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3. What are the steps to follow in order to determine which molecule has a higher bond order by drawing the Lewis structure? For example, with CO C O and COX3X2− C O X 3 X 2 −? 4 4 3 3.
The bond order is the average, "4 electron pairs" / "3 bonds" = "4/3 electron pairs per bond". The Lewis structure of nitrate ion is unusual because it features 3 formal charges in that the nitrogen is quaternized and formally cationic, and 2 of the oxygen atoms have formal negative charges. Of course, the sum of the charges +1+2xx(-1)=-1 for a uninegative ion. See this old answer.
In that case, SOX3 S O X 3 contains one double bond and two single bonds, which is why people tend to list the overall bond-order as 1.33. The actual bonding structure of SOX3 S O X 3 is a little more complicated than that, as J. LS points out, so you might need to brush up on molecular-orbital theory to get into the nitty-gritty of its bonding ...
The "$2$" on the bottom of the equation is the number of resonance structure being considered. Thus, for benzene the bond order is $1.5$. Applying this same method to the carbonate ion, we have 3 resonance structures with bond orders of 2, 1, 1 when considering the bond between carbon and a single oxygen.
By the same token, the "double". bonds at the ends of the molecules should both have some single bond character. These are examples of fractional bond order. The real world consequences of fractional bond order is a higher barrier to rotation about the central "single" bond in these two molecules than we might have expected, and a lower barrier ...
It depends on your definition of bond order. If you are using the simple formula $$\text{Bond order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2}$$ then, yes, it won't be able to capture the subtleties that you describe.
$\begingroup$ Thanks for commenting @Alchimista, Indeed its also weird to me that O2 its more stable than O2+ but when calculating bond order O2 its a 2 while O2+ its a 2.5, since they have the same number of protons in the nucleus I think this is because O2 has 2 unpaired electrons while O2+ only has one therefore making it more stable.
In short: The bond order of CO C O is not exactly 3 and removing an electron will not increase the bond order to 3.5. In both cases, the observed bond order is probably closer to 2.5, while experiments suggest that the bond is stronger in COX+ C O X +. An orbital with bonding character has no node perpendicular to the bond axis; an orbital with ...
Paramagnetism is a consequence of having one or more un-paired electrons in the outer electronic configuration. e.g. O$_2$, and NO.
For a single C-C bond in benzene, the total BO = σ + π = 1 + 0.5 = 1.5. Answer link. You draw the molecular orbitals. Then you add electrons and count the number of bonding and antibonding electrons. The bond order of a bond is half the difference between the number of bonding and antibonding electrons. BO = ½ (B – A) The C-C σ Bonds Each ...