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The Periodic Table - Transition elements

Grade 12IGCSEChemistry

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

🔑Concepts

A transition element is defined as a d-block element that forms at least one stable ion with a partially filled dd subshell. This explains why ScSc (forms Sc3+Sc^{3+} with 3d03d^0) and ZnZn (forms Zn2+Zn^{2+} with 3d103d^{10}) are sometimes excluded from the strict definition.

Physical properties include high melting points, high densities, and high tensile strength compared to s-block metals due to the involvement of 3d3d and 4s4s electrons in metallic bonding.

Variable oxidation states occur because the energy levels of the 4s4s and 3d3d subshells are very close, allowing varying numbers of electrons to be lost. For example, Manganese (MnMn) can exhibit oxidation states from +2+2 to +7+7.

Complex ions consist of a central metal cation bonded to ligands via coordinate (dative covalent) bonds. Ligands like H2OH_2O, NH3NH_3, and CNCN^- possess at least one lone pair of electrons.

The characteristic colors of transition metal complexes arise from the splitting of dd orbitals into two non-degenerate energy levels (t2gt_{2g} and ege_g). Electrons absorb specific frequencies of visible light to jump from lower to higher energy levels (dd-dd transitions).

Transition metals act as catalysts because they can provide a surface for adsorption (heterogeneous) or change oxidation states to facilitate electron transfer (homogeneous). Examples include FeFe in the Haber Process and V2O5V_2O_5 in the Contact Process.

📐Formulae

ΔE=hf=hcλ\Delta E = h f = \frac{h c}{\lambda}

[Ar]3dn4s2 (General configuration for Period 4 transition metals)[Ar] 3d^n 4s^2 \text{ (General configuration for Period 4 transition metals)}

Cr:[Ar]3d54s1 (Anomaly due to half-filled stability)Cr: [Ar] 3d^5 4s^1 \text{ (Anomaly due to half-filled stability)}

Cu:[Ar]3d104s1 (Anomaly due to fully-filled stability)Cu: [Ar] 3d^{10} 4s^1 \text{ (Anomaly due to fully-filled stability)}

Kstab=[Complex][Metal Ion][Ligand]nK_{stab} = \frac{[Complex]}{[Metal\ Ion][Ligand]^n}

💡Examples

Problem 1:

Determine the electron configuration of the Fe3+Fe^{3+} ion (Z=26Z=26 for Iron).

Solution:

1s22s22p63s23p63d51s^2 2s^2 2p^6 3s^2 3p^6 3d^5 or [Ar]3d5[Ar] 3d^5

Explanation:

Neutral FeFe is [Ar]3d64s2[Ar] 3d^6 4s^2. When forming ions, electrons are lost from the 4s4s subshell first, then the 3d3d subshell. Losing three electrons removes two from 4s4s and one from 3d3d.

Problem 2:

Explain why [Cu(H2O)6]2+[Cu(H_2O)_6]^{2+} is blue while [Zn(H2O)6]2+[Zn(H_2O)_6]^{2+} is colorless.

Solution:

Cu2+Cu^{2+} has a 3d93d^9 configuration, while Zn2+Zn^{2+} has a 3d103d^{10} configuration.

Explanation:

In Cu2+Cu^{2+}, there is a vacancy in the upper dd orbitals, allowing for dd-dd transitions where an electron absorbs yellow-red light to jump levels, leaving blue as the complementary color. In Zn2+Zn^{2+}, the dd subshell is full, so no dd-dd transition is possible.

Problem 3:

Calculate the oxidation state of Manganese in the permanganate ion MnO4MnO_4^-.

Solution:

x+4(2)=1x=+7x + 4(-2) = -1 \Rightarrow x = +7

Explanation:

Let xx be the oxidation state of MnMn. Oxygen usually has an oxidation state of 2-2. The sum of oxidation states must equal the overall charge of the ion (1-1).

Transition elements - Revision Notes & Key Formulas | IGCSE Grade 12 Chemistry