Download RULES FOR ASSIGNING OXIDATION NUMBERS 1. All elements in

RULES FOR ASSIGNING OXIDATION NUMBERS
http://www.csun.edu/~hcchm001/IntroChemHandouts.html
1.
All elements in their free state (uncombined with other elements) have an oxidation number of zero (for
example, Na, Cu, Mg, H2 , O2 , Cl2 , N2 ).
2.
H is +1, except in metal hydrides, where it is -1 (for example, NaH, CaH2 )
Na : H
Electrons in the bond are assigned to H, the more electronegative atom.
Na appears to have “lost” one electron, so its oxidation number is +1.*
H appears to have “gained” one electron, so its oxidation number is -1.*
3.
O is -2, except in peroxides, where it is -1, and in OF2 , where it is +2.
H :
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O :
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O :
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H
Electrons in the H-O bonds are assigned to the O.
Electrons in the O-O bond are divided equally between the two O atoms.
H appears to have “lost” one electron, so its oxidation number is +1.*
Each O appears to have “gained” one electron, so its oxidation number is -1.*
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F
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:
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O :
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F :
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Electrons in the F-O bonds are assigned to the F.
Both atoms of F appears to have “gained” 1 electron each, so the oxidation number for each is -1.*
The oxygen appears to have “lost” 2 electrons, so its oxidation number is +2.*
*When compared to the electrically neutral atom.
4. (a)
(b)
The metallic element in an ionic compound has a positive oxidation number. For monoatomic
cations, the oxidation number is equal to the charge on the ion. For example, Na+ , Ca2+ , Al3+ , Fe3+ ,
etc.
The nonmetallic element in an ionic compound has a negative oxidation number. For monoatomic
anions, the oxidation number is equal to the charge on the ion. For example, Cl-, S2-, N3-, etc.
5. In covalent compounds, the negative oxidation number is assigned to the most electronegative atom:
F > O > N > Cl > Br > I > S > C > H
>>>
metals
6. The algebraic sum of the oxidation numbers of elements in a compound is zero.
CaF2
+2 + 2(-1) = 0
H2 SO4
2(+1) + S + 4(-2) = 0
Find S = +6 for ox. no. of S
7. The algebraic sum of the oxidation numbers of the elements covalently bound into a polyatomic ion is
equal to the charge of the ion.
NH4 +
-3 + 4(+1) = +1
HPO3 2+1 + P + 3(-2) = -2
Find P = +3 for ox. no. of P