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

\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)?

• iron (II) -> iron (III); sodium oxides are not really very stable; not comparable reaction conditions. I am quite certain, that you are not looking for reaction mechanisms in any of your recent questions. A reaction mechanism is a detailed step-by-step description in terms of elementary reactions (also including transition states). May 26, 2016 at 8:22
• @Martin-マーチン - then I'm not sure which tag to pick. I'm looking for something to help me memorize, something logically explaining the differences. I wonder why can't we have Fe(NO2)2, for instance. My latest questions are from USE Exam Sample Question No.37, considered one of the hardest in the exam. The question contains a textual description of a chain of reactions which you should write down, guessing the products. I've already memorized some interesting reactions, like SO2 + H2S (occurs in volcanoes) and am plowing further. May 26, 2016 at 8:49
• For the thermal decomposition of sodium nitrate see Eli S. Freeman, J. Phys. Chem. 1956, 60 (11), 1487–1493. May 26, 2016 at 11:13

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.
• Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I disagree that $\mathrm{[Ar]\,3d^5\,4s^1}$ has a lower symmetry than $\mathrm{[Ar]\,3d^5}$. May 26, 2016 at 11:11