How does the electron from the 2s orbital "jump" to the 2p, thus leaving 4 unpaired electrons to form four covalent bonds?

Also, does the octet rule not apply to carbon? If it does, how can it bond with only two electrons? It would lack two more.

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The octet rule does apply to carbon. In order for carbon to form a complete octet, it must either gain 4 electrons, lose 4 electrons or form 4 covalent bonds. Placing 4 units of charge on such a small nucleus would be exceedingly difficult, so carbon typically binds in a covalent manner.

Let's imagine an experiment where we place carbon atoms in a container of hydrogen. This becomes a chemical reaction, just like any other. $\ce{CH4}$ will be the final product because we need 4 bonds around the central carbon atom in order for the carbon to have an octet. As the carbon and hydrogen begin to react, a chemist would say that the atoms begin to move along a reaction coordinate that connects starting materials with products. As the atoms move along this trajectory they explore many possible configurations. For example, carbon could form 4 bonds to hydrogen using its unhybridized atomic orbitals. If this pathway produced $\ce{CH4}$ it would have 2 C(2p)-H(1s) bonds and 2 C(2s)-H(1s) bonds. The two C-H bonds formed from the carbon p orbitals would have an H-C-H angle around 90 degrees. The 2 bonds formed using the carbon 2s orbitals would be rather floppy as they could travel over the surface of an s orbital sphere. The most stable configuration would probably involve these 2 C(2s)-H(1s) bonds arranging themselves as far away from the other 2 hydrogens as possible. So we'd probably wind up with a distorted tetrahedron with one H-C-H angle around 90 degrees and another H-C-H angle much wider than the 109 degree tetrahedral angle. Note that the 2 hydrogens involved in the H-C-H 90 degree angle would be very close together resulting in a destabilizing steric interaction.

Let's now consider a second possible path along our reaction coordinate. In this case let's mix our carbon atomic orbitals (the 2s and 2p orbitals) into a new set of atomic orbitals. We will "raise" the energies of the two 2s electrons and "lower" the energies of the two 2p electrons such that the 4 new orbitals we generate will have the same energy as the average energy of our initial s and p orbitals. Equal mixing of the s and p orbitals results in the formation of 4 equivalent new atomic orbitals, they are often referred to as $\ce{sp^3}$ hybridized orbitals. These 4 equivalent orbitals point towards the vertices of a tetrahedron. Reaction with hydrogen will produce a $\ce{CH4}$ molecule that has 4 equivalent C(sp3)-H(1s) bonds and all of the resulting bond lengths and angles will be equivalent.

In the case of carbon, the reaction coordinate follows this last pathway. Why? Because it produces the lowest energy (most stable) version of $\ce{CH4}$ possible. This lowest energy version of $\ce{CH4}$ does not suffer from unfavorable H-H steric repulsions due to small (90 degree) H-C-H angles. All of the C-H bonds in this stable version of $\ce{CH4}$ also have stronger (more stable) bonds. The C(sp3)-H bonds are much more directed and have better overlap that the diffuse, non-directed C(2s)-H bonds we discussed earlier.

If we move from carbon to phosphorous, we find that in the molecule $\ce{PH3}$ the bond angles are close to 90 degrees and the central phosphorous uses its original atomic orbitals (pure P in this case) to form bonds to hydrogen - no further mixing of orbitals occurs in this case.

The bottom line is that nature is smart. Nature knows how to mix (or not mix) atomic orbitals in order to produce the lowest energy compound possible, and it's method of achieving this is to travel along a reaction coordinate sampling all possible pathways and then letting energetics drive the reaction along the lowest energy pathway.

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Look up "orbital hybridization". It says that atomic orbitals can be combined to make new orbitals for bonding purposes. Carbon can take 1 s and 3 p orbitals to make 4 sp3 orbitals that can make 4 bonds.

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    $\begingroup$ Orbital hybridisation does not exist in the way you suggest it here. No atom can make hybrid orbitals. These are a mere mathematical treatment of a certain bonding pattern in a molecule. $\endgroup$ – Martin - マーチン Jul 18 '14 at 18:38
  • $\begingroup$ Agreed. Orbital hybridizations were a workaround to make early MO theory play nice with observed geometries. The four $\ce{C-H}$ bonds in methane $\ce{CH4}$ are not degenerate, and thus we do not require the electrons involved in the bonding be described by the same types of orbitals. $\endgroup$ – Ben Norris Jul 18 '14 at 18:48

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