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A Level Biology: Structure and Bonding 5 – Chemical Formulae

Hi! Welcome to my fifth video on the series about Structures and Bonding. Today, we are

going to look at the formula of ionic compounds.

So we are going to look at how we determine what the formula of an ionic compound might

be. By formula, I mean something like NaCl or MgCl2. This is a five-stage step that will

help you apply this to any situation and we are going to use the example of calcium bromide.

The first thing that you will do is you write down the symbols in this case: Ca and Br.

You then write down the charges on the ion. Calcium is in Group 2 and hence, it must lose

two electrons, so that’s a 2+ charge. Bromine, a non-metal in Group 7, it must gain an electron

has a 1- charge.

Therefore, we can establish that we are going to need more bromine than calcium. The ratio

of the ions, you need one of calcium and two of bromines. Therefore, the formula for that

is CaBr2 and remembering that the 2 on the end of the Br only refers to the bromine,

so that means two lots of bromine. If we are to draw a diagram of this, it will show the

single calcium donating its two electrons because it loses the two electrons and each

of the bromines take on a single electron, hence you end up with a calcium with 2+ charge

and a bromine with a 1- charge. Hence, that’s why you have a formula of CaBr2.

Another example might be aluminum bromide. We will go through the five steps. The symbols

are Al and Br. The charges on the ions: aluminum is a metal and is in Group 3 and has 3+ charges;

bromine, as you will remember from the last example, has a 1-. So therefore, you are going

to definitely need more bromine in the formula and so that the ratio of the ions is for every

aluminum, you need 3 bromines and that’s why the formula would be AlBr3. Again, the

3 that is on the bottom right referring just to the bromine. The diagram of this looks

like this for instance, so you got aluminum which has 3+ and it’s going to donate its

three electrons to these bromines, so that each one of the bromines take on a single

electron. That gives us our ions, so Al is 3+ and each bromine is a 1-. That’s why

AlBr3.

Now with aluminum oxide, this is a bit trickier one because aluminum and oxygen, neither of

those has a single charge and neither of those is a single positive or a single negative.

That’s what makes this one is a little bit trickier.

First, we write down the symbols. So, it is Al and O. The charges on the ions are 3+ in

the case of aluminum, 2- on the oxygen. Therefore, you are going to need more oxygens. As to

the ratio of the ions, what you are looking to do is to get a whole number that they can

both divide into. Now, if you look to this, what you could do is 3/2 = 1.5. You can’t

have a formula saying AlO1.5.

The numbers you multiply each ion must be whole numbers and therefore, we go through

a number that both 3 and 2 can divide into and that’s 6. So, you will need two aluminums

and three oxygens. That comes out with our formula of two aluminums and three oxygens.

That looks like this. So you have aluminum atoms here. Each of those will donate 6 electrons

in total and each oxygen will take on two electrons in total, giving it a negative charge

each. You’ll then be left with aluminums with a 3+ charge and oxygens with a 2-. That

will all be attracted because opposite charges and hence, you got the formula of Al2, so

the two aluminums here and O3.

So in summary, when we look into ionic bonding and we are looking at writing up formulas

for ionic bonding, the number of positive charges must equal the number of negative

charges. So an example of this, we looked at aluminum oxide and in which aluminum has

a 3+ charge and oxygen has a 2- and what you need to do is you need each number to get

to 6. If you multiply the 3+ on the aluminum by 2, that gets you to the 6 and if you multiply

the 2- by 3, that gives you 6- and hence, they balance.

[end of audio – 04:54] A Level Biology: Structure and Bonding 5 – Chemical

Formulae Page…1

Hi! Welcome to my fifth video on the series about Structures and Bonding. Today, we are

going to look at the formula of ionic compounds.

So we are going to look at how we determine what the formula of an ionic compound might

be. By formula, I mean something like NaCl or MgCl2. This is a five-stage step that will

help you apply this to any situation and we are going to use the example of calcium bromide.

The first thing that you will do is you write down the symbols in this case: Ca and Br.

You then write down the charges on the ion. Calcium is in Group 2 and hence, it must lose

two electrons, so that’s a 2+ charge. Bromine, a non-metal in Group 7, it must gain an electron

has a 1- charge.

Therefore, we can establish that we are going to need more bromine than calcium. The ratio

of the ions, you need one of calcium and two of bromines. Therefore, the formula for that

is CaBr2 and remembering that the 2 on the end of the Br only refers to the bromine,

so that means two lots of bromine. If we are to draw a diagram of this, it will show the

single calcium donating its two electrons because it loses the two electrons and each

of the bromines take on a single electron, hence you end up with a calcium with 2+ charge

and a bromine with a 1- charge. Hence, that’s why you have a formula of CaBr2.

Another example might be aluminum bromide. We will go through the five steps. The symbols

are Al and Br. The charges on the ions: aluminum is a metal and is in Group 3 and has 3+ charges;

bromine, as you will remember from the last example, has a 1-. So therefore, you are going

to definitely need more bromine in the formula and so that the ratio of the ions is for every

aluminum, you need 3 bromines and that’s why the formula would be AlBr3. Again, the

3 that is on the bottom right referring just to the bromine. The diagram of this looks

like this for instance, so you got aluminum which has 3+ and it’s going to donate its

three electrons to these bromines, so that each one of the bromines take on a single

electron. That gives us our ions, so Al is 3+ and each bromine is a 1-. That’s why

AlBr3.

Now with aluminum oxide, this is a bit trickier one because aluminum and oxygen, neither of

those has a single charge and neither of those is a single positive or a single negative.

That’s what makes this one is a little bit trickier.

First, we write down the symbols. So, it is Al and O. The charges on the ions are 3+ in

the case of aluminum, 2- on the oxygen. Therefore, you are going to need more oxygens. As to

the ratio of the ions, what you are looking to do is to get a whole number that they can

both divide into. Now, if you look to this, what you could do is 3/2 = 1.5. You can’t

have a formula saying AlO1.5.

The numbers you multiply each ion must be whole numbers and therefore, we go through

a number that both 3 and 2 can divide into and that’s 6. So, you will need two aluminums

and three oxygens. That comes out with our formula of two aluminums and three oxygens.

That looks like this. So you have aluminum atoms here. Each of those will donate 6 electrons

in total and each oxygen will take on two electrons in total, giving it a negative charge

each. You’ll then be left with aluminums with a 3+ charge and oxygens with a 2-. That

will all be attracted because opposite charges and hence, you got the formula of Al2, so

the two aluminums here and O3.

So in summary, when we look into ionic bonding and we are looking at writing up formulas

for ionic bonding, the number of positive charges must equal the number of negative

charges. So an example of this, we looked at aluminum oxide and in which aluminum has

a 3+ charge and oxygen has a 2- and what you need to do is you need each number to get

to 6. If you multiply the 3+ on the aluminum by 2, that gets you to the 6 and if you multiply

the 2- by 3, that gives you 6- and hence, they balance.

[end of audio – 04:54] A Level Biology: Structure and Bonding 5 – Chemical

Formulae Page…1