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 electron configuration 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.