In case of hybridisation like $\mathrm{sp^3}$ hybridisation of carbon or nitrogen why do we excite electrons before hybridisation? Why couldn't we just mix orbitals without exciting electrons?

  • 9
    $\begingroup$ The way I see it, it doesn't make any difference. Bear in mind that hybridisation is just a mathematical concept, not a physical process. $\endgroup$
    – orthocresol
    Sep 30 '15 at 8:11
  • 4
    $\begingroup$ There is nothing like sp3 hybridised isolated carbon. But you can describe the carbon in CH4 as sp3 hybridised. But you cannot say: I take elemental carbon in ground state, excite and hybridise it (whatever it means), attach four hydrogen radicals and get methane. Neither you can dissect the methane in reverse process. All this stuff is just headology. $\endgroup$
    – ssavec
    Sep 30 '15 at 11:40

Actually number of bonds formed by an atom is generally equal to number of unpaired electrons it has. Like oxygen and nitrogen make two and three bonds because they have these many unpaired electrons in their valence shell. But, in its ground state, carbon got only two unpaired electrons, so it should make two bonds. But in reality carbon make four bonds. Two satisfy four bonds we need four unpaired electrons. This is possible to consider the excited state. Hope its clear.

  • 1
    $\begingroup$ The excited $sp^3$ state does not exist on its own. See the comments on the question. $\endgroup$
    – bon
    Oct 25 '15 at 10:31

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