What is the ground-state electronic configuration of the iron(III) ion, Fe^{3+}? (Atomic n… — JAMB UTME

What is the ground-state electronic configuration of the iron(III) ion, $Fe^{3+}$ ? (Atomic number of $Fe = 26$ )

The correct ground-state electronic configuration of the iron(III) ion is

[Ar] 3d^5

Neutral iron (

Z=26

) has the ground-state electron configuration

[Ar] 3d^6 4s^2

. When transition metal atoms form cations, electrons are removed first from the outermost principal energy level (

4s

) before removing electrons from the inner

3d

subshell. Ionization to

Fe^{3+}

involves removing the two

4s

electrons and one

3d

electron, resulting in a stable half-filled

d

-subshell configuration of

[Ar] 3d^5

Determine the electronic configuration of the neutral iron atom (

Fe

Z = 26

Neutral

Fe

has 26 electrons:

1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2

or abbreviated as

[Ar] 3d^6 4s^2

The

4s

orbital is filled before

3d

in neutral atoms according to the Aufbau principle.

Apply the cation ionization rule for transition metals to form

Fe^{3+}

Remove 3 electrons in total: first remove 2 electrons from the outermost

4s

orbital, then remove 1 electron from the

3d

orbital.

Electrons in the outermost shell (

n=4

) are lost before inner

(n-1)d

electrons during ionization.

Write the resulting electronic configuration for

Fe^{3+}

[Ar] 3d^5

Removing two

4s

electrons leaves

[Ar] 3d^6

, and removing one more

3d

electron yields

[Ar] 3d^5

Electronic configuration of transition metal cations

Question