How do you determine oxidation states?

1 Answer
Jul 29, 2017

Answer:

#"Oxidation state"# and #"oxidation numbers"# are VERY contrived concepts.

Explanation:

We know from when we first introduced to chemistry that ionic bonding results from the TRANSFER of electrons to form ions, and covalent bonding results from the sharing of electrons to form electrons clouds, to which the participating nuclei are strongly attracted such that nucleus-nucleus repulsion is negated, and a net attractive force results.

Oxidation states assigns a number to each participating atom in a molecule or an ion, which represents the number of electrons that atom has accepted (i.e. in the case of #"reduction"#) or donated, (i.e. in the case of #"oxidation"#).

AS regards rules of assignment of oxidation, I append the following rules (which of course are from a prior answer). I would read these in conjunction with your text.

#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."#

And how do we use them? Well, let us apply to two common ions, i.e. #ClO_4^-#, and #CrO_4^(2-)#. Following rule 10, the sum of the oxidation numbers equals the charge on the ion......for #ClO_4^-# we have #stackrel(-II)O# (rule 5), and thus #4xx-II+Cl^(ON)=-1#; clearly we have #Cl(VII+)#. For #CrO_4^(2-)#, #4xx-II+Cr^"ON"=-2#, i.e. we gots #stackrel(VI+)"Cr"#. Anyway, read your text, and certainly seek clarification here.