What are oxidation numbers?

1 Answer
Aug 2, 2017

Answer:

This is formally the charge left on a central atom when all the bonding pairs of electrons are REMOVED, with the charge assigned to most electronegative atom......

Explanation:

The rules for assignment of oxidation numbers are given here.

As given, #"oxidation number"# and #"oxidation state"# are FORMALISMS, they are convenient fictions that may nevertheless have a practical use. And when we assign oxidation numbers, we use them to solve, i.e. balance, a chemical equation, with respect to mass and charge, #"redox equations"#, straightforwardly and systematically (says he, #"modelling a modern major general"#.)

And so we can interpret a given chemical reaction with the use of oxidation numbers. Consider the oxidation of ammonia to give nitrate ion..........in terms of formal oxidation state this is the transition, #stackrel(-III)N# to #stackrel(+V)N#, an 8 electron oxidation, which we formally represent in the equation.......

#NH_3(aq) +3H_2O rarr NO_3^(-) +9H^(+) + 8e^(-)# #(i)#

Is this balanced with respect to mass and charge? It must be if we purport to represent physical reality.

And, inevitably, something must be reduced to effect the oxidation; let's say it is oxygen.

#stackrel(0)O_2+4e^(-) rarr2O^(2-)# #(ii)#

We add the individual redox equations together in a way to eliminate the electrons, which are particles of convenience......And so we take #(i) + 2xx(ii)#:

#NH_3(aq) +3H_2O +2O_2+8e^(-)rarr NO_3^(-) +underbrace(9H^(+) +4O^(2-))_(4H_2O+H^+) + 8e^(-)#

And so we cancel out what we can.....

#NH_3 +cancel(3H_2O) +2O_2+cancel(8e^(-))rarr NO_3^(-) +underbrace(9H^(+) +4O^(2-))_(cancel(4)H_2O+H^+) + cancel(8e^(-))#

....to give finally.........

#NH_3 +2O_2rarr NO_3^(-) +H_2O+H^+#

....or........

#stackrel(-III)NH_3 +2stackrel(0)O_2rarr Hstackrel(+V)NO_3 +H_2stackrel(-II)O#

I acknowledge that is a lot of pfaff.......but your question was rather open-ended.