How many d electrons are there in the Fe+3 ion

Please explain how to find the number of d electrons.

Well, the d electrons start being added in the d-block (the bit with no groups). So just count along from the beginning of the d-block. Fe on its own has 6 d electrons. Fe3+ would lose the outermost s electrons (4s electrons in this case) first, and then begin to lose the d electrons. There are two 4s electrons in Fe, as it is to the right of Group 2 and within Period 4, so it loses those first, and then loses its d electrons. It has lost 3 electrons, so 1 d electron is lost. Therefore, as Fe has 6, Fe3+ has 5.

For reference, the electron configuration of Fe is: 1s2 2s2 2p6 3s2 3p6 3d6 4s2 or Ar 3d6 4s2

Fe is Group 8. It has 8 valence e⁻s: [Ar] 3d^6 4s^2 The 4s^2 e⁻s are lost first (Fe^2+) and then one of the d e⁻s to give the 3+ ion. So Fe^3+ is [Ar] 3d^5, that is, five d e⁻s. And since there are five d AOs and Hund's Rule is obeyed the configuration is (↑) (↑)(↑)(↑)(↑).