How do you find the oxidation number of an element? Is there an equation?

1 Answer
Jul 22, 2017

Answer:

Well, the oxidation number of an element is a big fat ZERO........

Explanation:

The oxidation number of an element INVOLVED in a chemical bond is (usually) non-zero, and it is conceived to be the charge left on the element of interest when all the bonding pairs of electrons are broken, and the electrons assigned to the MORE electronegative atom.

Here a few rules.....

#1.# #"The oxidation number of a free element is always 0."#

#2.# #"The oxidation number of a mono-atomic ion is equal"# #"to the charge of the ion."#

#3.# #"For a given bond, X-Y, the bond is split to give "X^+# #"and"# #Y^-#, #"where Y is more electronegative than X."#

#4.# #"The oxidation number of H is +1, but it is -1 in when"# #"combined with less electronegative elements."#

#5.# #"The oxidation number of O in its"# compounds #"is usually -2, but it is -1 in peroxides."#

#6.# #"The oxidation number of a Group 1 element"# #"in a compound is +1."#

#7.# #"The oxidation number of a Group 2 element in"# #"a compound is +2."#

#8.# #"The oxidation number of a Group 17 element in a binary compound is -1."#

#9.# #"The sum of the oxidation numbers of all of the atoms"# #"in a neutral compound is 0."#

#10.# #"The sum of the oxidation numbers in a polyatomic ion"# #"is equal to the charge of the ion."#

I grant that these rules look intimidating, but you will soon be assigning oxidation numbers with the best of them, and you certainly do not have to commit the rules to memory....

For simple molecule such as #HF#, we split the beast up to give #H^+# and #F^-#, and thus allocate oxidation numbers of #H(+I)# and #F(-I)#. I urge you to consult your text and notes. For a more complicated molecule or ion such as #ClO_4^(-)# we get #Cl(+VII)# and #O(-II)#; does the sum of the oxidation numbers equal the charge on the ion?