 2p orbitals and one non-directed bond using 2s orbital. Whereas in fact, it is known that the 4 C-H bonds are disposed at an angle of 109°28'. This can be accounted for if we use the concept of hybridization in functions defining the orbitals.

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  • $\begingroup$ The first two sentences started out so well, and then you go on about a hypothetical excited state … sorry, $-1$ for proposing that that formal excitation happen in the real world. $\endgroup$ – Jan Nov 25 '15 at 15:24
  • $\begingroup$ No offence, but it is all in chapter 1 of peter sykes' book. I have just referred to it and answered the question. Please feel free to edit it. $\endgroup$ – ShankRam Nov 25 '15 at 15:26
  • $\begingroup$ @Jan Could you please tell me what's wrong with my answer?I am in high school and it would really help me. $\endgroup$ – ShankRam Nov 25 '15 at 15:29
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    $\begingroup$ It is wrong to assume excitation before bond formation if hybridisation is not done; see my answer. So sentences 4 and 5 of your first paragraph are wrong. I cannot edit without conflicting with the author’s (i.e. your) interest, though, since you clearly build up on that ‘lost energy’. $\endgroup$ – Jan Nov 25 '15 at 16:24
  • $\begingroup$ So, we need to use MOT to explain bonding instead of that excitation stuff I said earlier? Which book can I refer to understand this concept? $\endgroup$ – ShankRam Nov 25 '15 at 17:02

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