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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.

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  • $\begingroup$ 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. . $\endgroup$ – RJM Nov 29 '17 at 0:35
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    $\begingroup$ 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 $\endgroup$ – Ahmed S. Attaalla Nov 29 '17 at 0:57
  • $\begingroup$ 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 $\endgroup$ – Ahmed S. Attaalla Nov 29 '17 at 1:07
  • $\begingroup$ I think what you are doing is nice. I did not work it out, but i believe it will deviate again, when you consider more than p-electrons. What valency do they teach you to apply for sulfur? What would you get for $SO_4$$^{-2}$? Maybe more generally, it should include the difference of the signed charge. $\endgroup$ – RJM Nov 29 '17 at 1:11
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    $\begingroup$ Your statement "(in the molecule form)" suggests the charges are covered. You do not have to add or subtract a factor from H, according to the proof, as it is built into the electrons on one or more "atoms." The charge excess or deficit is assigned to an atom. In practice with your final equation, you use the typical number of valence electrons, then add or subtract the charges. Your proof covers simple charged molecules. $\endgroup$ – RJM Nov 29 '17 at 5:38
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It is nice that you can do rigorous math with electron counting rules but I should advise you that they are only heuristic rules used to rationalize the structure of molecules, reactivity etc. Well, they are quite accurate but to know the distribution of electrons in the molecule one has to perform a measurement or apply a quantum mechanical model, both depict a more complicated pattern. Moreover, you would not know how to assign electrons because you would not know where is the boundary between one atom and others, not to mention what is the bonding region.

Nevertheless, many people has worked in that direction because it is important to rationalize the properties of molecules in terms of their constituents. Some researchers have come to the conclusion that the best way is to partition the real space where the molecule "lives" and then count electrons integrating the electronic distribution $|\Psi|^2$ in these regions. So, instead of applying the inclusion principle in the model you brought to the table I suggest you to find out that the probability of finding exactly an integer number of electrons $\nu$ in a region of the Euclidean space $\Omega$ (an atom for instance) is given by the formula $$ p(\nu) = \frac{1}{v!} \sum_{i=0,N-\nu} \frac{(-1)^i}{i!} \int_\Omega \Gamma^{(\nu+i)} $$ where $N$ is the total number of electrons in the molecule, that is broken in two regions ($\mathbb{R}^3 = \Omega + \Omega^c$), and $$\Gamma^{(m)} = \frac{N!}{(N-m)!} \int |\Psi|^2 \, d\pmb r_{m+1} \cdots d\pmb r_{N} $$ is a reduced density matrix of order $m$, whose integration $$ \int_\Omega \Gamma^{(m)} $$ means the average number of electrons in $\Omega$ when $m=1$, average number of electron pairs when $m=2$, and so on up to $N$.

You will find the equation and the details here. I am pretty sure that it follows applying the inclusion principle. I did it once but not totally rigorous.

If you like this topic you will enjoy another article that describes the same probabilities with a more mathematical treatment.

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I arrived at the same conclusion using another approach. My approach is to let $G(X,E)$ be a graph whose vertices are the atoms $X_i$, and the edges are the bonds $b_{ij}=(X_i,X_j)$. And using your definition of $H_i$ and $N_i$ , we observe that for all vertices:$$H_i +\deg(X_i) = N_i$$ Since degree of an atom corresponds to the number of bonds that atom made, number of electrons needed in the final configuration is just the number of electrons already existing in the atom and the ones that are gained by bonds and since each bond brings another electron this statement holds. Then, using Handshaking lemma:$$ \sum_{X_i\in{X}}\deg(X_i)=2|E|=2B \\ \sum_{X_i\in{X}}\deg(X_i)=\sum_{X_i\in{X}}(N_i-H_i)=N-H \\ B=\frac{N-H}{2}$$

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  • $\begingroup$ $N_i$ is the amount of electrons on the $i$th atom right? What is $H_i$? $\endgroup$ – Ahmed S. Attaalla Nov 29 '17 at 2:23
  • $\begingroup$ @Ahmed S. Attaalla $H_i$ is the number of electrons that atom has(in the valence shell) before bonding, e.g. 6 for oxygen and 1 for hydrogen $\endgroup$ – 2chromatic Nov 29 '17 at 2:41
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    $\begingroup$ Sorry for so many questions, does your approach also work for a charged molecule? $\endgroup$ – Ahmed S. Attaalla Nov 29 '17 at 2:59
  • $\begingroup$ The molecular charge would have to be incorporated into one or more of the $X_i$. Technically, those $X_i$ would not be considered atoms, but ions. Either way it does not matter. In using $B = \frac{N - H}{2}$, H is calculated by adding the valence electrons of each atom and adding negative charges or subtracting positive charges. Therefore, it is not necessary in the proof, unless the number of electrons for each "atom" had to be strictly unionized. $\endgroup$ – RJM Nov 29 '17 at 5:45

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