What are the metal oxidation states in Pb(NO_3)_2 and NaNO_3?

Sep 11, 2017

We gots $P b \left(I I +\right)$.....

Explanation:

Oxidation state is the charge left on the central atom, when all of the bonding pairs of electrons are broken, with the charge assigned to the most electronegative atom. For salts, such sodium nitrate, or lead nitrate, we break up the compound into its constituent ions, which are here easy to see:

$P b {\left(N {O}_{3}\right)}_{2} \rightarrow P {b}^{2 +} + 2 N {O}_{3}^{-}$

$N a N {O}_{3} \rightarrow N {a}^{+} + N {O}_{3}^{-}$

And thus here, clearly, we gots $P {b}^{2 +}$ and $N {a}^{+}$. What about for $F e {\left(N {O}_{3}\right)}_{3}$?

We can go farther than this, and drill down to the oxidation states of nitrogen, and oxygen in the nitrate ion; and oxygen is still marginally more electronegative than nitrogen. As always, the sum of the oxidation numbers is equal to the charge on the ion, here $- 1$. The oxidation number of oxygen is certainly $- I I$ in the oxides...and thus...

$N {O}_{3}^{-} : {N}_{\text{oxidation number}} + 3 \times \left(- 2\right) = - I$

${N}_{\text{oxidation number}} = - 1 + 6 = + V$

What are the nitrogen oxidation numbers in $N O$, $N {O}_{2}$, and ${N}_{2} {O}_{4}$?