The confirmatory test for the ion involves its reaction with potassium hexacyanidoferrate(II), yield — Qualitative and Quantitative Analysis Chemistry Question
Question
The confirmatory test for the $Fe^{3+}$ ion involves its reaction with potassium hexacyanidoferrate(II), yielding a characteristic intense Prussian blue precipitate. The chemical formula for Prussian blue is $Fe_4[Fe(CN)_6]_3$. What is the formal oxidation state of the iron atom that is located OUTSIDE the square brackets (the coordination sphere) in this formula?
💡 Solution & Explanation
The formula for Prussian blue is $Fe_4[Fe(CN)_6]_3$. The complex ion inside the brackets is the hexacyanidoferrate(II) ion, $[Fe(CN)_6]^{4-}$, where the inner iron is in a $+2$ oxidation state. To balance the $3 \times (-4) = -12$ charge from the three complex anions, the four iron atoms outside the brackets must contribute a total charge of $+12$. Therefore, the oxidation state of each iron atom outside the coordination sphere is $+12 / 4 = +3$.