Electronic configuration of d and f Block Elements

d and f Block Elements

Electronic configuration

The general electronic configuration for first and second transition series can be written as ns²(n − 1)d^x

• First = [Ar]4s²3d^x; second = [Kr]5s²4d^x

For third and fourth transition series, the general electronic configuration isns²(n − 2)f¹⁴(n − 1)d^x

Some elements deviate from the general pattern by “promoting” one or more s electrons into the d orbital to complete the subshell. Because of sublevel splitting, the 4s sublevel is lower in energy than the 3d; therefore, the 4s fills before the 3d. However, the difference in energy is not large.The electron configuration of d block elements across the 1^st row of the d block (Sc to Zn) is given in the table below. Each element in this row has 1 more electron and 1 more proton where each “additional” electron enters the 3d sub-shell and the electronic configuration of argon is Ar: 1s²2s²2p⁶3s²3p⁶

Element	Core electrons	3d electrons	4s electrons
Sc	[Ar]	3d1	4s2
Ti	[Ar]	3d2	4s2
V	[Ar]	3d3	4s2
Cr	[Ar]	3d5	4s1
Mn	[Ar]	3d5	4s2
Fe	[Ar]	3d6	4s2
Co	[Ar]	3d7	4s2
Ni	[Ar]	3d8	4s2
Cu	[Ar]	3d10	4s1
Zn	[Ar]	3d10	4s2

In the electronic configuration, the 4s orbital is filled before the 3d orbitals. This is because before filling orbitals, 4s orbitals have lower energy than 3d orbitals. The reversed order of the 3d and 4s orbitals only applies to build the atom up in the first place.

Scandium has the electronic structure [Ar] 3d¹4s². In ionic form, it loses the 3 outer electrons to make an argon structure. However, the Sc³⁺ ion does not have d electrons and so doesn’t meet the definition of transition metal. Similarly, Zinc has the electronic structure [Ar] 3d¹⁰4s². Loss of two 4s electrons gives a 2+ ion with the electronic structure [Ar] 3d¹⁰. This zinc ion has full d orbitals and doesn’t meet the definition of transition metals either.

Copper, [Ar] 3d¹⁰4s¹,can form two ions. The Cu⁺ ion: [Ar] 3d¹⁰ while Cu²⁺ ion :[Ar] 3d⁹, this means that Copper is definitely a transition metal because the Cu²⁺ ion has an incomplete d orbitals. Cr and Cu are electron configuration exceptions because they prefer half-filled and filled d-orbitals for stability. At Cr the orbital energies such that putting one e- into each 3d and 4s orbital gives lower energy than having 2 e- in the 4s orbital. At Cu, putting 2 e- into the 4s orbital would give higher energy than filling the 3d orbitals

To write the electronic structure for Co²⁺:

The 2+ ion is formed by loosing two 4s electrons.

To write the electronic structure for V³⁺:

The 4s electrons are lost first and then 3d electrons are lost.

Some of the transition metals have irregular electron configurations in which the ns only partially fills before the (n − 1)d or doesn’t fill at all. Therefore, their electron configuration must be found experimentally.

The transition elements show more than one oxidation state and these states are related to patterns in successive ionization energies. The first-row d-block elements form ions as they lose the 4s electrons first in order to make the 2+ ions. To make ions of oxidation state higher than 2+, they start losing the 3d electrons. Because the 3d and 4s orbitals are close in energy, the electrons can be removed without a huge jump in energy as you would see from the s and p orbitals.

Consider the 2 examples:

Ca: 1s²2s²2p⁶3s²3p⁶4s²

Ti: 1s²2s²2p⁶3s²3p⁶3d²4s²

Calcium will lose the 4s² electrons and then it would take a huge amount of energy to pull off the electrons in the 3p orbital. Titanium will lose the 4s² electrons to make Ti⁺², then one of the 3d electrons to make Ti⁺³, then the other 3d electron to make Ti⁺⁴. It refrains from making a +5 ion because it takes too much energy to pull off the p electrons.