I was asked to draw the lewis-dot structure for ClO and I drew this:

   ..  ..
  :Cl-  O.
   ..  ..

The correct answer is apparently this:

   .   ..
  :Cl-  O:
   ..  ..

How would I know that? In the first example, all the formal charges are zero, but in the second, Oxygen has -1 and Chlorine has +1.

  • $\begingroup$ 1. Lewis structures are models that help us predict things and understand things about ions and molecules. 2. In some problems there is no "right" answer, or be.tter to say, there are several answers, neither of which are really "right". I like your reasoning about formal charge, by the way. I would ask an additional question: What about CLO can we learn, predict, or understand better with one of those structures vs the other? $\endgroup$ – Brinn Belyea Jul 27 '14 at 20:26
  • $\begingroup$ intermolecular forces - would they be higher with the second due to the formal charge? $\endgroup$ – Jakob Weisblat Jul 27 '14 at 20:32
  • $\begingroup$ Maybe they would be weaker. The dipole moment of CO is .1D or so since C has a -1 and O a +1 formal charge. One might expect a C-O triple bond to be quite polar but it isn't. $\endgroup$ – Brinn Belyea Jul 27 '14 at 20:34

In fact the unpaired electron belongs neither to Cl nor to O, but to a molecular orbital spread over both atoms [but not equally: see below]. The Lewis structure is simply not a complete representation of what is going on. (And for what it's worth wikipedia gives the same Lewis structure you drew: http://en.wikipedia.org/wiki/Chlorine_monoxide)

Edit: Actually it seems the highest occupied molecular orbital (where the unpaired electron is) is mostly localized on the oxygen atom.

MO Diagram (taken from here): MO diagram for ClO-

That is to say, your structure is more correct than the one in the answer key. However the justification involves molecular orbital arguments outside the scope of a general chemistry course.

Another edit: I too like your argument about formal charges and would accept it as a justification that the unpaired electron belongs to O rather than to Cl.

  • $\begingroup$ You can come to the same generalization using electronegativity. O is more EN than Cl; O preferentially takes the remaining electron. $\endgroup$ – Dissenter Jul 27 '14 at 21:50

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