Question

Oxidation number of N in $$N{H}_{4}N{O}_3$$is:
A.-3
B.+5
C.-3 and +5
D.+3 and -5

Answer 1

Hint: To find oxidation number of N in $$N{H}_{4}N{O}_3$$, we have to break it into ion and find the oxidation number of N of both ions and it is going to be different for both N. We have to break them into ion because $$N{H}_{4}N{O}_3$$is an ionic compound.

Complete answer:

When we break this into ions, we will see $$N{H}_4^{+}$$ and $$N{0}_3^{-}$$.
For $$N{H}_4^{+}$$ oxidation state = +1
N+4H=+1
N+4(+1) =+1 …. (Oxidation number of H =+1)
N+4=+1
N=-3
For $$N{0}_3^{-}$$ oxidation state = -1
N+3(O)=-1
N+3(-2) =-1 … (Oxidation number for O=-2)
N-6=-1
N=-1+6
N=+5
Therefore, Oxidation state of N in $$N{H}_4^{+}$$is -3 and in $$N{0}_3^{-}$$ is +5.

Note: As for$$N{H}_4^{+}$$ ion, N nominally gains one electron from three of the four hydrogen atoms it is bound to; the fourth hydrogen, that nominally is an H+ ion, is bound by nitrogen by means of its lone pair (dative bond), but, owing to the higher electronegativity of N with respect to H, this pair of electrons is closer to N, so that, as regards this dative bond, there is no gain or loss of electrons. Therefore, as a whole, N nominally gains 3 electrons, hence its oxidation number -3. Of course, in the real $$N{H}_4^{+}$$ ion, the positive charge is evenly distributed over all of the four hydrogen atoms, owing to resonance.
As for$$N{0}_3^{-}$$ion, one of the three oxygen atoms nominally have a -1 charge (one electron more than neutrality), so that it has only one unpaired electron. The N atom employs one of its three unpaired electrons to bind that “charged” oxygen atom, and the other two unpaired electrons to bind a neutral oxygen atom. As for the third, neutral, oxygen atom, nitrogen binds it by means of its lone pair (dative bond), having the oxygen atom brought its two formerly unpaired electrons in the same orbital, thus providing an empty orbital to accommodate the lone pair of the nitrogen. Since oxygen is more electronegative than nitrogen, the latter has nominally lost 5 electrons (the previously unpaired three, and the lone pair), hence its +5 oxidation number. Of course, in the real $$N{0}_3^{-}$$ion, the three oxygen atoms are perfectly identical, as regards charge and kind of bond, owing to resonance.