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Oxidation States
Chemistry 1050
Oxidation States:
A bookkeeping device used to tell whether a neutral atom has gained or lost electrons
when it forms a compound. They are not real charges except for simple ions. Also called
an oxidation number.
Background
Atoms in molecules and ions are assigned oxidation numbers by assuming transfer of all
shared bonding electrons to the more electronegative atom. The resulting net “charge”
on each atom is its oxidation number.
Electronegativity: A relative measure of the strength of attraction of an atom for bonding
electrons (i.e. electrons the atom shares with other atoms in chemical
bonds).
Periodic Trends:
Electronegativity increases access a period left to right and from the
bottom to the top of a family or group of elements. The most
electronegative elements are the non-metals in the top right hand
corner of the periodic table and the least electronegative elements are
the metals in the bottom left hand corner. Top 6: EN(F) > EN(O) >
EN(Cl) > EN(N) > EN(Br) > EN(I).
Rules for Assigning Oxidation Numbers (X)
Note: The oxidation number of most elements can vary considerably from one substance
to another. The oxidation number of such elements can be determined in a given
compound or ion using rule #8 if, as is usually the case, the oxidation number of the
other elements present can be positively determined from the other rules.
1.
In free (uncombined) elements, each atom has an oxidation number of zero.
e.g. all atoms in Fe, O2, P4, S8, Ca, Li have oxidation number = 0
2.
For an ion composed of only one atom, the oxidation number of the atom is equal
to the charge on the ion.
e.g. Fe3+ (XFe = +3), S2G (XS = !2)
3.
The oxidation number of fluorine in all its compounds with other elements is equal
to !1. This is because each fluorine atom makes only one single bond and fluorine
is the most electronegative element.
e.g. in CF4, OF2, SF6, XeF2, XF = !1
4.
The oxidation number of hydrogen (H) in most compounds is equal to +1.
e.g. in H2O, NH3, C2H5OH, XH = +1
exception: metal/hydrogen compounds where it is equal to !1
e.g. in NaH, LiAlH4, XH = !1
5.
The oxidation number of oxygen in most compounds is equal to !2. It is the second
most electronegative element and almost always forms two bonds to less
electronegative atoms.
e.g. in H2O, SO2, H2C2O4, KMnO4, XO = !2
exceptions:
1.
The oxidation number of oxygen can be +1 or +2 only in binary compounds
oxygen forms with fluorine. Fluorine is more electronegative than oxygen.
1
e.g. OF2 : XO = +2
2.
O2F2 : XO = +1 (see rules # 3 and # 8)
The oxidation number of oxygen is !1
(a)
in hydrogen peroxide (H2O2) which contains an oxygen-to-oxygen
bond. The Lewis structure of H2O2 is
(b)
in Alkali metal and Alkaline Earth metal peroxides, which contain
the peroxide ion (O22G).
H O O H
e.g. Na2O2, BaO2
3.
The oxidation number of oxygen is !½ in the more reactive Alkali metal
and Alkaline Earth metal superoxides, which contain the superoxide ion
(O2G)
e.g. KO2
6.
In binary ionic compounds, the oxidation numbers of the atoms are equal to ionic
charges.
e.g. CrBr3 XBr = !1, XCr = +3
7.
In binary compounds of non-metals, the oxidation number of the more
electronegative atom is identical to the charge on its common anion. e.g. in CI4, XI
= !1 (common anion of I is IG). The more electronegative atom is almost always the
second atom in the formula.
8.
The algebraic sum of the oxidation numbers of the elements in a neutral compound
must equal zero. In a polyatomic ion the sum must equal the ionic charge.
e.g. identify the oxidation numbers of all elements in each species given below:
9.
HNO3
XH + XN + 3XO = 0
+1 + XN + 3(!2) = 0
XN = +5
SO42G
XS + 4XO = !2
XS + 4(!2) = !2
XS = +6
PH3
XP + 3XH = 0
XP + 3(!1) = 0
XP = !3
VO2+
XV + XO = +2
XV + (!2) = +2
XV = +4
O 2G
2XO = !1
XO = ! ½
CI4
Xc + 4XI = 0
Xc + 4(!1) = 0
Xc = +4
In all compounds, the oxidation number of Alkali metals is equal to +1, of Alkaline
earth metals is equal to +2 and, of Aluminum is equal to +3.
The oxidation number of Na in Na2O2 or K in KO2 must be +1. Hence it is used to
find the oxidation number of oxygen which can vary. See the examples below.
e.g. in Na2O2, XNa = +1
: 2XNa + 2XO = 0
2(+1) + 2XO = 0
XO = !1
in KO2, XK = +1
: XK + 2XO = 0
+1 + 2XO = 0
2XO = !1
XO = !½
Developed by Dr. Chris Flinn
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