Why thermal decomposition of iron(II) nitrate and sodium nitrate yields different products?

Question

\begin{align} \ce{4Fe(NO3)2 &-> 2Fe2O3 + 8NO2 + O2}\tag{1} \\ \ce{2NaNO3 &-> 2NaNO2 + O2}\tag{2} \end{align}

Why do nitrates of different metals decompose in different ways?

Why do we get a nitrite with sodium, but an oxide with iron(II)?

Answer 1

Iron(II) cation is moderately reducing- an iron(II) cation has a valence electron configuration of $\mathrm{3d^5 4s^1}$, and by giving up an electron in its $\mathrm{4s}$ orbital the cation will transform into an iron(III) cation with $\mathrm{3d^5}$ valence electron configuration, which, with every $\mathrm{d}$ orbital half filled, has a higher degree of symmetry over $\mathrm{3d^5 4s^1}$ electron configuration and hence is more stable (thermodynamically favored).

Sodium(I) cation, on the contrary, has a highly stable valence electron configuration of $\mathrm{2s^2 2p^6}$, with all its $\mathrm{2s}$ and $\mathrm{2p}$ orbitals filled, identical to the electron configuration of the chemically inert noble gas neon, and either removing or adding one electron to the sodium(I) ion will lead to an electron configuration much less stable and hence thermodynamically unfavored.

You may notice that in the decomposition reaction, the iron(II) cation is oxidized into iron (III), which is favored, and the sodium cation remain intact. Put it in another way, we may imagine a first step thermal decomposition reaction of iron(II) nitrate resembling decomposition of sodium nitrate and yields iron(II) nitrite and oxygen gas as products. Then we easily notice that iron(II) nitrite is reducing, and oxygen gas is a strongly oxidant, hence the two products of the first step decomposition can further react with each other to produce iron(III) oxide and nitrogen dioxide.