# What makes an atom favor another one when bonding, and should there be enough atoms for both parts?

I'm learning chemistry through Khan Academy and I have 2 pretty basic questions.

Oxygen (O) has 6 electrons on it's last shell, so it needs 2 more to become stable. Hydrogen (H) has 1 electron, so it needs 1 more. O might combine with 2 H's, creating H2O, and it might combine with another O, creating O2. It can also create H2O2.

QUESTION 1: So, given a hypothetical solution (or environment, or whatever), with only O and H atoms, what will make O to favor a bond with another O instead of a bond with 2 H's (or vice-versa)? Both situations will make it stable.

QUESTION 2: Also, if we assume that, in that solution, there's a limited number of both O and H atoms, and after all bonds are made, there's only one O atom and one H atom left, will that H atom bond with the one O atom left, since that will make the H stable (albeit not the O atom), creating a HO molecule ? So, I guess what I'm asking is, will bonds only be made when there's enough atoms for both parts to become stable?

Thanks.

The answer to your question boils down to one thing, bond energies. When a bond is formed, energy is released. When a bond is broken, energy must be absorbed. Elements prefer to be in a low energy state, so if you just added H and O into a container, the reaction with the most negative net bond energy will take place. To calculate the overall bond energy of the reaction, the equation is:

$E_{total}$= (Energy of bonds broken) - (Energy of bonds formed).

(As a side note, the answer is in joules of energy per mole) For naturally occurring hydrogen and oxygen the bond energy is:

$2H_2 + O_2$ ->$2H_2O$

= $(436*2 + 493*1)-(463*4)= -487 \frac{kj}{mol}$

(2 H-H bonds broken + 1 O=O bond broken) - (4 O-H bonds formed)

This reaction is favored and would occur under typical conditions if oxygen was in the presence of hydrogen and would release -487 joules of energy per mole. The bond energies for each bond can be found by simply looking them up online. The bond energy for the formation of $H_2O_2$ I calculated turned out to be -149kj/mol. Therefore the formation of $H_2O$ would be favored over $H_2O_2$. In your question, you asked about single O and H atoms, not $H_2$ and $O_2$, which would not occur naturally, but is still very possible. This result however would be the same. The atoms would first bond to form $O_2$ and $H_2$, then would proceed to form water as noted above.

Finally, to answer your last question, if there is one hydrogen and one oxygen left, they will indeed bond to make an $OH$ molecule.

• What is the "bond energy for the formation of H2O2"? – orthocresol Feb 18 '16 at 21:09
• Thank you for your answer. So, first the reactions with the most negative net bond energy would be formed, then the reactions with the most negative net bond energy, and so on ? If you want to assure reactions with higher energy state, without using any artificial methods, could you do so by some kind of counting of atoms involved in the process ? For instance, is there like a cuttof: "ok, all H2O are formed, now it will proceed to form H2O2." Sorry if it's a dumb question. – JohnnyJohnny Feb 18 '16 at 21:32
• It is better to ask a seemingly dumb question than to remain with some ingrained misperception. See, chemistry is all about "artificial" methods and not about counting (not in this particular sense, at least). By the time all H2O is formed, you will run out of H, or O, or maybe both, so you won't be able to form any H2O2. – Ivan Neretin Feb 18 '16 at 21:39
• There is nothing like an OH molecule. An OH ion, however, is present. (OH-) As per my knowledge, an ion does not exist in free state. It has to be part of another molecule. So no, an OH molecule (so-called) will not be formed. And as Ivan Neretin said: See, chemistry is all about "artificial" methods and not about counting (not in this particular sense, at least). By the time all H2O is formed, you will run out of H, or O, or maybe both, so you won't be able to form any H2O2 or OH. – Abhirath Anand Feb 19 '16 at 1:54
• There indeed is such a particle as $\ce{OH}$ (without minus). Surely it is unstable in that it would readily react with many things, but other than that, it is stable. If a lone H in the middle of nowhere encounters a lone O, they would not wait for another H. Instead, they would stick together to form that particle, also called radical: $\ce{\cdot OH}$. – Ivan Neretin Feb 19 '16 at 5:53