In k4[fe (cn) 6], how many 3d electrons does the iron atom have?

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.