A i i Oxidation
Assigning
O id ti States
St t
An oxidation state or oxidation number is a number that we
assign to an element in a compound to show whether or not
the element is missing electrons, has excess electrons, or has
just as many electrons as it needs.
Oxidation numbers are useful numbers because they help us to
k
keep
t k off electron
track
l t
flow
fl in
i oxidation-reduction
id ti
d ti reactions.
ti
If the oxidation number of an element increases, the element
has been oxidized, if the oxidation number of an element
decreases, the element has been reduced.
Rules for assigning oxidation
states
t t
(From Table 4.2,
4 2 page 167 of your text,
text I changed the order
and added Rule 6.)
1. An atom in an element always has an oxidation state of 0.
2. For monoatomic ions the oxidation state is the same as the
charge on the ion.
3. Oxygen
yg is usuallyy assigned
g
the oxidation state of -2. ((The one
exception is peroxides that contain O22-, here the oxidation
state is -1)
4 Hydrogen has a +1 oxidation state in covalent compounds.
4.
compounds
(But may be -1 in ionic compounds like NaH).
5. Fluorine is always assigned the oxidation state of -1.
6. The sum of the oxidation states of all atoms in a compound
or ion must equal the total charge on that ion or molecule.
P ti with
Practice
ith the
th first
fi t five
fi rules
l
Determine the oxidation every atom in the following compounds.
Pause video if yyou need to.
A. N2
B. Na+(aq)
C HF
C.
D. H2O
E. Na(s)
2 (aq)
F S2F.
( )
G. S8
(Answers on next slide)
P ti with
Practice
ith the
th first
fi t five
fi rules
l
A N2 Oxidation state of N=0,
A.
N=0 N2 is elemental form of N
B. Na+(aq) Oxidation state of Na+ = +1, Monoatomic ion.
C. HF
H= +1 (rule 4) , F= -1 (Rule 5)
D. H2O
H= +1(rule 4) O= -2 (Rule 3)
E Na(s) Na= 0 Na(s) is elemental form (Rule 1)
E.
F. S2-(aq) S = -2 Monoatomic ion (Rule 2)
G. S8 S=0 S8 is an elemental form (Rule 1)
M
More
complicated
li t d molecules
l l
The first five rules only apply to a few atoms in a few special
compounds. When you have other atoms, or atoms in more
complicated molecules,
molecules that
that’ss when you use Rule 6: The sum
of the oxidation states of all atoms in a compound or ion
must equal the total charge on that ion or molecule.
When yyou combine this rule with the other rules,, you
y now
have a way to figure out the oxidation states of every atom
in a compound.
E
Example
l 1:
1 CH4
The H part is easy. We can assign the oxidation number
of H in this compound by Rule 4,
4 hydrogen has a +1 state
in covalent compounds so….
Oxidation # for H = +1
E
Example
l 1:
1 CH4
What about the C? Well we don’t have a rule for C,
C so this is
when we use Rule 6. CH4 does not have superscript so this
molecule has no charge. If the charge is zero, then the sum of
the oxidation numbers for all the atoms is also equal to zero.
T i ll we make
Typically
k Rule
R l 6 into
i t an equation.
ti
Sum of Oxidation numbers = 0
= (number of C atoms × Oxidation number of C)
+ (number of H atoms × oxidation number of H)
E
Example
l 1:
1 CH4
If we make X = oxidation number for C and plug the rest of in
the numbers from this example into the above equation we have:
0 = (1×X) + [4×(+1)]
=X +4
X=oxidation state for C = -4
In summary for CH4:
C = -4
H = +1
E
Example
l 2:
2 SF6
By rule 5, the oxidation number of F = -1
SF6 is another uncharged
g covalent molecule so is we call
X the oxidation state of S we can set up the equation:
0 = (1×X) + [6×(-1)]
[6×( 1)]
0=X-6
X= oxidation number of S = +6
So for SF6….
the oxidation number of F is -11,
and the oxidation number of S = +6
E
Example
l 3:
3 NO3While this is a charged ion, it works the same way:
Oxidation number of O is -2 by Rule 3
Here the charge of the ion is -11 so if we use
X for the oxidation state of N, our Rule 6 equation
looks like this:
-11 = 1(X) +[3(-2)]
+[3( 2)]
-1 = X -6
5=X
The oxidation state of N is +5 in this ion.
y In NO3- the oxidation state of O is -2,,
Summary:
and the oxidation state of N is +5
E
Example
l 4:
4 NaCl
N Cl
First of all what kind of compound is this, ionic or covalent?
It is ionic. - Whenever you are given an ionic compound, the
first thing you do is to break it into it into its component ions,
i this
in
thi case, Na
N + andd ClOnce you do that , the rest is easy
Na+ is a monoatomic ion, so by rule 2,
Oxidation state of Na+ = +1
Cl- is also a monoatomic ion,, so byy rule 2,,
Oxidation number of Cl- = -1
E
Example
l 5:
5 KMnO
KM O4
This is ionic so again we break it into ions, K+ & MnO4(You should recognize the covalent ion permanganate)
K+ is a monoatomic ion, so by Rule 2,
Oxidation state of K+ = +1
What about the MnO4- ? Well you just did a covalent ion
in Example 3. You apply Rule 6 to the complex ion.
E
Example
l 5:
5 KMnO
KM O4
Applying Rule 6 to MnO4By Rule 3 the oxidation state of O is -2
If we use X for the oxidation state of Mn, the Rule 6
equation for the ion is:
-11 = 1(X) + 4(
4(-2)
2)
-1=X-8
X O
X=
Oxidation
id ti state
t t off Mn
M = +7
E
Example
l 5:
5 KMnO
KM O4
In summary for
f KMnO4
Oxidation state of K+ is +1
Oxidation state of O is -2
Oxidation state of Mn is +7
E
Example
l 6:
6 Fe
F 3O4
One last note. In most compounds these rules work fine,
and we get nice integral oxidation numbers, but there are a
few funny compounds like this example, Fe3O4
By Rule 3 Oxidation state of O is -2
Using
U
i X as the
h oxidation
id i state off Fe,
F we can make
k the
h Rule
R l 6
equation:
0 = 3(X)
( ) + 4(-2)
( )
0 =3X-8
8=3X
X = 8/3,
8/3 not a nice whole number but the correct
number for the correct application of the rules given here.
E
Example
l 6:
6 Fe
F 3O4
In reality what has happened is that this compound has two
Fe’s in the +3 state and one in the +2 state. Compounds like
this do exist,
exist but there is no way you could figure this out
with our simple rules, so don’t worry about it.
My advice to you is if you get a non-integer number for an
oxidation state, work through your calculations a second time,
and see if they check. Non-integer oxidation numbers are
fairly rare, but there is no guarantee that I won’t give you
one on a test,
test just to see if you are awake.
awake
P bl
Problems
to
t try
t
Assign the oxidation numbers for all the atoms in the following
compounds:
Pause video if you need to.
CO2
CH3OH
CaBr2
KNO2
H2SO4
HClO4
(The answers are on the next slide)
P bl
Problems
to
t try
t
Assign the oxidation numbers for all the atoms in the following
compounds:
CO2 O= -2, C= +4
CH3OH O= -2, H= +1, C= -2
CaBr2 Ca= +2, Br = -1
KNO2 K= +1, O= -2, N= +3
H2SO4 H= +1,, O= -2,, S= +6
HClO4 H= +1, O= -2, Cl= +7