First let's find out the oxidation number of Fe in K₄[Fe(CN)₆] compound.
The oxidation number of cation, K is +1. Hence, the total charge of the anion, [Fe(CN)₆] is -4. CN has charge has -1. There are 6 CN in anion. Let's assume the oxidation number of Fe is 'a'.
Sum of the oxidation numbers of each element = Charge of the compound a + 6 x (-1) = -4 a -6 = -4 a = +2
Hence, oxidation number of Fe in [Fe(CN)₆]⁴⁻ is +2.
Now Fe has the atomic number as 26. Hence, number of electrons in Fe at ground state is 26. Electron configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² = [Ar] 3d⁶ 4s²
When making Fe²⁺, Fe releases 2 electrons. Hence, the number of electrons in Fe²⁺ is 26 - 2 = 24. Hence, the electronconfiguration of Fe²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ = [Ar] 3d⁶
Hence, the number of 3d electrons of Fe in K₄[Fe(CN)₆] compound is 6.
Four molecules have carbon as the central atom and hydrogen atoms around it. The shape that would give rise to a polar molecule is a bent shape. <span>The bent shape will not be symmetrical which will result to being polar. Hope this answers the question.</span>
