Oxidation States

Question 1

Oxidation state of a pure element

Answer 1

0
Ex: 
O2 = 0
H2= 0
Zn=0

Question 2

Oxidation state of fluorine

Answer 2

-1 whenever it is within a compound.
Oxygen has an oxidation state of -2 unless it is in a compound with Fluorine- in this case oxygen will have a charge of +2

Question 3

Oxidation state of hydrogen

Answer 3

It can have a value of -1 or +1

When hydrogen is bonded to a non-metal, hydrogen has a charge of +1 and when bonded to a metal, it has a charge of -1.

Question 4

Oxidation state of carbon

Answer 4

It can vary between -4 and +4.

Question 5

Oxidation states of SO3

Answer 5

According to the periodic table, both elements would like a charge of -2. However, since oxygen is more electronegative than sulphur, the oxidation state of oxygen will be -2 and we will have to solve the oxidation state for sulphur.

S-> x
O-> 3x (-2)
X -6= 0
X= 6

Question 6

Oxidation state of oxygen difluoride

Answer 6

Fluorine is more electronegative than oxygen therefore it will have a charge of -1 whereas oxygen will have a charge of +2

2 + (-1x2) = 0

Question 7

Oxidation state of NO2

Answer 7

Oxygen is more electronegative therefore nitrogen will have a charge of -2 whereas nitrogen will have a positive charge. We can solve it using an equation.

Question 8

Oxidation states of hydrogen (examples)

Answer 8

CH4- carbon is more electronegative than hydrogen which is why it has a charge of +1
NaH- this is a metal hydride; Hydrogen is more electronegative than sodium which is why it has a charge of -1
H2SO4- sulphate has an oxidation state of -2 since it is more electronegative than hydrogen. This is why it has a charge of +1

Question 9

Oxidation state of oxygen (peroxide)

Answer 9

When oxygen is in a peroxide, it has a charge of -1
In a peroxide, two oxygen atoms are linked together by a single covalent bond.
Examples:
O2 -2 (-1x2)
Na2O2 - sodium peroxide
H2O2 - hydrogen peroxide

Question 10

The oxidation state of an ion

Answer 10

The oxidation states of ions made from a single atom equal to its charge.
Ex: 
Na+ -> +1
Mg2+ -> +2
O-2 -> -2

Question 11

Oxidation states of compounds

Answer 11

Oxidation states of compounds always equal to 0
Ex:
H2SO4- [(1x2)+ (-2)]= 0
( we consider the sulphate ion as one; not each element separately)
CH4- [(-4) + (1x4)]= 0

Question 12

Polyatomic ions

Answer 12

The oxidation states of each atom in the ion add up to the overall charge of the ion
Ex: 
SO4 -2 
Sulphur -> x
Oxygen ->  -2
x + (4 x -2) = -2
x + (-8) = -2
x = -2 +8
x = +6

Question 13

Definition of oxidation state

Answer 13

Oxidation states are assigned to atoms in molecules or ions to show how many electrons the atoms has used in bonding.

Question 14

Electronegativity

Answer 14

The more electronegative species will have the negative value
Electronegativity is the ability of an atom to attract electrons from other atoms.
Electronegativity increases across a period and decreases down a group

Question 15

Covalent molecules

Answer 15

Covalent molecules are assumed to be ionic
Ex:
H2O- H+ and O2-

Question 16

The oxidation state of S in S4O6 2-

Answer 16

Sulphur here, has an oxidation state of 2.5

An oxidation state must be a whole number, so this is an average value

Question 17

Theoretical maximum

Answer 17

The highest possible state for nonmetals is equal to the total number of electrons present the outermost shell of the atom.

Question 18

Oxidation states of metals

Answer 18

Have positive oxidation states in compounds
The oxidation state is usually that of the Group Number
Where the metal can have multiple oxidation states, this value can never go higher than the Group Number
Mn- +2,+4, +6 and +7

Question 19

Non-metals

Answer 19

The oxidation states are mostly negative

Can have values that are higher than their Group Number

Question 20

Naming compounds

Answer 20

Ex: PbO2
We know that the oxidation state of oxygen is -2.
So -2x2= -4
So the oxidation state of Pb should be +4.
So the name of this compound should be Lead (IV) oxide

Ex: Potassium Manganate (VII)
The VII means 7 and this is the oxidation state of manganese in this compound
The ‘ate’ suggests that there is oxygen present in this compound and the oxidation state of this is -2.
The formula for this compound is KMnO4
The total oxidation state of O is -8 (-2x 4)
-8 +1 is -7
Therefore manganese should be +7