Wednesday, November 11, 2020

Neutral Molecules

General Chemistry Index

Where are we going with this? This page will assist in developing the ability to predict products of simple reactions as listed in of reactions: synthesis (i.e., combination), decomposition, single displacement, double displacement, acid/base, and combustion.

Neutral Molecules 
What happens if I mix this with that?

Predicting products of chemical reactions is a process by which potential reactants are scrutinized to determine if they will react and if so what product(s) will be formed.

The products that are formed must be "neutrally" charged. Since some of the elements have positive charges only not everything can combine into a compound with everything else. To get a "neutrally charged" molecule, some part of the molecule has to be positive and the other part negative.

Once you figure out that it WILL react, the next thing to do is to find the subscripts that will result in a neutrally charged molecule. With ionic compounds, its fairly easy…

The valences will sum to zero. In other types of compounds, there's a similar thing going on at the conceptual level with covalent bonds. There are oxidation numbers, and they have to end up… let's just say they also add up to zero.

But what does that even mean!!!!

Let's go with explanation by example…

Let's take some Al and combine it with Cl.

So, we are trying to end up with a molecule that is neutrally charged. That is to say that the sum of the positive charges and the sum of the negative charges adds up to zero.

Suppose you have 3 things with a charge of +2. That means you have a charge of +6.
You need a charge of -6 to even that out. If the negative thing you are dealing with has a charge of -3, then you need 2 of them to get to -6.

In chemistry, you can specify how many of a molecule's parts you have with subscript. (The coefficient tells how many molecules you have.)

Okay, if we are going to combine Al and Cl we'll end up with some number of Al atoms and some number of Cl atoms so that the sum of the charges is zero.

Thus… we need to find out what the charges of one atom is. We get to go find the charges on these two things. Luckily, there is a chart. And also the periodic table.

So, here we go:

Al has a charge of +3
That means it has 3 electrons it will give up.

Cl has a charge of -1
That means it will accept 1 electron.

We can write it like this.



So, the goal is to get a zero charge and all we have to work with is Al and Cl…

Al+3 = +3      

Cl-1 = -1    

AlCl  is this:

Al+3 Cl-1= +2         <--Not 0

So one Al and one Cl does NOT become neutral. Bring on the subscripts!

Al2Cl  is this:     Al+3 Al+3 Cl-1 = +5         <--Not 0  That made it worse!

AlCl is this:     Al+3 Cl-1 Cl-1 = +1         <--Not 0  But closer!

AlCl is this:     Al+3 Cl-1  Cl-1 Cl-1 = 0        <--THIS IS 0  It takes 3 -1 Cl to become neutral with 1 +3 Al!

It works the same way with polyatomic ions.

If you have:



You need 3 minus ones to neutralize 1 plus 3.

Hence, you get this:



There is a "trick" called the criss-cross method:

  1. Take the charge number without the sign of the negative ion and use it as the subscript of the positive ion.

  2. Take the charge number without the sign of the positive ion. and use it as the subscript of the negative ion.

  3. Reduce to "lowest terms."

Tricks are nice. Let's try that!

N is -3

Mg is +2

So let's see… step 1… step 2… 


Step 3… reduce… not needed in this case. You might see occasionally a 4 and 2… 

Woo! Easy!


Let's try another one… just for fun?

Be and O

We can write it like this.



So, the goal is to get a zero charge and all we have to work with is Be and O…

Be+2 O-2= +2         <--THIS IS 0  It takes 1 -2 O to become neutral with 1 +2  Be

So you would write it as…



If you use the criss-cross method, you would take the 2 from O and put it on the Be and then the 2 from Be and put it on the O.

That would be:


In this case, you have the SAME subscripts on both elements, so you reduce to "lowest terms" and end up with the same thing as above.


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