Why do $Zr$ ($4d$) and $Hf$ ($5d$) have almost identical atomic radii ($160 \text{ pm}$ and $159 \text{ pm}$)?

This occurs due to **Lanthanoid Contraction**. Before $Hf$ ($Z=72$), the $4f$ subshell is completely filled with $14$ electrons. Since $f$-orbitals have highly diffused shapes, their mutual shielding power is extremely poor. Consequently, the nuclear charge increases by $+14$ units with negligible increase in shielding, pulling the outer electrons inward and neutralizing the expected shell expansion from $4d$ to $5d$.

Why is $Cu^+$ diamagnetic and unstable in aqueous solution, whereas $Cu^{2+}$ is paramagnetic and stable?

$Cu^+$ has a fully filled $d^{10}$ configuration ($n=0$, diamagnetic), while $Cu^{2+}$ is $d^9$ ($n=1$, paramagnetic). In aqueous solution, $Cu^+$ undergoes disproportionation ($2Cu^+_{(aq)} \rightarrow Cu^{2+}_{(aq)} + Cu_{(s)}$) because the hydration energy released by the smaller, highly charged $Cu^{2+}$ ion is much more negative than that of $Cu^+$, easily overriding the second ionization enthalpy of copper.

How should I memorize all the specific redox reactions of $KMnO_4$ and $K_2Cr_2O_7$ for NEET and JEE?

Do not memorize individual reactions mechanically. Focus on the core half-cell reduction reactions ($MnO_4^- \rightarrow Mn^{2+}$ in acid; $Cr_2O_7^{2-} \rightarrow 2Cr^{3+}$ in acid) and learn the standard oxidation products of common reductants ($I^- \rightarrow I_2$, $Fe^{2+} \rightarrow Fe^{3+}$, $S^{2-} \rightarrow S$, $C_2O_4^{2-} \rightarrow CO_2$). Use the ion-electron method to balance them on the fly.

d- and f-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 transition metal ions exhibits the highest spin-only magnetic moment value?

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Step-by-Step Solution & Concept

Step 1

Write down electronic configurations for each gaseous ion: $Fe^{3+}$ is $[Ar] 3d^5$ , $Cr^{3+}$ is $[Ar] 3d^3$ , $Mn^{3+}$ is $[Ar] 3d^4$ , and $Co^{3+}$ is $[Ar] 3d^6$ .

Step 2

Determine number of unpaired electrons ( $n$ ): For $Fe^{3+}$ ( $3d^5$ ), $n = 5$ ; for $Cr^{3+}$ ( $3d^3$ ), $n = 3$ ; for $Mn^{3+}$ ( $3d^4$ ), $n = 4$ ; for high-spin $Co^{3+}$ ( $3d^6$ ), $n = 4$ .

Step 3

Calculate magnetic moment using $\mu = \sqrt{n(n+2)} \text{ B.M.}$ . For $Fe^{3+}$ ( $n=5$ ), $\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92 \text{ B.M.}$ , which is the maximum achievable in the $3d$ series.

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Question #2JEE Main 2023

When acidic $KMnO_4$ solution reacts with oxalic acid ( $(COOH)_2$ ), the change in oxidation state of manganese and the total number of electrons involved per mole of $KMnO_4$ are respectively:

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Step-by-Step Solution & Concept

Step 1

Write the balanced ionic half-reaction for the reduction of permanganate ion in acidic medium: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ .

Step 2

Calculate the oxidation state change of $Mn$ : In $MnO_4^-$ , $x + 4(-2) = -1 \implies x = +7$ . In product $Mn^{2+}$ , the oxidation state is $+2$ .

Step 3

Conclude that the transformation is $+7 \rightarrow +2$ , requiring $5$ electrons transferred per mole of $KMnO_4$ .

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Mastering d- and f-Block Elements: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Spin-Only Magnetic Moment Formula

Effective Nuclear Charge ($Z_{eff}$)

Standard Redox Free Energy Relation

Equivalent Weight in Redox Titrations ($KMnO_4$)

Lanthanoid Contraction Radius Relation

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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