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 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.

The reaction will shift to the right