# An application of the Inclusion Principle to Chemistry? (Proof Verification)

Background

I’m taking chemistry and one thing they have us do is draw Lewis structures for molecules. Guessing if there are going to be double or triple bonds is kind of annoying. I’d like to be able to come up with a formula to predict the total number of bonds a molecule will form. I count a double bond as $2$ bonds, and a triple bond as $3$.

What I’d like to prove

Count a single bond as $1$ bond, a double as $2$ bonds, and a triple as $3$.

Scrolling through YouTube one video seems to suggest that when octets (or a duet for hydrogen) are formed, the number of bonds a molecule will form is,

$$B=\frac{N-H}{2}$$

Where $N$ is the sum of the electrons needed on each atom. Ex: In $H_2O$, we have $N=2+2+8=12$.

And $H$ is the number of electrons we have to distribute. In our case $1+1+6=8$ hence, $B=\frac{12-8}{2}=2$ and water forms $2$ bonds as expected (two single bonds).

I’d like to be sure this holds in general for the conditions mentioned to be sure I’m not using some nonsense. I think I came up with a proof:

My proof

Suppose we have $n$ atoms $X_1,X_2,\dotsc,X_n$ part of a molecule. Think of each atom $X_i$ as a set consisting of it’s electrons (in the molecule form). $X_1 \cup X_2 \cup \dotsb \cup X_n$ is what you get when you combine these elements/electrons and overlapping may occur. $X_1 \cup \dotsb \cup X_n$ is the molecule, which has a cardinality of $H$ electrons by definition.

By inclusion exclusion,

$$H = \left| \bigcup_{k=1}^{n} X_k \right| = \sum_{k=1}^{n} |X_k| - \sum_{1 \leq i < j \leq n} |X_i \cap X_j| + \dotsb + (-1)^{n-1} \left| \bigcap_{k=1}^{n} X_k \right|$$

On the right most part of the equation above, everything vanishes except $|\sum_{k=1}^{n} |X_k|-\sum_{1 \leq i < j \leq n} |X_i \cap X_j|$ since each electron can only share $2$ atoms at most.

Finally, $\sum_{k=1}^{n} |X_k|=N$ since the cardinality of each atom is exactly how many electrons it needs, if we give it what it needs. And $\sum_{1 \leq i < j \leq n} |X_i \cap X_j|$ is the some of the bonding electrons (no double counting), between each element. That is $\sum_{1 \leq i < j \leq n} |X_i \cap X_j|=2B$.

So, $H=N-2B$ and $\frac{N-H}{2}=B$.

Question

Is the above proof correct.

• I believe this would break down for the sulfate dianion. For that matter, I am not sure it works for the ammonium ion. N= 2+2+2+2+8, H = 5+1+1+1+1, B = (16-9)/2 = 3.5. Mathematically, it seems reasonable, but chemically, it does not seem general. . – RJM Nov 29 '17 at 0:35
• For the ammonium ion, $NH_4^+$ the ion “needs” a sum of $N=8+4(2)=16$ but has $H=5+1(4)-1=8$ electrons, so $B=\frac{16-8}{2}=4$ as expected. I didn’t prove it for ions, but the video seems to suggest this is true. I’m trying to think of a way to easily extend/modify the proof I wrote to include ions but am struggling. Will add that in the question...@RJM – Ahmed S. Attaalla Nov 29 '17 at 0:57
• Also the formula appears to break down for the sulfate ion because the Lewis structure in that case is weird in that the octet rule on sulfur is not obeyed. However if you tell the formula ahead of time that you are allowing $12$ electrons for sulfur, then it will correctly get you $B$ because $N=12+4(8)=44$ whereas $H=6+4(6)+2=32$ and $\frac{44-32}{2}=6$ bonds as expected. That has more to do with chemistry, (stabilization, formal charge,etc) than math. Accounting for that I think would be to difficult. @RJM – Ahmed S. Attaalla Nov 29 '17 at 1:07