Transcript Oxidation Numbers (Ox #’s)

Oxidation Numbers (Ox #’s)
What are they used for? Why do you need to learn them?
• to write chemical names and chemical formulas
• to balance redox equations for analytical, organic and inorganic chemistry
What is an oxidation number?
Ox # is a charge assigned to an ion or an atom.
There are several cases to consider…
1. Elements, both monatomic, e.g., Cu and polyatomic, e.g., H2, Cl2, etc.
2. Ionic compounds, e.g., NaCl
3. Covalent compounds, e.g., HCl, H2O, etc.
1
1. a) Ox # of a monatomic atom = 0 (because its net electric charge = 0)
Consider a hydrogen atom, 1H.
It has one proton (+) and one electron (-).
A hydrogen atom is neutral (-1 + 1 = 0).
Its Ox # = 0.
Consider a helium atom, 2He.
It has two protons (+2) and two electrons (-2).
Its Ox# = 0. (-2 + 2 = 0)
The same is true for all 118 monatomic atoms.
They’re all neutral.
They all have an Ox # = zero.
1H
2He
2
1. b) Ox # of a polyatomic element = 0 (because these atoms are neutral).
Consider the diatomic hydrogen molecule, H2.
In the Lewis structure of H2, the line drawn between
the H atoms represents a covalent bond made of two
shared e’s.
e‘s are always shared equally between identical atoms.
To determine H’s Ox #, divide the shared e’s and give
one e’ to each atom.
This produces two neutral H atoms, each owning one e’.
Each H atom is now the same as its Lewis symbol.
Recall that Lewis symbols represent neutral atoms by
showing one dot for each e’ in the outermost shell
(valence shell) of the atom.
Since the atoms in H2 are neutral, their Ox # = 0.
A line bond represents a shared
pair of e’s in a covalent bond
Ox # = 0
Ox # = 0
Lewis Symbols
3
The Ox # of a polyatomic element = 0 (continued).
Consider the diatomic oxygen molecule, O2.
The Lewis structure of O2 shows two shared e’ pairs
(a double bond) represented by two lines.
Dividing the shared e’s and giving two e’s to each atom,
produces two neutral O atoms, each owning 6 valence e’s.
Each O atom is now the same as its Lewis symbol.
Since the atoms in O2 are neutral, their Ox # = 0 in O2.
The same reasoning applies to diatomic nitrogen, N2
Dividing its triple bond gives three e’s to each N atom,
producing two neutral N atoms each owing 5 valence e’s
N’s Ox # in N2 = 0
The same is true for all polyatomic elements, e.g.,
P4, S8, F2, Cl2, Br2, I2, etc. Their atoms all have Ox # = 0.
Ox # = 0
Ox # = 0
Ox # = 0
Ox # = 0
Lewis Symbols
4
2. In ionic compounds, the Ox # of an ion is the same as its electric charge
Ox # = +2
Ox # = -1
Ox # = +3
Ox # = -2
Calcium chloride, CaCl2, is an ionic compound, made of one calcium cation, Ca+2,
and two chloride anions, 2Cl-.
The Ox # of Ca+2 cation in CaCl2 is the same as its charge, +2.
The Ox # of each Cl- anion in CaCl2 is the same as its charge, -1.
Practice: State the Ox #’s of both atoms in Al2O3
Answer: Since Al2O3 is an ionic compound, the Ox #s of Al+3 and O-2 ions are the
same as their electric charges, +3 and -2, respectively.
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3. Calculating the Ox # of atoms in a covalent compound
By definition, compounds are made of different types of atoms.
Unlike ionic compounds (e.g., NaCl) that contain charged ions, covalent compounds
(e.g., HCl) and covalent molecules (e.g. Cl2) are contain no ions.
+
Na
Cl
-
NaCl, ionic compound
doesn’t share e’s.
H
Cl
HCl, polar covalent compound
shares e’s unequally.
Cl
Cl
Cl2, nonpolar covalent molecule
shares e’s equally.
In Cl2, bonded e’s are shared equally, but in HCl, e’s are shared unequally because different
atoms hold their electrons more or less strongly.
In HCl, the shared e’s spend most of their time near Cl, as indicated by the larger e’-cloud
around Cl and less time around H indicated by the smaller e’-cloud around H.
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Linus Pauling’s Table of Electronegativities (EN)
More EN atoms hold e’s more strongly than less EN atoms. More EN atoms have
higher EN values, nonmetals being the highest and metals being the lowest.
H
2.1
Li
1.0
Be
1.5
B
2.0
C
2.5
N
3.0
O
3.5
F
4.0
Na
1.0
Mg
1.2
Al
1.5
Si
1.8
P
2.1
S
2.5
Cl
3.0
K
0.9
Ca
1.0
Sc
1.3
Ti
1.4
V
1.5
Cr
1.6
Mn
1.6
Fe
1.7
Co
1.7
Ni
1.8
Cu
1.8
Zn
1.6
Ga
1.7
Ge
1.9
As
2.1
Se
2.4
Br
2.8
Rb
0.9
Sr
1.0
Y
1.2
Zr
1.3
Nb
1.5
Mo
1.6
Tc
1.7
Ru
1.8
Rh
1.8
Pd
1.8
Ag
1.6
Cd
1.6
In
1.6
Sn
1.8
Sb
1.9
Te
2.1
I
2.5
Cs
0.8
Ba
1.0
La
1.1
Hf
1.3
Ta
1.4
W
1.5
Re
1.7
Os
1.9
Ir
1.9
Pt
1.8
Au
1.9
Hg
1.7
Tl
1.6
Pb
1.7
Bi
1.8
Po
1.9
At
2.1
EN increases
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Calculating the Ox # of atoms in a covalent compound (continued)
To calculate the Ox # of atoms in covalent compounds, all shared electrons
(in covalent bonds) are assigned to the more electronegative (EN) atom
(and taken away from the less EN atom).
Consider HCl
Since Cl (EN = 3.0) is more EN than H (EN = 2.1), we assign
both shared e’s in the covalent bond to Cl.
We ascribe all 8 e’s in the structure to Cl, thus giving Cl one
more e’ than its neutral atom has (see the Lewis symbol of Cl)
Thus Cl in HCl has Ox # = -1.
This leaves the H atom in the structure without any e’s, one
less e’s than its neutral atom has (see the Lewis symbol of H).
Thus H in HCl has Ox # = +1.
Be Careful. No ions are actually present in HCl.
The Ox #’s do not represent charges in covalent compounds.
Ox # = -1
Cl owns 8 e’s
H owns 0 e’s
Ox # = +1
Lewis Symbols
8
EN O = 3.5
EN H = 2.1
Calculating the Ox # of atoms in a covalent compound (continued)
Ox # = -2
Look at the Lewis structure of a H2O molecule.
O has two pairs of unshared (nonbonded) e’s.
Nonbonded e’s belong solely to the O atom.
There are also two shared e’ pairs (the covalently bonded e’s)
These 4 bonded e’s are not shared equally. They spend more
time around O, because O (EN=3.5) is more EN than H (EN=2.1)
O owns 8 e’s
Ox # = +1
H owns 0 e’s
Lewis Symbols
So all 8 valence e’s are assigned to O when calculating its Ox #.
Thus O is assigned 8 e’s and the H’s are not assigned any e’s.
Each H in H2O has one less e’ than its Lewis symbol, so H is assigned an Ox # of +1
The O in H2O has two more e’s than its Lewis symbol, so O is assigned an Ox # of -2.
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Oxidation Numbers
Ox # = -2
Note that in almost all its compounds, O has Ox # = -2.
But recall that H and O are not truly charged in H2O.
H2O is not ionic, but is considered ionic only to calculate Ox #’s.
Ox # = +1
Peroxides are exceptions. Peroxides have one more O in their
formula than is normally present.
Hydrogen peroxide, H2O2, is a simple example of a peroxide.
The O atoms in peroxides have Ox # = -1, because they are
assigned 7 e’s (one more than a neutral O atom).
Ox # = -1
Ox # = +1
Ox # = -1
Ox # = +1
Ox # = +1
H can also have an Ox # of -1, when it is bonded to a less EN
element, such as a reactive metal. Lithium hydride, LiH, is an
example.
In LiH, H owns one more e’ than its Lewis symbol so its Ox # = -1.
Note that the ‘ide’ ending in hydride indicates that H is more EN
and is named as an anion.
Ox # = +1
Ox # = -1
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Oxidation Numbers
Now that you understand how Ox #’s are determined, you can relax in the knowledge
that you don’t have to draw Lewis structures every time you need to find an Ox #.
There is a simple method for learning Ox #’s, that will require a little memory and a
little understanding of the periodic table.
Lewis Symbols
Group 1A metals (with 1 valence e’) are always Ox # +1 (so is Ag+)
Group 2A metals (with 2 valence e’s) are always Ox # +2 (so are
Zn+2 and Cd+2)
Group 3A elements (with 3 valence e’s) B, Al and Ga are always +3.
Recall that the A-Group metals tend to lose all their valence e’s to
become isoelectronic with the nearest noble gas.
So if you know the group number of an A-Group metal, you usually
know its Ox #.
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Oxidation Numbers
In Group 7A, F always has Ox # = -1.
The other halogens may have ⊕ Ox #’s but will always be -1
when they are the more EN atom in a compound, e.g.,
HCl (Cl is more EN), NaBr (Br is more EN), KI (I is more EN).
So Cl, Br and I all have Ox #’s = -1 in these compounds.
Lewis Symbols
In Group 6A, Ox # of O is always -2 (except in peroxides).
The other chalcogens will also have Ox # of -2 when they are
the more EN atom in a compound, e.g., H2S. (Ox # S = -2)
The Group 5A atoms will always have Ox # of -3 when they
are the more EN atom in a compound, e.g., NH3 (Ox # N = -3)
Recall that the A-Group nonmetals tend to gain enough e’s to fill their valence
shell to become isoelectronic with the nearest noble gas.
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Atoms with fixed Ox #’s are shown. Metals cannot have ⊖ Ox #’s (they never gain e’s).
Most nonmetals can have ⊖ or ⊕ Ox #’s, e.g., ICl (Ox # Cl = -1), ClF (Ox # Cl = +1, Ox # F = -1)
Note that the nonmetal with higher EN always uses its ⊖ Ox #
1A
s1
1H
5e + 3e = 8
6e + 2e = 8
Fixed Ox #’s of Atoms
(2.1)
1
-1
3Li
7e + 1e = 8
symbol & proton number
hydrogen
Key
2A
s2
(1.0)
lithium
4Be(1.5)
1
---
2
---
3A
s2p1
electronegativity (EN)
7N
beryllium
(3.0)
nitrogen
2,3,4,5
possible ⊕ Ox #’s
-3
possible ⊖ Ox #’s
5B (2.0)
boron
3
---
11Na (1.0) 12Mg (1.2)
13Al
(1.5)
sodium
magnesium
aluminum
1
---
2
---
+3
---
19K
(0.9)
3B
s2d1
20Ca (1.0) 21Sc (1.3) 22Ti
potassium calcium scandium
1
---
4B
s2d2
2
---
(1.4)
5B
s2d3
23V (1.5)
6B
s1d5
7B
s2d5
8B
s2d6
8B
s2d7
8B
s2d8
8A
s2p6
1B
s1d10
(1.6) 25Mn (1.6) 26Fe (1.7) 27Co (1.7) 28Ni (1.8) 29Cu (1.8)
titanium vanadium
chromium manganese iron
cobalt nickel copper
24Cr
2B
s2d10
30Zn (1.6) 31Ga (1.7)
zinc
gallium
2
---
3
---
4A
s2p2
5A
s2p3
6A
s2p4
7A
s2p5
7N (3.0) 8O (3.5) 9F (4.0)
(2.5)
nitrogen oxygen fluorine
carbon
6C
-3
(1.8) 15P (2.1)
phosphorus
silicon
14Si
-3
32Ge (1.9)
germanium
---2
---1
16S
(2.5) 17Cl (3.0)
sulfur chlorine
-2
1
---
2
---
10Ne
neon
----18Ar
argon
36Kr
(2.4) 35Br (2.8)
arsenic selenium bromine krypton
33As (2.1) 34Se
-3
-2
-----
-1
47Ag (1.9) 48Cd (1.6)
(1.0) 39Y (1.2) 40Zr (1.3)
42Mo (1.6)
51Sb (1.9) 52Te (2.1) 53I (2.5)
Rh (1.8) 46Pd (1.8)
50Sn (1.8)
rubidium strontium yttrium zirconium 41Nb (1.5) molybdenum 43Tc (1.7) 44Ru (1.8) 45
cadmium 49In (1.6)
antimony tellurium
silver
iodine
technetium ruthenium rhodium palladium
niobium
indium
tin
2
---
-----
-----
-1
37Rb (0.9) 38Sr
1
---
2He
helium
-3
-2
-1
13
54Xe
xenon
-----
Oxidation Numbers
Once you memorize the Ox #’s of those elements that have fixed Ox #’s, then all
other Ox #’s are calculated from the chemical formula.
Important Rule:
The sum of all the Ox #’s in a formula equals the total charge of the formula.
Examples:
The charge on hydroxide, OH-, is -1, so the sum of the Ox #’s of O and
H must = -1.
-2
-2
The charge on H2O is 0. A charge of zero is never written in a formula.
The sum of Ox #’s of O and H must = 0
The charge on ammonium ion, NH4+, is +1.
The sum of Ox #’s of N and H must = +1, so Ox # N = -3
-3
14
Oxidation Numbers
Example 1: Calculate the Ox # of the manganese atom in MnO2
Process: Use the known Ox # of O to calculate the unknown Ox # of Mn
we don't know 'yet'
we will calculate
O anion is always -2
-4
Mn? (O-2)2
The # of -'ve charges = 4,
so the # of +'ve charges must also be 4.
The sum of charges = 0; [(+4) + (-4) =0]
The charge of Mn = +4
the total negative charge = (-2 × 2) = -4
manganese(IV) oxide
Ox # of ‘O’ = -2. Multiply this by 2 because there are 2 ‘O’ atoms in the formula: (2 × -2 = -4)
Since the formula shows no charge, you know it is neutral. This means that the Mn atom must
have an Ox # = +4.
You can think of it as a simple math equation, where the sum of all Ox #’s = total formula charge
Mn + (-4) = 0

Mn = 0- (-4)

Mn = +4
This compound is named manganese(IV) oxide, where ‘IV’ is the Ox # of Mn in Roman numerals
Note that the Ox # of an atom is stated in a name only when the atom can have > 1 Ox #
15
Oxidation Numbers
Example 2: Calculate the Ox # of the manganese in Mn2O7
Process: Use the known Ox #’s of O to calculate the unknown Ox # of Mn
we don't know 'yet'
we will calculate
-14
(Mn+?)2
The # of -'ve charges = 14,
so the # of +'ve charges must also be 14.
The sum of charges = 0; [(+14) + (-14) =0]
The charge of each Mn ion = 14/2 = +7
(O-2)7
O is always -2
the total negative charge = (-2 × 7) = -14
manganese(VII) oxide
Ox # of ‘O’ = -2. Multiply this by 7 because there are 7 ‘O’ atoms in the formula: (7 × -2 = -14)
Since the formula shows no charge, you know it is neutral. This means that both Mn atoms
together must contribute a charge of +14. Divide this by 2, to find the Ox # of each Mn atom.
You can think of it as a simple math equation, where the sum of all Ox #’s = total formula charge
2Mn + (-14) = 0 
2Mn = 0- (-14) 
2Mn = +14

Mn = +14/2 = +7
This compound is named manganese(VII) oxide, where ‘VII’ is the Ox # of Mn in Roman numerals
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Oxidation Numbers
Practice: Calculate the Oxid # of the underlined atom in each formula
N2O
N2O5
SO3-2
2N + (-2) = 0
2N + (-10) = 0
S + (-6) = -2
Cd(BrO3)2
SiF4
N= +1
N= +5
S= +4
Na2CrO4 2 + Cr -8 = 0
P2O7-4
HCO3-
Cr2S3
2 + 2(Br -6) = 0
(Br -6) = -1
Br= +5
Si -4 = 0
Si= +4
Fe3P2
2P + -14 = -4
2P = +10
1 + C -6 = -1
C = -1 -1 +6
2Cr +3(-2) = 0
2Cr = 6
3Fe +2(-3) = 0
3Fe = 6
Cr= +6
P= +5
C= +4
Cr= +3
Fe= +2
17
Oxidation Number Summary:
example
electric charge
Ox #
Monatomic element
Cu, Al, Zn, etc.
0
Ox # = 0
Polyatomic element
H2, O2, P4, S8, etc.
0
Ox # = 0
Ionic compounds
NaCl, CaCl2, Al2O3
variable charge
but ≠ 0
Ox # = ion charge
Covalent compounds
HCl, CO2
no charges
Shared e’s assigned to more
EN atom
Ox # ≠ charge
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A List of Common Ox #’s
1A
s1
1H
(2.1)
2He
symbol & proton number
hydrogen
1
-1
8A
s2p6
2A
s2
beryllium
1
---
2
---
3A
s2p1
electronegativity (EN)
3Li (1.0) 4Be(1.5)
lithium
helium
Key
7N
(3.0)
nitrogen
2,3,4,5
possible ⊕ Ox #’s
-3
possible ⊖ Ox #’s
5B (2.0)
boron
3
---
11Na (1.0) 12Mg (1.2)
sodium
magnesium
1
---
2
---
4A
s2p2
5A
s2p3
6A
s2p4
-----
7A
s2p5
6C
(2.5) 7N (3.0) 8O (3.5) 9F (4.0)
carbon nitrogen oxygen fluorine
2, 4 2,3,4,5
-----
-4
-3
-2
-1
10Ne
neon
-----
(1.5) 14Si (1.8) 15P (2.1) 16S (2.5) 17Cl (3.0) 18Ar
aluminum silicon phosphorus sulfur
chlorine argon
3,
4,
5
+3
2, 4, 6 1,3,5,7
4
---
13Al
3B
s2d1
4B
s2d2
5B
s2d3
6B
s1d5
7B
s2d5
8B
s2d6
8B
s2d7
8B
s2d8
1B
s1d10
2B
s2d10
---
---
-3
-1
-2
---
(1.6) 25Mn (1.6) 26Fe (1.7) 27Co (1.7) 28Ni (1.8) 29Cu (1.8) 30Zn (1.6) 31Ga (1.7) 32Ge (1.9) 33As (2.1) 34Se (2.4) 35Br (2.8) 36Kr
potassium calcium scandium titanium vanadium chromium manganese iron
cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton
4
1
3
3, 4 2,3,4,5 2, 3, 6 2,3,4,6,7 2, 3
4, 6
3, 5
2
3
1, 5
--2, 3
2, 3
1, 2
2
19K
(0.9)
---
20Ca (1.0) 21Sc (1.3) 22Ti
---
---
37Rb (0.9) 38Sr (1.0) 39Y
rubidium strontium
1
---
2
---
(1.2)
yttrium
3
---
(1.4)
---
23V (1.5) 24Cr
---
---
---
---
---
---
---
---
---
---
-3
-2
-1
40Zr (1.3) 41Nb (1.5) 42Mo (1.6) 43Tc (1.7) 44Ru (1.8) 45Rh (1.8) 46Pd (1.8) 47Ag (1.9) 48Cd (1.6) 49In (1.6) 50Sn (1.8) 51Sb (1.9) 52Te (2.1) 53I
zirconium niobium
4
---
3,5
---
molybdenum technetium ruthenium
2,3,4,5,6
---
-----
2,3,4
---
rhodium
2, 3, 4
---
palladium
2, 4
---
silver cadmium indium
2
1, 3
1
-------
tin
2, 4
---
antimony
tellurium
3, 5
4, 6
-2
-3
---
(2.5)
iodine
1, 5, 7
-1
54Xe
xenon
-----
19