RULES OF OXIDATION NUMBER
ASSIGNMENT
STEPS IN ASSIGNNING OXIDATION NUMBERS
1) IF THE PERIODIC TABLE GIVES ONLY ONE OXIDATION STATE,
USE THAT STATE. EXAMPLE Zn+2 .
2) APPLY THE RULES OF THIS POWERPOINT TO SOLVE FOR
OXIDATION NUMBERS OF ELEMENTS WITH MULTIPLE OXIDATION
STATES. THIS IS VALUABLE WITH TRANSITION METALS.
3) IF A MULTIPLE OXIDATION STATE ELEMENT HAS NO RULE,
ASSIGN IT A VARIABLE AND SOLVE USING THE ALGEBRA
RULE(SLIDE 3).
IF AN ELEMENT HAS MULTIPLE
OXIDATION STATES, AS IN
CARBON, YOU WILL HAVE TO
USE THE ALGEBRA RULE TO
SOLVE FOR THE STATE IN THE
PARTICULAR COMPOND YOU
ARE INVESTIGATING.
SLIDE-3 ALGEBRA RULE
THE SUM OF THE OXIDATION NUMBER (STATE ) OF
EACH ATOM IN A COMPOUND EQUALS THE
CHARGE OF THAT COMPOUND.
KMnO4
GIVEN: K IS +1(GROUP 1 RULE ), O IS -2 (OXYGEN RULE)
Mn HAS MULTIPLE STATES(PERIODIC TABLE)
+1 + X + 4(-2) = 0
ALGEBRA RULE EXAMPLES.
Use group one as a known for
muli oxidation stare elements
Na2SO4
2(+1) + (X) + 4(-2)= 0
X = S = +6 each S
Use group one to find other
elements.
note: caution with
oxygen in binary
compounds
Li2O2
2(+1) + 2(X) = 0
X=O (oxygen) = -1
Each O is 2- rule
Each Na is 1+,
OXYGEN RULE
GROUP 1 RULE
Each H is 1+, rule
HYDROGEN RULE
NaOH
1 + -2 + 1 = 0
EACH O is 2OXYGEN RULE
Cr HAS MULTIPLE
OXIDATION
STATES AND HAS
NO RULE, SOLVE
FOR IT AS THE
VARIABLE X
Cr2O7
2-
THE CHARGE OF
THIS POLYATOMIC
ION IS -2, ALL THE
SUM OF THE
OXIDATION STATES
MUST EQUAL -2.
2X + 7(-2) = -2
X = OXIDATION STATE OF Cr = 6+
SLIDE-6 GROUP 1 RULE
GROUP ONE ELEMENTS ARE ALWAYS 1+ IN
COMPOUNDS OR AS FREE (AQ) IONS.
Ex in NaCl EACH Na has an oxidation state of 1+.
in K2SO4 EACH K has an oxidation state of 1+.
A Li+1 (aq) free ion has an oxidation number of 1+.
GROUP ONE RULE EXAMPLES.
Use group one as a known for
Use group one to find other
elements.
muli oxidation stare elements.
Na2SO4
Li2O2
2(+1) + (X) + 4(-2)= 0
X = S = +6 each S
2(+1) + 2(X) = 0
X=O (oxygen) = -1 each
SLIDE-7 GROUP 2 RULE
GROUP TWO ELEMENTS HAVE AN OXIDATION STATE OF 2+
IN COMPOUNDS AND AS FREE (AQ) IONS.
Ex in1) BaCl2 EACH Ba has an oxidation state of 2+
JJIBGO 2) In Ca3(PO4)2 EACH Ca has an oxidation state of 2+.
A free Mg+2 ION (AQ) has an oxidation state of 2+.
GROUP ONE RULE EXAMPLES.
Use group TWO as a known for
Use group TWO to find other
muli oxidation stare elements.
elements.
Ca3(PO4)2
3(+2) + 2(x) + 8(-2)=0
X = P = +5 each P
BaO2
1(+2) + 2(X) = 0
X=O (oxygen) = -1 each
SLIDE-8 ELEMENTAL STATE RULE
ALL UNCOMBINED NEUTRAL (ELEMENTAL
STATE) ATOMS HAVE AN OXIDATION STATE
OF 0. THE NONPOLAR COVALENT
DIATOMOIC MOLECULES OF THE PERIODIC
TABLE ALSO HAVE AN OX# OF 0.
EX: Au0, Fe0, Na0, etc.
EX: H2, O2, N2, Cl2, Br2, I2, F2 are all in the 0
oxidation state
SLIDE-9 IONIC CHARGE RULE
THE OXIDATION NUMBER OF IONS IS THE
IONIC CHARGE.
EX: Au+2 = 2+, Fe+3 = 3+, Na+ = 1+, etc. --- USE THE
CHARGE WRITTEN ON THE ION…NOT OTHER
STATES ON P TABLE.
EX: NO3 -
HAS AN OXIDATION NUMBER OF 1- FOR THE (EACH) POLY
ATOMIC ION GROUP.
SLIDE-10 OXYGEN RULE
THE OXYGEN RULES: Each oxygen USUALLY
has an
oxidation state of 2- in compounds, 0 in O2. USE CAUTION IN BINARY
COMPOUNDS OF OXYGEN
EX, in H2O oxygen is 2AND HYDROGEN -- CHECK
In H2SO4 EACH oxygen is 2FOR RULE EXCEPTIONS.
IMP -- EXCEPTION #1– when oxygen is contained in a PEROXIDE, it
has an oxidation state of 1-.
PEROXIDES - contain the peroxide ion (O2-2), EACH oxygen is 1-.
- peroxides are when 2 group one elements combine with the peroxide ion
H2O2, Li2O2, K2O2 are peroxides, each O is 1BaO2, CaO2, MgO2 are peroxides, each O is 1-
IMP -- EXCEPTION #2 – when oxygen is combined with FLOURINE,
its oxidation state is 1+ or 2+. Fluorine is electronegative enough to
oxidize oxygen.
EXAMPLE :
in FO the oxygen is +1, in F2O each oxygen is 2+
Fluorine is always 1in its compounds
SLIDE-11 HYDROGEN RULE
THE HYDROGEN RULE:
THE OXIDATION STATE OF HYDROGEN IS USUALLY 1+
EX in H2O each hydrogen is 1+
IMP-EXCEPTION – in GROUP 1 METAL HYDRIDES the
hydrogen is 1-, hydrogen is electronegative enough to oxidize
group one metals.
EX: NaH, KH etc the hydrogen is 1USE CAUTION IN BINARY
COMPOUNDS OF OXYGEN AND
HYDROGEN -- CHECK FOR RULE
EXCEPTIONS.
SLIDE-12 HALOGEN RULE
THE HALOGEN RULE:
THE OXIDATION STATE OF HALOGENS IS USUALLY 1- IN
BINARY COMPOUNDS.
EX in MgF2 each F is 1- AS INDICATED BY THIS RULE
IMP-EXCEPTION – HALOGEN IN POLYATOMIC IONS WILL
EXHIBIT MULTIPLE OXIDATION STATES…USE ALGEBRA RULE
TO SOLVE FOR THE HALOGEN. SEE NEXT SLIDE…
EACH O is 2OXYGEN RULE
Cr HAS MULTIPLE
OXIDATION
STATES AND HAS
NO RULE, SOLVE
FOR IT AS THE
VARIABLE X
ClO4
2-
THE CHARGE OF
THIS POLYATOMIC
ION IS -2, THE SUM
OF THE OXIDATION
STATES MUST
EQUAL -2.
X + 4(-2) = -2
X = OXIDATION STATE OF Cl IS 6+
SLIDE-15 HALOGEN RULE
THE HALOGEN RULE:
THE OXIDATION STATE OF HALOGENS IS USUALLY 1- IN
BINARY COMPOUNDS.
EX in Na2S each SODIUM is 1- AS INDICATED BY THIS RULE
IMP-EXCEPTION – HALOGENS IN POLYATOMIC IONS WILL
EXHIBIT MULTIPLE OXIDATION STATES…USE ALGEBRA RULE
TO SOLVE FOR THE HALOGEN.
THE SOLUBLE SALT RULE:
1) MANY TERNARY SALTS CONTAIN 2 ELEMENTS WITH MULTIPLE
OXIDATION STATES AND THAT HAVE NO RULE DEFINNING THE
OXIDATION NUMBER IN THAT PARTICULAR SALT.
2) AS YOU CANNOT HAVE TWO VARIABLES AT ONCE YOU CAN SOLVE
ONE AT A TIME, THE METAL BY DISSOCIATION AND THE NONMETAL BY
ALGEBRA.
FOR EXAMPLE THE SALT Ni(NO3)3 HAS Ni AND n, BOTH WITH
MULTIPLE STATES. THE SHORT FORM FOR GETTING THE Ni IS:
Ni (NO3)3
+3 (-1) * 3
+3 -3
CHARGE -1 CHARGE ON
NITRATE ISFROM TABLE E
NI HAS MULTIPLE
OXIDATION
STATES AND HAS
NO RULE, SOLVE
FOR IT AS THE
VARIABLE X
NITROGEN HAS MULTIPLE
OXIDATION STATES, SOLVE
FOR THIS ALSO.
OXYGEN IS 2- IN
MOST
COMPOUNDS
…OXYGEN RULE.
Ni(NO3)3
THIS SALT CONTAINS 2 ELEMENTS WITH MULTIPLE OXIDATION
STATES (Ni AND N). YOU HAVE TWO WAYS TO SOLVE.
1) USE THE DISSOCIATION REACTION AND BALANCE FOR
CHARGE TO GET THE METAL (Ni) OXIDATION STATE. THEN
SOLVE FOR NITROGEN.
2) OR SOLVE FOR NITROGEN IN THE NITRATE ION FIRST, THEN
SOLVE FOR Ni. SEE SOLUTIONS IN NEXT TWO SLIDES…
THIS SALT CONTAINS 2 ELEMENTS WITH MULTIPLE OXIDATION STATES
(Ni AND N).
1) WRITE THE DISSOCIATION REACTION AND BALANCE FOR CHARGE
TO GET THE METAL (Ni) OXIDATION STATE. THEN SOLVE FOR
CHARGE
NITROGEN.
0 = X
+
X = Ni = +3
3 (-1)
Ni(NO3)3  Ni + 3
FROM
TABLE E
NO3
NITROGEN HAS MULTIPLE OXIDATION STATES
Ni HAS A
OXIDATION STATE
OF 3+ FROM THE
DISSOCIATION
ABOVE
Ni(NO3)3
OXYGEN IS 2- IN
MOST
COMPOUNDS
…OXYGEN RULE.
+3 + 3(X) + 9(-2) = 0
X = OXIDATION STATE OF N = 5+
THIS SALT CONTAINS 2 ELEMENTS WITH MULTIPLE OXIDATION STATES
(Ni AND N).
1) OR SOLVE FOR NITROGEN IN THE NITRATE ION FIRST, THEN SOLVE
FOR Ni. SEE SOLUTIONS IN NEXT TWO SLIDES… CHARGE
NO3
FROM
TABLE E
-
X + 3(-2) = -1
X = OXIDATION STATE OF N = 5+
NITROGEN HAS AN OXIDATION STATE OF 5+
FROM THE DISSOCIATION CALCULATION
Ni HAS MULTIPLE
OXIDATION
STATES, SOLVE
FOR IT.
Ni(NO3)3
OXYGEN IS 2- IN
MOST
COMPOUNDS
…OXYGEN RULE.
X + 3(+5) + 9(-2) = 0
X = OXIDATION STATE OF NI = 3+