How many electrons are in the p-orbital of the methyl radical (CH3^.)

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  1. 6C is 1s2 2s2 2p2
    Unpair one of the two electrons in the 2s orbital and promote it to the 2p. It now looks like this:
    1s2 2s1 2p3
    Now hybridize the 2s and 2p orbitals to give sp3 hybrid orbitals (that's four places to bond with hydrogen so the H adds one electron to each of the s and 3p hybrid orbitals. The compound now looks like this:
    1s2 [2s2 2p6] with the brackets denoting sp3 hybridization. The extra four electrons have come from 4 H atoms; i.e., 6 from C and 4 from 4 hydrogen atoms = 10 electrons total AND there are now 8 electrons in the sp3 hybrid orbitals (or 2 in the s and 6 in the p if you are separating them). So if we remove one of the H atoms from the p orbitals, that will leave a +1 charge on the CH3^+ ion and I still count 8 electrons total or 2 in the s and 6 in the p if you are separating them. We had 1 e in the s orbital and 3 in the p before the C combined with anything; we now have 2 in the s and 6 in the p. Note: I don't think this agrees with an earlier answer I gave because I counted only electrons from C in that earlier response. Check my thinking.

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