Summary of Key Chemistry Formulae: d- and f-Block Elements

1. Stoichiometry and Chemical Reactions

1.1 Oxidation States and Stoichiometry

Formulae:

• Oxidation State Calculation: $\text{Oxidation state}=(\text{Number of bonds to more electronegative atoms})-(\text{Number of bonds to less electronegative atoms})$

Explanation: Oxidation states in d- and f-block elements can vary widely due to the involvement of d-orbitals. For example, transition metals such as manganese can exhibit oxidation states ranging from +2 to +7.

Common Mistake: Students often confuse oxidation states with charges on the ion. Remember, oxidation states refer to the hypothetical charge an atom would have if all bonds were ionic.

Example: Manganese in potassium permanganate ($KMn{O}_4$) has an oxidation state of +7.

1.2 Balancing Redox Reactions

Formulae:

• Nernst Equation for Redox Potential: $E=E^{\circ }-\frac{0.0591}{n}\log \left(\frac{[\text{oxidized form}]}{[\text{reduced form}]}\right)$

Explanation: The Nernst equation helps in calculating the cell potential under non-standard conditions. It is crucial for understanding the redox behavior of transition metals.

Example Application: Calculate the cell potential for a reaction where $E^{\circ }=1.10,V$, the concentration of the oxidized form is 1 M, and the reduced form is 0.1 M.

2. Thermodynamics

2.1 Enthalpy of Atomisation

Formulae:

• Enthalpy of Atomisation: $\Delta H_{\text{atomisation}}=\sum \Delta H_{\text{bond dissociation}}$

Explanation: Enthalpy of atomisation is crucial for understanding the strength of metallic bonds in transition metals. Higher enthalpy indicates stronger metallic bonds.

Example: Calculate the enthalpy of atomisation for iron, knowing its bond dissociation enthalpy is high due to its strong metallic bonding.

Common Mistake: Forgetting to consider all bond dissociations when calculating enthalpy of atomisation can lead to incorrect results.

2.2 Standard Electrode Potentials

Formulae:

• Standard Electrode Potential: $E^{\circ }=-\frac{\Delta G^{\circ }}{nF}$

Explanation: This formula relates the Gibbs free energy change to the standard electrode potential, which is key in understanding the feasibility of redox reactions.

Example Application: Calculate the standard electrode potential for the reduction of ${\text{Fe}}^{3+}$ to ${\text{Fe}}^{2+}$.

NEET Tip: Focus on the trends in electrode potentials across the d-block elements, as they are often tested in NEET.

3. Physical Chemistry

3.1 Magnetic Properties

Formulae:

• Magnetic Moment: $\mu =\sqrt{n(n+2)},\text{BM}$ Where $n$ is the number of unpaired electrons.

Explanation: Magnetic properties of transition elements are due to unpaired electrons in d-orbitals. The magnetic moment provides insight into the number of unpaired electrons, which is useful in determining the electronic configuration of ions.

Example: Calculate the magnetic moment for a ${\text{Fe}}^{3+}$ ion with 5 unpaired electrons.

Common Mistake: Ignoring the orbital contribution to the magnetic moment, especially for elements in the lower rows of the d-block.

Quick Recap:

• Transition metals exhibit a variety of oxidation states due to the involvement of d-orbitals.
• The Nernst equation is essential for calculating cell potentials in redox reactions.
• Enthalpy of atomisation reflects the strength of metallic bonds, which are stronger in transition metals.
• Standard electrode potentials are linked to Gibbs free energy changes and help predict redox reaction feasibility.
• Magnetic properties are determined by the number of unpaired electrons and are calculated using the magnetic moment formula.

Practice Questions:

1. Calculate the oxidation state of manganese in ${\text{MnO}}_4^{-}$.
2. Balance the redox reaction: ${\text{Cr}}_2{\text{O}}_7^{2-}+{\text{Fe}}^{2+}\to {\text{Cr}}^{3+}+{\text{Fe}}^{3+}$.
3. Calculate the enthalpy of atomisation for nickel given bond dissociation energies.
4. Determine the magnetic moment for a ${\text{Co}}^{2+}$ ion.
5. Use the Nernst equation to find the potential for a cell with $[{\text{Cu}}^{2+}]=0.01M$ and $[\text{Cu}]=1M$.

This summary provides a comprehensive guide to key formulae and concepts in the chapter "d- and f-Block Elements" from your NCERT textbook, focusing on stoichiometry, thermodynamics, and physical chemistry relevant for NEET preparation.