Question

Oxidation States tutorial
Creator of Tutorial: Somy

Answer 1

First of all lets understand what are/or what these oxidation states show

- basically saying all they show is 'how many electrons this particular element or compound- is able to give/take or already gave/took'

So now we move one to periodic table.
These are maximum oxidation states each group can reach:

Group : 1
Oxidation state : +1 (loose 1 electron)

Group : 2
Oxidation state : +2 (loose 2 electrons)

Group : 3
Oxidation state : +3 (loose 3 electrons)

Group : 4
Oxidation state : +4 (loose 4 electrons)

Group : 5
Oxidation state : -3 (gain 3 electrons)

Group : 6
Oxidation state : -2 (gain 2 electrons)

Group : 7
Oxidation state : -1 (gain 1 electron)

Group : 8
Oxidation state : 0 (neither gain nor loose any electron)

Now, I'll explain reasoning.
Usually each element always want to reach its 'full noble state', I'd say
meaning have filled last shell and thus have electronic configuration which is same as of one of the noble gases

So
Group 1 : has 1 electron in its last shell
To get to its 'noble' state, group 1 has to gain 7 more electrons
BUT it also can loose that 1 electron in the last shell and reach the noble state

Imagine it this way, you are going to the mall lets say and you know about 2 ways to get there, meaning 2 routes:
one longer
one shorter

Which one you'd want to use more?
Shorter, right? :)

Same thing happens here for group 1 its easier to Loose that 1 electron rather than gain 7 more electrons, thus maximum oxidation state of group 1 elements is always +1

The same concept can be applied to Group 2, 3 and 4.

Answer 2

Now as we move to group 5-7, this one is a bit different though we can use same idea

Now why do group 5-7 gain rather then loose electrons?
well look
lets say group 7
Element of group 7 has 7 electrons in its last shell
to reach 'noble state' it can loose all 7 electrons OR gain 1 more electron

Which one is more convenient?
Gain 1 more electron, right?

So, that is why group 7 usually gain 1 electron rather than loosing 7 electrons.

That's why their maximum oxidation state is -1

This concept also applies to Groups 5 and 6.

Answer 3

Next there are some oxidation states you have to always remember
these elements have 'fixed' oxidation state that is ALWAYS same

H : +1
O : -2
F : -1
OH : -1

and group 1 and 2 are included

Group 1 and 2 (maybe 3 also) and 7 have fixed oxidation states. Other groups tend to have different oxidation states depending on a compound.

Answer 4

Well, that's pretty much it about higher oxidation states :) Next, we will move on to Oxidation states in Pure element and in Compounds.

Always remember pure element has oxidation state of 0, because it neither lost or gained any electron (not involved in reaction yet).

For example:

If in reaction you are given
Al + Cl2 --->
Both Al as well as Cl2 have oxidation states of 0.

Same goes to Compounds, meaning if you are given a compound (lets say HCl) its SUM of oxidation states has to ALWAYS be equal to 0.

This idea is TRUE when you are not given anything like this:


OR



In these cases, that ' - ' sign in Cl shows that Cl has already gained 1 more electron, and as for SO4 with ' -2 '

This -2 tells thats SUM of oxidation states in this compound (meaning SO4) has to be equal to NOT 0 but -2, which means when you add up all oxidation states of each element in a compound - the answer you should get has to be ' -2. '

Answer 5

Alright, so now we are done with basics of oxidation state. We are going to move on to actual deal:

How to use these rules?
How to find oxidation state of an element in a compound?

Example number 1

H2SO4 - this is going to be our compound
our target is S

we know fixed oxidation states for
H its always +1
for O its always -2

now look at one thing
there are 2 H atoms in that compound
thats means EACH H atom will have oxidation state of +1
thus +1 *2 = +2 is total oxidation state of 2 H atoms

now O
there are 4 O atoms in this compound
thus EACH O atom will have oxidation state of -2
so -2 *4= -8 is total oxidation state of all O atoms

now we need to find oxidation state of S

we know that if the compound is given like H2SO4 with no 'oxodation state signs' or so thats means H2SO4 's total oxidation state HAS to be EQUAL to 0


so now lets do pure math :)

H : +2
S : X
O : -8

so

(+2) + X + (-8) = 0
so
+2+X-8 = 0
X= 0-2+8
X= +6

That means oxidation state of S is +6 (in this compound)

Answer 6

Example 2
P2O5

our target is P

we know oxidation state of O is -2
and we have 5 O atoms
thus -2 *5= -10

now look at this
there are 2 P atoms in this compound
BUT that means if you try finding oxidation state using the rule that we use awhile back, you will get TOTAL oxidation state of 2 P atoms
but when question is asking you for oxidation state of some element in a compound
they mean oxidation state of 1 atom of that element!

so
O : -10
P : 2X
2 - is because there are 2 P atoms

now we can use same rule

2X + (-10) = 0
2X- 10 = 0
2X= 0 +10
2X= 10
X= 10/2
so X= +5

That's how we get oxidation state of P as +5 :)

Answer 7

Example 3

NO3-
target is N oxidation state
this ' - ' sign shows that SUM of oxidation states of this compound is -1

the rest is same as previous examples
so

oxidation state of O is -2 and there are 3 O atoms, so
-2 * 3 = -6 total oxidation state of all O atoms

now
O : -6
N : X
so using same rule

X+ (-6) = 0 -> THIS IS ABSOLUTELY WRONG for this compound!

X+ (-6) = -1 -> now this one is ✓

so

X -6 = -1
X= -1 + 6
X= +5

Oxidation state of N in this compound is +5 :)

Answer 8

Example 4
SO4 ^-2
target S
-2 shows that SUM of oxidation states of all atoms in this compound has to be equal to -2!
again we do same thing as before

4 O atoms each with -2 oxidation state so
-2* 4= -8
so
O : -8
S : X

X + (-8) = 0 WRONG!

X+ (-8) = -2 ✓

so

X-8= -2
X= -2+8
X= +6

Thus we found oxidation state of S to be +6 in this compound :)

Answer 9

Well, now you must have gotten a good idea about this part. Now lets talk about how reaction occurs

Let's say you are given a compound
H2SO4 + NaOH-->

It's not like you straightaway get a product...
Nope!

First what happens is that 'members' of compound break into 'cations' and 'anions'

cations are those that lost electron thus + charge
anions are those that gained electron thus - charge

So that means H2SO4 ions that we will get would be
H^+ cation
H^+ cation
SO4^ -2 anion

NaOH
Na^+ cation
OH^- anion

So what happens here is that cation of one compound takes anion of the other compound while cation of other compound takes anion of this compound
so
H+ and H+ take OH^- from NaOH and thus form H2O

and
Na+ take SO4^-2 from H2SO4 and so form Na2SO4

so
H2SO4 + NaOH --> Na2SO4 + H2O
and balance it
H2SO4 + 2NaOH ---> Na2SO4 + 2H2O

That's how the products form! :))

Answer 10

hmm i guess thats pretty much it about oxidation :)
Hope it helps!

Answer 11

thanks