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In $\ce{CH4}$, $1 s$ and $3 p$ orbitals are hybridised into four $sp^3$ orbitals. But on homolytic bond cleavage an unpaired electron remains in an unhybridised $p$ orbital?

Similarly in heterolytic bond cleavage of $\ce{CH4}$, the methyl carbocation has an unhybridised empty $p$ orbital?

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    $\begingroup$ We have had this confusion many times before. See here for some detail. Hybridisation is simply a mathematical quirk to make valence bond theory work better. It doesn't exist in the real world. In methane, the atomic orbitals overlap to form a set of molecular orbitals. $sp^3$ orbitals do not come into the final bonding description of methane (at least in the MO theory version). $\endgroup$
    – bon
    Jun 28, 2015 at 16:04

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