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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.
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 ...
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.
The 3 resonance forms of naphthalene are shown below however, these resonance forms are inconsistent with the bond lengths displayed by Clayden. From looking at these resonance forms I would not expect the C−C C − C bond fusing the benzene rings to be the shortest since it has less double bond character compared to some of the other C−C C ...
For HNO3, in order to satisfy the octet rule, the nitrogen atom would form 1 double bond and 2 single bonds. Based on octet rule alone, there are 3 possible resonance structures that are favorable. However, the first two resonance structures are significantly more favorable than the third, because they have smaller amount of formal charges.
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.
These can not only be seen as bonding with respect to the $\ce{M\bond{<-}L}$ bond but also antibonding with respect to the $\ce{C=O}$ bond. The maximum we can fill in here is 6 electrons. These six electrons reduce the $\ce{C#O}$ bond order as can be seen in the formula:
The electronegativity difference serves as a measure of percentage at which a bond is ionic.Roughly speaking, electro negativity difference of 1.7 is equivalent to 50 ℅ ionic character;.(calculated ionic character in your question ) Thus, ionic character of a given compound is 50% ×∆ (E.N)/1.7
This is the system of writing chemical formulas. In this system the carbon atoms are first, then hydrogen atoms and then other in alphabetical order. When the formula contains no carbon or hydrogen atoms, then all the elements are sorted alphabetically. So it should be: $\ce {CH4}$. However, the Hill system is not the only method of ordering ...
Basically, at least some states can be viewed as an $\ce{H3+}$ ion working as a nucleus in a bigger hydrogen-like atom. If said 'nucleus' catches the electron, the molecule immediately dissociates. So, let's focus for a bit on $\ce{H3+}$. In this ion a two-electron three-center bond is formed.