What is the oxidation number of O in the ion OH-?

2 Answers
Apr 20, 2018

Answer:

We got #O^(-II)#

Explanation:

The oxidation number is a fictitious charge of atoms involved in chemical bonds that reflects electronegativity. The bond, the two electrons binding an element or ion are broken, with the charge, the electron assigned to the most electronegative atom. The sum of the oxidation numbers is equal to the charge on the ion or the molecule, and of course this charge might be ZERO...

Here we gots....#H-O^(-)#...this is split to give #H^+# and #O^(2-)#...and thus we get oxidation numbers of #O(-II)# and #H(+I)#....of course the sum of the oxidation equals the charge on the ion. Does it?

Here are some rules taken from a prior answer...they are included as reference. I do not propose that you learn them verbatim...

#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."#

Apr 20, 2018

Answer:

#-2#

Explanation:

In a hydroxide #(OH^-)# ion, we see that the total charge of the ion is #-1#. That means, the total sum of the oxidation numbers of the elements present in the ion totals out to be #-1#.

The ion has one oxygen atom and one hydrogen atom. Hydrogen is less electronegative than oxygen, and so will possess its usual #+1# state.

Let #x# be the oxidation number of oxygen. Then we have,

#x+1=-1#

#x=-1-1#

#=-2#

So, oxygen will have an oxidation number of #-2# in this ion, and this is also its preferred oxidation state.