Oxidation Numbers
 Oxidation Numbers: a.k.a. oxidation
states, indicates the general distribution
of electrons among the bonded atoms in
a molecular compound or a polyatomic
ion.
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Oxidation Numbers ~ Rules
1) Atoms of a pure element have an ox. # of zero
2) More electroneg. elements in binary molecular compounds
have an ox. # equal to their charge as an anion, the less
electroneg. elements have an ox. # equal to the cation.
3) Fluorine has an ox. # of -1.
4) Oxygen has an ox. # of -2 (exceptions are peroxides, -1,
or when bonded Fluorine, +2)
5) Hydrogen has an ox. # of +1 when bonded with more
electronegative elements, -1 when bonded with metals.
6) The algebraic sum of the ox. # in a neutral compound is
zero.
7) The algebraic sum of the ox. # in a polyatomic ion is
equal to the charge of the ion.
8) Ox. # can be assigned to atoms in ionic compounds.
9) A monatomic ion has an ox. # equal to the charge of the
ion.
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Oxidation Numbers ~ Examples
 What are the oxidation numbers for the
atoms in the following compounds:
 H2O: O is -2, H is +1
 HNO3: H is +1, O is -2 (x3 = -6), N must
be +5
 P4O10: O is -2 (x10 = -20), P must be +5
(x 4 = +20)
 ClO3
-: O is -2(x3 = -6), Cl must be +5
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Practice
9) What are the oxidation numbers for the
atoms in the following compounds?
a) PCl3:___________________________
b) CF4:_____________________________
c) PbO2:____________________________
_
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Practice
9) What are the oxidation numbers for the
atoms in the following compounds?
a) PCl3 :__ P +3, Cl -1_________
b) CF4:____F -1, C +4___________
c) PbO2 :__Pb +4, O -2__________
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Molecular Formulas
Molecular Formulas give the actual
number of atoms of each element in a
molecular compound. This is always a
whole number multiple of the empirical
formula.
By dividing the molar mass by the empirical
formula mass, you get the molecular
formula.
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Practice
12) Find the molecular formula for a compound that contains
4.90 g N and 11.2 g O with a molar mass of 92.0 g/mol.
4.90 g N x 1.0 mol N = 0.350 mol N
14.01 g N
11.2 g O x 1.0 mol O = 0.700 mol O
16.00 g O
N: 0.350 = 1.00 mol N O = 0.700 = 2.00 mol O
0.350 0.350
NO2 is the empirical
formula.
Molar mass = 92.0g/mol = 2
Empirical formula mass 46.01g/mol
Molecular formula = 2 X empirical formula = N2O4
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The
End!