In order to determine the number of bonds that an element will form, you must determine its charge. Once you know the charge, you know the bonds because for the most part the number of bonds is equal to the absolute value of the charge. So, lets learn how to determine the charge or oxidation number of an atom.
First, you will need to gather a few tools to help you with your task. You will need a periodic table, the octet rule and a couple of physics concepts. You can get a periodic table from your book, your teacher or online and the octet rule is below.
The Octet Rule:
An element will tend to gain or lose electrons in order to obtain a Noble Gas valence.
Nature tend to move toward a state of lower energy, which basically means nature takes the easy way. Also, you need to remember that opposites attract an likes repel. These two concepts will help you decide whether an element will gain or lose electrons to obtain the Noble gas valence.
Alright, ready to put these tools to work? Good! Lets determine the charge for lithium.
- Find lithium on the periodic table and determine the valence electrons.
- Lithium has 1 valence electron.
Now remember the octet rule and ask, "Is it easier to lose 1 electron and go back to helium or is it easier to gain 7 electrons and move toward neon?"
I hope you answered that it is easier to lose one and go back. So, lithium will tend lose one electron and become a (+1) cation.
Remember, if you lose electrons you become more positive and if you gain electrons you become more negative because electrons are negative .
If the charge of lithium is (+1) then it will have one bond. It really is just that easy.
Should we try another? How about chlorine?
- Find chlorine on the periodic table and determine the valence electrons.
- Chlorine has 7 valence electrons.
Now ask, "Is it easier to gain 1 electron and go towards argon or is it easier to lose 7 electrons and move back to neon?"
That's right it is easier to gain one and move toward argon. So, chlorine will have a (1) charge and form one bond.