To draw the structure of $\ce{ClO2}$, I tried to find the hybridization of the central atom i.e $\ce{Cl}$. I figured out that there would be two $\ce{Cl - O}$ double bonds, which means $\ce{Cl}$ would have 4 bonding electrons. But this assigns 3 electrons as nonbonding residing on the $\ce{Cl}$ atom. So, $\ce{ClO2}$ is an odd electron species.

Then, how do I find its hybridization and draw its structure?

Can I not just count the number of σ bonds and the number of lone pairs as usual, i.e, considering the 3 non bonding electrons as 1 lone pair, and find the number of hybrid orbitals and thus, find its hybridization? This gives the answer of $\ce{ClO2}$ being $\ce{sp^2}$ hybridized and it's geometry being trigonal planar. Is this the correct answer? Am I right in my determination of bonding and nonbonding electrons of the atom or is there something special in this compound $\ce{ClO2}$?

  • $\begingroup$ The $\ce{Cl-O}$ bond is a 3 electron bond. $\endgroup$
    – Yashas
    Apr 18, 2017 at 3:01
  • $\begingroup$ @Yashas, what is a 3 electron bond? Never heard of that. How is it formed? Can you please explain it. $\endgroup$ Apr 18, 2017 at 8:37
  • $\begingroup$ related chemistry.stackexchange.com/questions/39001/… $\endgroup$
    – Mithoron
    Apr 18, 2017 at 20:10
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    $\begingroup$ The answers to all these questions are on the wikipedia page. But you're mostly right. I would call the structure bent, not trigonal planar. That is, it's in the $C_{2v}$ point group. $\endgroup$
    – jheindel
    Apr 19, 2017 at 5:57
  • $\begingroup$ @jheindel, I have mentioned that it's geometry is trigonal planar i.e including the configuration of lone pair. Indeed, it's shape would be bent. $\endgroup$ Apr 19, 2017 at 9:46