krit.club logo

d and f Block Elements - General introduction and electronic configuration

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Transition elements are defined as elements which have incompletely filled dd-orbitals in their ground state or in any of their common oxidation states.

•

The general electronic configuration of dd-block elements is (n−1)d1−10ns1−2(n-1)d^{1-10} ns^{1-2}.

•

ZnZn, CdCd, and HgHg of group 12 have a general configuration of (n−1)d10ns2(n-1)d^{10} ns^2. Since their dd-orbitals are completely filled in their ground state as well as in their common oxidation states, they are not strictly regarded as transition elements.

•

There are exceptions in electronic configurations due to the extra stability of half-filled and fully-filled orbitals. For example, Chromium (Z=24Z=24) is [Ar]3d54s1[Ar] 3d^5 4s^1 and Copper (Z=29Z=29) is [Ar]3d104s1[Ar] 3d^{10} 4s^1.

•

The ff-block elements (Inner Transition Elements) consist of two series: Lanthanoids (4f4f series) and Actinoids (5f5f series).

•

The general electronic configuration for ff-block elements is (n−2)f1−14(n−1)d0−1ns2(n-2)f^{1-14} (n-1)d^{0-1} ns^2.

•

In transition elements, the nsns electrons are lost first during ionization, followed by (n−1)d(n-1)d electrons.

📐Formulae

(n−1)d1−10ns1−2(n-1)d^{1-10} ns^{1-2}

(n−2)f1−14(n−1)d0−1ns2(n-2)f^{1-14} (n-1)d^{0-1} ns^2

Cr (Z=24):[Ar]3d54s1\text{Cr (Z=24)}: [Ar] 3d^5 4s^1

Cu (Z=29):[Ar]3d104s1\text{Cu (Z=29)}: [Ar] 3d^{10} 4s^1

Gd (Z=64):[Xe]4f75d16s2\text{Gd (Z=64)}: [Xe] 4f^7 5d^1 6s^2

💡Examples

Problem 1:

Write the electronic configuration of Fe3+Fe^{3+} (ZZ for Fe=26Fe = 26).

Solution:

[Ar]3d5[Ar] 3d^5

Explanation:

The ground state configuration of FeFe is [Ar]3d64s2[Ar] 3d^6 4s^2. To form Fe3+Fe^{3+}, three electrons are removed. Electrons are removed from the 4s4s orbital first, then from the 3d3d orbital: [Ar]3d64s2→[Ar]3d5[Ar] 3d^6 4s^2 \rightarrow [Ar] 3d^5.

Problem 2:

Why does Gadolinium (Z=64Z=64) have a configuration of [Xe]4f75d16s2[Xe] 4f^7 5d^1 6s^2 instead of [Xe]4f86s2[Xe] 4f^8 6s^2?

Solution:

Due to the extra stability of the half-filled ff-subshell (4f74f^7).

Explanation:

The 4f4f and 5d5d orbitals are very close in energy. By shifting one electron to the 5d5d orbital, the 4f4f subshell becomes exactly half-filled (4f74f^7), which provides extra exchange energy and symmetry, making the atom more stable.

Problem 3:

Explain why Sc3+Sc^{3+} is colorless while Ti3+Ti^{3+} is colored in aqueous solution.

Solution:

Sc3+Sc^{3+} has 3d03d^0 configuration, while Ti3+Ti^{3+} has 3d13d^1 configuration.

Explanation:

Color in transition metal ions usually arises from d−dd-d transitions. Sc3+Sc^{3+} ([Ar]3d0[Ar] 3d^0) has no electrons in the dd-orbital to undergo transition. Ti3+Ti^{3+} ([Ar]3d1[Ar] 3d^1) has one unpaired electron which can be excited, resulting in the absorption of specific wavelengths of light.

General introduction and electronic configuration Class 12 Notes & Examples