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In AP Chemistry, we learned two ways to determine bond order. The first method is: Bond order = (bonding electrons - antibonding electrons)/2. The second method is: number of bonds / number of bond locations.

Applying the first method for the bond order of CO gives an answer of (8-2)/2 = 3. Using the second method, CO has one triple bond which gives an answer of 3/1 = 3.

Both methods also yield the same answer for CO^2+. One important characteristic of this bond is that carbon and oxygen make a double bond instead of a triple bond because the CO molecule is electron deficient. Since there is one double bond and one bonding location, the answer is 2/1=2. The result is the same for the other method as well.

In CO^+, the methods give different results. The first method gives an answer of (7-2)/2 = 2.5. I am unsure of the exact structure of the CO^+. Using a triple bond and leaving either the oxygen electron deficient by one or the carbon electron deficient by one gives an answer of 3. I am not sure how else the molecule can be depicted, perhaps with a delocalized bond somehow? I do not know much about delocalized bonds, so I do not know their rules and limitations. Am I incorrectly drawing the structure and is there a way to draw it so that it can satisfy the second method? Feel free to make edits on my formatting, I am new to this.

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  • $\begingroup$ Your second method is essentially valence bond theory. If you used a valence bond approach to describe the bond in $\ce{CO+}$ you would probably have to invoke a number of resonance structures, for example: i.imgur.com/2ZCkFOx.png There are more possibilities e.g. putting a 2+ charge on oxygen and a 1- charge on carbon but that's kind of pointless. So, if you just take an "average" of those two resonance structures, you get a "bond order" of 2.5. $\endgroup$ Oct 29, 2015 at 23:16
  • $\begingroup$ Could you add some paragraph breaks to your post to make it easier to read? Also, you mention nitrogen somewhere but you are talking about carbon monoxide ions aren’t you? $\endgroup$
    – Jan
    Oct 30, 2015 at 12:46

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