1.1 OXIDATION NUMBERS

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Hi again! Let's start with the most important definition in this post which is:

Oxidation number- is the number of electrons that need to be added to (or taken away from) an element to make it neutral.

Example: if we have an oxide ion, it has a -2 charge, this means it needs to lose 2 electrons to become a neutral atom.

Example: if we have an iron(III) ion, it has a +3 charge, which means it has to gain 3 electrons to become a neutral atom.

RULES!!! They are very important here, and you have to get to know them before you start doing any of the questions.

1. The oxidation number of any uncombined element is 0 e.g. Sodium has na oxidation number of 0 and the same oxygen and the same as any single element. As log as it is just itself and uncombined, the o.n. of it will be 0.

2. When you have ionic compounds present, the oxidation number is equal to the charge on the ion. Let's do an example, we have a sulfate ion. Its oxidation number is -2 and we can prove that by doing a calculation: __+(4x-2)=-2 The missing number is the oxidation number of sulfur. (If you're wondering why this equation looks like this, look at rule number 4).

3. The next rule is that the sum of oxidation numbers in a compound is 0. Example: if we get NaCl for the example, the rule is that the oxidation number of the compound is 0. We can prove this with equation: +1(for sodium)+(-1)(for chlorine)=0.

4. In compounds:

--> Group 1 elements have an oxidation number of +1 (always!!) e.g. Lithium oxide, the o.n. of lithium is +1 and the o.n. of oxygen is -2, so the equation is: 2x(-1)+(-2)=0 (The answer to this equation is 0 because this is a compound).

--> Group 2 elements have an oxidation number of +2 (always!!) e.g. Magnesium hydroxide, the o.n. of magnesium is +2 and the o.n. of hydroxide is -1, so the equation is: +2+(2x-1)=0. The answer is again 0 because this is a compound.

--> In any compound, the oxidation number of oxygen is -2. So whenever you want to find the oxidation number of any element in a compound and oxygen is there you can sraight away write -2 for the oxygen. The only exceptions are when oxygen is with fluorine, superoxides and peroxides such as hydrogen peroxide. (Example: Water (2x+1)+(-2)=0 )

--> In any compound, the oxidation number of hydrogen is +1. The only exception is with metal hydrides such as NaH. (Example: Hydrochloric acid +1+(-1)=0 )

5. The last rule for calculating oxidation numbers is that for all covalent compounds, the most electonegative element has a '-' oxidation number. (Example: Carbon tetrachloride +4+(4x-1)=0 ).

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The next thing we are going to focus on is redox equations. There are a lot of questions in the exams about redox equations and showing which elements have been oxidised and which reduced. So, this is what we are going to do right now. But let's start with some definitions:

Oxidation- is the loss of electrons (OIL- oxidation is loss)

Reduction- is the gain of electrons (RIG- reduction is gain)

Oxidising agent- the one who oxidises something else, so it is itself reduced

Reducing agent- the one who reduces something else, so it is itself oxidised.

So, here I have prepared 3 examples for you to see my point:

You have to check the oxidation number of the element at the beginning and then at the end. 1.START- because you now know that the o.n. of oxygen is -2, so the equation looks llike this __+(-10)=0 and there are two V atoms so we have to divide the number by two which gives us a +5 o.n. for Vanadium.

END- this is an ion, so the oxidation number is equal to the charge which is +3.

Therefore, the oxidation number goes from +5 to +3, this means the electrons are gained, so it is a REDUCTION.

2. As with the previous one, you just have to apply the rules of calculating oxidation numbers to find out if it is a reduction or oxidation reaction or neither.

START- the charge on the ion is -2, so we put this in the equation after the equal sign. Oxygen is also present and we know the o.n. of oxygen which is -2. Now, we just have to put everything in the equation and see what the oxidation number of Cr is. ( __+(-14)=-2 ) There are two Chromiums, so each of them have an o.n. of +6.

END- applying the same technique as above, you can find out that the o.n. of chromium is the same as at the beginnig

Therefore, the reaction is NEITHER an oxidation or reduction.

3. START- the oxidation number of sulfur at the beginning is +4,( __+(-4)=0 ) The answer to the euqation is 0 because this is a compound.

END- The charge on the ion is now -2, so this is the answer to the equation. We know the o.n. of oxygen, so it works out that the o.n. of sulfur is +6, ( __+(-8)=-2 ).

Therefore, the o.n. goes from +4 to +6 and it's an OXIDATION.

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The only other thing that you may be asked in the exam regarding to oxidation numbers is to find unambiguous names of compounds. Many elements can comine in many different compounds. I think that as always it is best to try an example, so that you can visualise what I am trying to tell you.

So, as always the first thing that you have to do whn you're findng oxidation number is to write everything that you sraight away know into an equation.

1. The oxidation number of bromine is -1 because it is in the 7th group on the periodic table. However, there are two of them so it is -2. This is a compound, so the answer to the equation is 0. When we put all of this into one, we get __+(-2)=0. So the missing number (oxidation number of iron) is +2. Giving us iron (II) bromide.

2. The oxidation number of hydroxide is -1 and there are two hydroxides which is -2. As before you put everything in the equation and get +2 as the oxidation number of copper ( __+(-2)=0 ). Giving us copper(II) hydroxide.

That's all you have to know about oxidation numbers, if you want to practise some more questions related to this topic, please visit my youtube channel.

PS. Please remember, I am only a student, and as anyone, I can make mistakes. If you think you can see one, don't hesitate and comment (either here on on my youtube channel) Thank you!