What is the oxidation number with respect to nitrogen in nitric acid, #HNO_3#?

2 Answers
May 24, 2017

For nearly every compound (except superoxides such as #"KO"_2#), oxygen will always have an oxidation state of #2-#, so we can eliminate options (a) and (d) .

In most all cases, #"H"# will have an oxidtaion state of #1+# when bonded to nonmetals, and #1-# when bonded with metals (such as alkali hydrides). Since all these elements are indeed nonmetals, the #"H"# will have a #1+# oxidation state, so that eliminates option (b).

Our only option left is (c) (or none of these options), let's take a look:

#"HNO"_3# is a neutral compound, as it has no net ionic charge, so the sum of the oxidation states will be zero. We know that #"O"# has an oxidation state of #2-#, and that of #"H"# is #1+#. If we multiply these by how many of that atom we have in the compound, the number left is

#3(2-) + 1(1+) = -5#

Therefore, the only element remaining, #"N"#, must equalize this negative value by having an oxidation state of #5+#, which means option (c) is indeed correct.

May 24, 2017

Answer:

Well, the weighted sum of the oxidation numbers ALWAYS equals the charge on the starting species...........

Explanation:

Oxygen is typically #stackrel(-II)O#, and it is so here........Hydrogen is typically #stackrel(+I)H#, and so..........

#N_"oxidation number"+1+3xx(-2)=0#,

#N_"oxidation number"-5=0#,

#N_"oxidation number"=+5#, i.e. #N(V)#.

And we remember our definition of oxidation numbers. Oxidation number is the charge left on the atom of interest when all the bonding pairs of electrons are broken, with the charge going to the MOST electronegative atom.

What are the oxidation numbers of #N# in #N_2O_5#, #NO_2#, and #NH_3#?