Why would fluorine have a positive oxidation state in #"HOF"# even though it is more electronegative?

2 Answers

It is due to the oxidation numbers of hydrogen & oxygen.

Explanation:

Hydrogen has a #+1# oxidation number.
Oxygen generally has a #-2# oxidation number.

But, fluorine is a more electronegative element than that of oxygen.

Hence, the oxidation state of oxygen must be higher than that of fluorine in positive sense.

But in order to make the charge of the entire molecule #0#, though being more electro negative atom relative to oxygen, #F# has to get to an oxidation state of #+1#, while that should not be the case.

Hope it Helps:)

Oct 14, 2017

DISCLAIMER: CONTROVERSIAL!

As it turns out, the coordinates of #"HOF"# are given here:

#" "" "color(white)(...)x (Å)" "" "" "y(Å)" "" "color(white)(.)z(Å)#
#"O"" "color(white)(..)0.0529" "" "color(white)(.)0.7143" "color(white)(..)0.0000#
#"H"" "-0.8995" "color(white)(.....)0.8346" "color(white)(..)0.0000#
#"F"" "color(white)(..)0.0529" "color(white)(.)-0.7277" "color(white)(..)0.0000#

And since the dipole moment of #"HOF"# is

#vecmu = << mu_x, mu_y, mu_z >> = << 2.200, 0.370, 0 >>#,

an electron density map would then indicate that oxygen is slightly more negative than fluorine in this compound, while both oxygen and fluorine are significantly more negative than hydrogen.

So in that regard, apparently, an oxidation state of #+1# on fluorine is reasonable on the grounds of the strict oxidation rules of choosing oxygen as #-2# arbitrarily.

It makes no sense on electronegativity grounds, i.e. this is one case where blindly following IUPAC rules trumps electronegativity.


The oxidation state of fluorine in #"HOF"#, based on electronegativities, should be negative. Fluorine is the most electronegative non-noble gas, and thus obtains a negative oxidation state by hogging the electron density, hypothetically speaking...

Then, treating this as an acid (as anyone should!),

#"HOF"(aq) rightleftharpoons "H"^(+)(aq) + "OF"^(-)(aq)#,

and one would see that to practice conservation of charge AND electronegativity for #"OF"^(-)# anion,

  • #"F"# has a #-1# oxidation state.
  • #"O"# has a #0# oxidation state.

On the other hand, since #"Cl"# is less electronegative than #"O"#, in #"ClO"^(-)# one would instinctively say

  • #"O"# has a #-2# oxidation state.
  • #"Cl"# has a #+1# oxidation state.

However, the fluorine oxidation state of #-1# doesn't agree with the CCCBDB reference above.

In the end, oxidation states are just an accounting scheme. They may or may not mean anything physical. Relying on electronegativity especially falls apart for the fluorine oxoacids, and one cannot rely on oxidation state assignments, using electronegativities as a guide, as indicative of true partial charges.

[In fact, there are no partial charges according to oxidation state schemes! That falls on dipole moment analysis!]

See here for a further discussion...