Transcript REDOX OXIDATION NUMBERS NAMING COMPOUNDS File

REDOX
A guide for A level students
O
X
I
D
A
T
I
O
N
+7
+6
+5
+4
+3
+2
+1
0
-1
-2
-3
-4
R
E
D
U
C
T
I
O
N
2015
SPECIFICATIONS
SPECIFICATION POINTS
REDOX
LEARNING OBJECTIVES:
• Definitions of oxidation and reduction
• Calculating oxidation state
• Use of H, O and F in calculating oxidation state
• Naming compounds
REDOX
Before you start it is necessary to…
 Recall the layout of the periodic table
 Know the charges on common ions
 Be able to write formulae from memory
OXIDATION & REDUCTION – Definition first met at KS3
OXIDATION
GAIN OF OXYGEN
2Mg + O2
——> 2MgO
magnesium has been oxidised as it has gained oxygen
REMOVAL (LOSS) OF HYDROGEN
C2H5OH
——> CH3CHO + H2
ethanol has been oxidised as it has ‘lost’ hydrogen
OXIDATION & REDUCTION – Definition first met at KS3
REDUCTION
GAIN OF HYDROGEN
C2H4 + H2
——> C2H6
ethene has been reduced as it has gained hydrogen
REMOVAL (LOSS) OF OXYGEN
CuO + H2 ——> Cu + H2O
copper(II) oxide has been reduced as it has ‘lost’ oxygen
However as chemistry became more sophisticated, it
was realised that another definition was required
OXIDATION & REDUCTION - Definition first met at GCSE
OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
Oxidation and reduction are not only defined as changes in O and H
...
OXIDATION
Removal (loss) of electrons ‘OIL’
species will get less negative or more positive
REDUCTION
Gain of electrons
‘RIG’
species will become more negative or less positive
REDOX
When reduction and oxidation take place
OIL - Oxidation Is the Loss of electrons
RIG - Reduction Is the Gain of electrons
OXIDATION STATES
What are they?
Oxidation states are assigned to atoms in molecules or ions
to show how many electrons the atom has used in
bonding.
This method has the advantage that it can also be applied
to covalent substances where complete transfer of
electrons does not occur.
OXIDATION STATES
Reduction
• Look at the equation
• Write out the oxidation numbers of
the atoms involved
• When the oxidation number of an
atom increases (gets more positive)
oxidation has occurred
• When the oxidation number of an
atom decreases (gets more negative)
reduction has occurred.
Oxidation
Recognizing oxidation and
reduction in reactions
+7
+6
+5
+4
+3
+2
+1
0
-1
-2
-3
-4
-5
-6
-7
OXIDATION STATES
Used to...
tell if oxidation or reduction has taken place
work out what has been oxidised and/or reduced
construct half equations and balance redox equations
ATOMS AND SIMPLE IONS
The number of electrons which must be added or removed to become neutral
atoms
Na in Na =
0
neutral already ... no need to add any electrons
cations Na in Na+= +1
need to add 1 electron to make Na+ neutral
anions Cl in Cl¯ =
need to take 1 electron away to make Cl¯ neutral
-1
Q. What are the oxidation states of the elements in the following?
a) C (0)
b) Fe3+ (+3)
c) Fe2+ (+2)
d) O2- (-2)
e) He (0) f) Al3+ (+3)f) Al3+
OXIDATION STATES
MOLECULES
The SUM of the oxidation states adds up to ZERO
ELEMENTS
COMPOUNDS
H in H2 =
0 both are the same and must
add up to Zero
C in CO2= +4
O in CO2= -2
OVERALL (1 x +4) + (2 x -2) = Zero
EXPLANATION:
• because CO2 is a neutral molecule, the sum of the oxidation states must be
zero
• for this, one element must have a positive OS and the other must be
negative
OXIDATION STATES
MOLECULES
The SUM of the oxidation states adds up to ZERO
ELEMENTS
COMPOUNDS
H in H2 =
0 both are the same and must
add up to Zero
C in CO2= +4
O in CO2= -2
OVERALL (1 x +4) + (2 x -2) = Zero
EXPLANATION:
HOW DO YOU DETERMINE THE VALUE OF
AN ELEMENT’S OXIDATION STATE?
• from its position in the periodic table and/or
• the other element(s) present in the formula
OXIDATION STATES
MOLECULES
The SUM of the oxidation states adds up to ZERO
ELEMENTS
COMPOUNDS
H in H2 =
0 both are the same and must
add up to Zero
C in CO2= +4
O in CO2= -2
OVERALL (1 x +4) + (2 x -2) = Zero
EXPLANATION:
HOW DO YOU DETERMINE WHICH IS THE POSITIVE ONE?
• the more electronegative species will have the negative value
- electronegativity increases across a period and decreases
down a group
e.g. O is further to the right than C in the periodic table so it has
the negative value
OXIDATION STATES
COMPLEX IONS
The SUM of the oxidation states adds up to THE CHARGE
e.g. NO3- sum of the oxidation states =
SO42- sum of the oxidation states =
NH4+ sum of the oxidation states =
Examples
in SO42- the oxidation state of :
S = +6
there is ONE S
O = -2
there are FOUR O’s
+6 + 4(-2) = -2
so the ion has a 2- charge
-1
-2
+1
WORKED EXAMPLE
What is the oxidation state (OS) of Mn in MnO4¯ ?
• the oxidation state of oxygen in most compounds is - 2
• there are 4 O’s so the sum of its oxidation states - 8
• overall charge on the ion is - 1
• therefore the sum of all the oxidation states must add
up to - 1
• the oxidation states of Mn four O’s must therefore equal - 1
• therefore the oxidation state of Mn in MnO4¯is:
OS of Mn + 4(-2) = - 1
OS of Mn + -8 = - 1
OS of Mn = - 1 + 8
OS of Mn = +7
OXIDATION STATES
CALCULATING OXIDATION STATE (1)
Many elements can exist in more than one oxidation state
In compounds, certain elements are used as benchmarks to
work out other values
HYDROGEN
+1
except
0
-1
atom (H) and molecule (H2)
hydride ion, H¯ in sodium hydride NaH
OXYGEN
-2
except
0
-1
+2
atom (O) and molecule (O2)
in hydrogen peroxide, H2O2
in F2O
FLUORINE
-1
except
0
atom (F) and molecule (F2)
Q.
Give the oxidation state of the element other than O, H or F in...
SO2
NH3
NO2
NH4+
IF7
Cl2O7
NO3¯
NO2¯
SO32-
S2O32-
S4O62-
MnO42-
What is odd about the value of the oxidation state of S in S4O62- ?
OXIDATION STATES
RULE 1: for simple ions, the oxidation state is the
charge on the ion
e.g. Br- & Cl- oxidation state 1
RULE 2: for atoms in elements, the oxidation state is
zero
e.g. Cl & Br oxidation state 0
RULE 3: for compounds, the sum of the oxidation states
is zero
e.g. KBr (+1) + (-1) = 0
RULE 4: Covalent molecules are assumed to be ionic
e.g. H2O : H+ and O2-
OXIDATION STATES
A. The oxidation states of the elements other than O, H or F are
SO2
NH3
O = -2
H = +1
2 x -2 = - 4
3 x +1 = +3
overall neutral
overall neutral
S = +4
N=-3
NO2
O = -2
2 x -2 = - 4
overall neutral
N = +4
NH4+
H = +1
4 x +1 = +4
overall +1
N=-3
IF7
F = -1
7 x -1 = - 7
overall neutral
I = +7
Cl2O7
O = -2
7 x -2 = -14
overall neutral
Cl = +7
NO3¯
O = -2
3 x -2 = - 6
overall -1
N = +5
NO2¯
O = -2
2 x -2 = - 4
overall -1
N = +3
SO32-
O = -2
3 x -2 = - 6
overall -2
S = +4
S2O32-
O = -2
3 x -2 = - 6
overall -2
S = +2
S4O62-
O = -2
6 x -2 = -12
overall -2
S = +2½ ! (10/4)
MnO42-
O = -2
4 x -2 = - 8
overall -2
Mn = +6
(14/2)
(4/2)
What is odd about the value of the oxidation state of S in S4O62- ?
An oxidation state must be a whole number (+2½ is the average value)
OXIDATION STATES
CALCULATING OXIDATION STATE (2)
The position of an element in the periodic table can act as a guide
METALS • have positive values in compounds
• value is usually that of the Group Number
Al is +3
• where there are several possibilities the
values go no higher than the Group No. Sn can be +2 or +4
Mn can be +2,+4,+6,+7
NON-METALS • mostly negative based on their usual ion
• can have values up to their Group No.
Cl
Cl
usually -1
+1 +3 +5 or +7
Q. What is the theoretical maximum oxidation state of the following
elements?
Na
P
Ba
Pb
S
Mn
Cr
What will be the usual and the maximum oxidation state in compounds of?
Li
Br
Sr
O
B
N
+1
Answers:
What is the theoretical maximum oxidation state of the
following elements?
Na P
Ba
Pb
S
Mn Cr
+1
+5
+2
+4
+6
+7
+6
What will be the usual and the maximum oxidation state
in compounds of?
Li
USUAL
+1
MAXIMUM +1
Br
-1
+7
Sr
+2
+2
O
-2
+6
B
+3
+3
N
-3 or +5
+5
OXIDATION STATES
Q. What is the oxidation state of each element in the following
compounds/ions ?
1.
CH4
2.
PCl3
3.
NCl3
4.
CS2
5.
ICl5
6.
BrF3
7.
PCl4+
8.
H3PO4
9.
NH4Cl
10. H2SO4
11. MgCO3
12. SOCl2
OXIDATION STATES
THE ROLE OF OXIDATION STATE IN NAMING SPECIES
To avoid ambiguity, the oxidation state is often included in the name of a species
manganese(IV) oxide shows that Mn is in the +4 oxidation state in MnO2
sulphur(VI) oxide for SO3
S is in the +6 oxidation state
dichromate(VI) for Cr2O72-
Cr is in the +6 oxidation state
phosphorus(V) chloride for PCl5
P is in the +5 oxidation state
phosphorus(III) chloride for PCl3
P is in the +3 oxidation state
Q. Name the following...
PbO2
lead(IV) oxide
SnCl2
tin(II) chloride
SbCl3
antimony(III) chloride
TiCl4
titanium(IV) chloride
BrF5
bromine(V) fluoride
REVISION CHECK
What should you be able to do?
 Recall the definitions for oxidation and reduction in terms of
oxygen, hydrogen and electrons
 Write correct formulae given the relevant oxidation states
and vice versa
 Use the trend in electronegativity across periods to work out
likely oxidation numbers
 Predict the oxidation state of elements in atoms, simple ions,
compounds and complex ions - QUIZ
 Name compounds using roman numeral notation to indicate
key oxidation numbers as part of the name
HOMEWORK
 Complete Qus 1 & 2 on pg 33 of textbook
 Complete Qu 18 on pg 35 of textbook
 Complete Qu 5 a)ii & b)v on pg 37 of textbook
 Work through PA sheet ensuring you are secure in naming
the compounds