Why is $Bi(V)$ a powerful oxidizing agent while $Bi(III)$ is exceptionally stable?

This phenomenon is caused by the Inert Pair Effect. In heavy elements like Bismuth ($Bi$, Period 6), the outer $6s^2$ electron pair experiences poor shielding by inner $4f$ and $5d$ subshells, drawing $6s$ electrons closer to the nucleus under effective nuclear charge $Z_{\text{eff}}$. Consequently, the $6s^2$ pair resists participating in bonding, rendering the $+3$ state far more stable than $+5$. Thus, $Bi(V)$ readily gains $2$ electrons to reduce to stable $Bi(III)$, acting as a potent oxidizing agent.

Why does $F_2$ have a lower bond dissociation energy than $Cl_2$?

The $F-F$ single bond length is exceptionally short ($\approx 142 \text{ pm}$) due to the small atomic size of Fluorine ($2p$). This proximity causes severe electrostatic repulsion between the non-bonding valence lone pairs on adjacent Fluorine atoms, weakening the $F-F$ covalent bond compared to $Cl_2$, where larger $3p$ orbitals significantly reduce lone-pair repulsions.

What is the best strategy to master the vast reaction mechanisms of the $p$-block for NEET and JEE?

Avoid memorizing isolated reactions individually. Classify reactions into structural patterns: (1) Redox disproportionation driven by oxidation state stability diagrams, (2) Hydrolysis steps dictated by central atom electrophilicity and vacant $d$-orbital availability, and (3) Thermal decomposition based on cation polarizing power and lattice energy stability.

p-Block Elements Notes, Formulas, Traps & PYQs | ExamiFly

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Solved High-Yield Sample Questions

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Question #1NEET 2024

Which of the following oxoacids of phosphorus contains a direct $P-P$ bond and features phosphorus in an oxidation state of $+4$ ?

Solution Locked

Step-by-Step Solution & Concept

Step 1

Calculate the average oxidation state of $P$ in $H_4P_2O_6$ : $4(+1) + 2(x) + 6(-2) = 0 \implies 2x - 8 = 0 \implies x = +4$ .

Step 2

Analyze the molecular structure of $H_4P_2O_6$ , which consists of two symmetric $PO(OH)_2$ units linked directly by a single $P-P$ covalent bond without a bridging oxygen atom.

Step 3

Compare with $H_4P_2O_7$ (which contains a $P-O-P$ linkage and $P$ in $+5$ oxidation state) to confirm that Hypophosphoric acid ( $H_4P_2O_6$ ) is the correct choice.

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Question #2JEE Main 2023

Complete hydrolysis of $XeF_6$ yields compound $X$ . Partial hydrolysis of $XeF_6$ with one equivalent of $H_2O$ yields compound $Y$ . What are the oxidation states of Xenon in $X$ and $Y$ , respectively?

Solution Locked

Step-by-Step Solution & Concept

Step 1

Write the chemical equation for complete hydrolysis: $XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$ . Here, compound $X$ is $XeO_3$ .

Step 2

Write the chemical equation for partial hydrolysis with $1$ equivalent of $H_2O$ : $XeF_6 + H_2O \rightarrow XeOF_4 + 2HF$ . Here, compound $Y$ is $XeOF_4$ .

Step 3

Calculate oxidation states: In $XeO_3$ , $x + 3(-2) = 0 \implies x = +6$ . In $XeOF_4$ , $y + (-2) + 4(-1) = 0 \implies y = +6$ . Thus, the oxidation state remains $+6$ in both compounds.

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Mastering p-Block Elements (Group 15, 16, 17 & 18): Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Thermal Stability of Group 15 Hydrides

Acidity of Halogen Oxoacids

Inert Pair Effect Oxidation Stability

Disproportionation Hydrolysis of Xenon Tetrafluoride

Group 16 Hydride Bond Angle Trend

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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