calculate the oxidation number of Cr²(S04) ²
Answers
Answer:Sulfate is a covalent ion. So it's more realistic to use formal charges, not oxidation states, to describe its atoms.
It's OK to give sulfate an overall oxidation state, but it doesn't make sense to do so for sulfur and oxygen in sulfate.
OXIDATION STATE APPROACH: IDEAL IONIC BONDS
Oxidation states are hypothetical charges for a compound that has ideal ionic bonds, or would form ideal ionic bonds, i.e. complete transfer of the electron pair to only one atom in the bond.
You wrote
SO
2
−
4
, which is the sulfate polyatomic ion. You wrote that its total charge was
−
2
.
Since sulfate (
SO
2
−
4
) is already ionized, its charge is its oxidation state. So it's kind of pointless to find the "oxidation state" for
SO
2
−
4
because it's already given.
The oxidation states for sulfur and oxygen atom are determined from:
electronegativities
total charge
SO
2
−
4
has a total charge of
−
2
, and oxygen is more electronegative than sulfur. So, oxygen polarizes the electron density towards itself more, and thus has the more negative oxidation state here.
Since oxygen is on column 16, its expected oxidation state is
−
2
. Therefore:
−
2
×
4
+
x
=
−
2
⇒
x
gives the oxidation state of sulfur, which is
+
6
.
⇒
+
6
S
−
2
O
2
−
4
But keep in mind that this is unrealistic. If you ever do this for redox balancing, it's only an accounting scheme, and nothing more.
FORMAL CHARGE APPROACH: IDEAL COVALENT BONDS
SO
2
−
4
is primarily a covalent ion. The electronegativity differences are only
1.0
. So, it's more practical to find the formal charges.