What is the ground-state electronic configuration of the iron(II) ion, Fe^{2+}? (Atomic nu… — JAMB UTME

What is the ground-state electronic configuration of the iron(II) ion, $Fe^{2+}$ ? (Atomic number of $Fe = 26$ )

The ground-state electronic configuration of

Fe^{2+}

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

(or

[Ar] 3d^6

The correct configuration is

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

. Neutral iron has 26 electrons with the configuration

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

. Upon forming the

Fe^{2+}

ion, two electrons are removed from the outermost shell (

n=4

, the

4s

orbital), leaving 24 electrons and a

3d^6

valence structure.

Determine the electronic configuration of the neutral iron atom (

Fe

Z = 26

The neutral configuration is

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

Electrons fill subshells in order of increasing energy according to the Aufbau principle.

Determine which electrons are lost when forming the

Fe^{2+}

cation.

Two electrons are removed from the principal energy level with the highest quantum number (

n = 4

), which is the

4s

subshell.

Transition metals lose their outermost

4s

electrons first upon ionization because

4s

electrons experience greater shielding and are higher in energy once subshells are populated.

Write the resulting configuration for

Fe^{2+}

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

Removing two electrons from

4s^2

leaves an empty

4s

orbital and retains six electrons in the

3d

subshell.

Cation formation in d-block transition elements requires losing

ns

electrons before

(n-1)d

electrons.

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