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Every time I saw a covalent bond structure, I notice that there are always lone pairs or no electrons left in the outer shell of the atoms. Is it compulsory for covalent bonds to have lone pairs or no electron in outer shell left to form ? Thank you.

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    $\begingroup$ Well, check out NO! $\endgroup$
    – Ed V
    Apr 21 at 16:21
  • $\begingroup$ There are odd-electron species. $\endgroup$ Apr 21 at 16:27
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    $\begingroup$ @NisargBhavsar So, you won’t take NO for an answer? ;-) $\endgroup$
    – Ed V
    Apr 21 at 16:29
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    $\begingroup$ @EdV I cordially invite you to Chemistry Chat, there's them puns. Sometimes. $\endgroup$ Apr 22 at 22:42
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    $\begingroup$ And $\ce{H-H}$? $\endgroup$
    – Buttonwood
    Apr 23 at 20:49
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For many of the compounds that you encounter in a first-year chemistry course, yes, this is the case, as satisfying the octet rule usually results in stability through full s- and p-orbitals. However, this is definitely not a guarantee

Some common examples include nitrogen compounds due to its odd number of electrons in its ground state, such as $\ce{NO}$ with a bond order of 2.5.

enter image description here

More generally, compounds with unpaired electrons fall into the class of radicals, which make up a deep set of reactions. Radicals are formed through the homolytic cleavage of bonds (one electron transferred to each bonding constituent), and this cleavage occurs most often through the input of light or heat. These cleavages are promoted when the product can stabilize the free-electron through resonance, similar to the stabilization of a carbocation.

Radicalization of chlorine through light (light promotes electron to antibonding orbital, which is unstable and causes a homolytic cleavage):

enter image description here

Radicalization of benzoyl peroxide (resonance stability of radical due to conjugation):

enter image description here

Radicalization of an azo compound (formation of an especially strong $\ce{N2}$ triple bond):

enter image description here

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