Why is $\text{Tl}^{3+}$ a strong oxidizing agent while $\text{Tl}^+$ is stable?

Due to the inert pair effect, poor shielding by inner $4f$ and $5d$ electrons leaves the valence $6s^2$ electrons tightly bound to the nucleus. Consequently, $\text{Tl}^+$ is far more stable than $\text{Tl}^{3+}$, prompting $\text{Tl}^{3+}$ to readily accept $2e^-$ and act as a potent oxidizing agent.

Why does $\text{BF}_3$ act as a Lewis acid despite experiencing $p\pi-p\pi$ back-bonding?

Although $p\pi-p\pi$ back-donation from Fluorine to Boron partially relieves electron deficiency, Boron still retains an incomplete octet ($6$ valence electrons in monomeric $\text{BF}_3$). Thus, it remains electron-deficient and acts as a Lewis acid by accepting lone pairs from donors like $\text{NH}_3$.

What is the shortcut formula to identify shared oxygens per tetrahedron in silicates?

Use the general formula $(\text{Si}_n\text{O}_{4n - \frac{k}{2}n})^{(4n-kn)-}$, where $k$ is the number of shared oxygens per tetrahedron. For Orthosilicates $k=0$ ($[\text{SiO}_4]^{4-}$), Pyrosilicates $k=1$ ($[\text{Si}_2\text{O}_7]^{6-}$), Cyclic/Chain $k=2$ ($[\text{SiO}_3]^{2-}$), and Sheet silicates $k=3$ ($[\text{Si}_2\text{O}_5]^{2-}$).

p-Block Elements (Group 13 & 14) Notes, Formulas, Traps & PYQs | ExamiFly

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p-Block Elements (Group 13 & 14) Notes, Formulas, Traps & PYQs | ExamiFly | Examifly

Key Formulas & High-Yield Equations

Lewis Acidic Strength of Boron Halides

$\text{BF}_3 < \text{BCl}_3 < \text{BBr}_3 < \text{BI}_3$

Governed by

p\pi-p\pi

back-bonding efficiency. Valid strictly for monomeric trihalides where back-bonding decreases as halogen size increases from

\text{F}

\text{I}

Inert Pair Effect & Oxidation State Stability (Group 13)

$\text{B}^{3+} > \text{Al}^{3+} > \text{Ga}^{3+} > \text{In}^{3+} > \text{Tl}^{3+} \quad \text{and} \quad \text{Tl}^+ > \text{In}^+ > \text{Ga}^+$

Applies strictly down Group 13 due to poor shielding by

4f

and

5d

subshells;

\text{Tl}^+

is the predominant oxidation state for thallium.

Inert Pair Effect & Oxidation State Stability (Group 14)

$\text{C}^{4+} > \text{Si}^{4+} > \text{Ge}^{4+} > \text{Sn}^{4+} > \text{Pb}^{4+} \quad \text{and} \quad \text{Pb}^{2+} > \text{Sn}^{2+} > \text{Ge}^{2+}$

Valid under standard conditions; makes

\text{Pb}^{4+}

a powerful oxidizing agent and

\text{Sn}^{2+}

a versatile reducing agent.

General Formula of Silicate Units

$(\text{Si}_n \text{O}_{3n+1})^{(2n+2)-} \quad \text{or general oxygen-sharing ratio formula}$

Valid for fundamental tetrahedral

[\text{SiO}_4]^{4-}

networks where shared corner oxygen atoms determine silicate geometry (ortho-, pyro-, chain, sheet).

Hydrolysis of Group 14 Tetrahalides

$\text{MX}_4 + 2\text{H}_2\text{O} \rightarrow \text{MO}_2 + 4\text{HX} \quad (\text{M} = \text{Si}, \text{Ge}, \text{Sn}, \text{Pb}; \text{M} \neq \text{C})$

Requires vacant

d

-orbitals on central atom

\text{M}

to coordinate water; strictly inert for

\text{CCl}_4

under normal conditions due to absence of

d

-orbitals.

Solved High-Yield Sample Questions

Practice Mode

Question #1NEET 2024

Which of the following statements is INCORRECT regarding Diborane ( $\text{B}_2\text{H}_6$ )?

Solution Locked

Step-by-Step Solution & Concept

Step 1

Analyze hybridization. Each Boron atom in $\text{B}_2\text{H}_6$ forms 4 bonds (two terminal $2c-2e^-$ bonds and two bridging $3c-2e^-$ bonds), adopting $sp^3$ hybridization.

Step 2

Examine spatial geometry. The two Boron atoms and four terminal Hydrogen atoms lie together in one horizontal plane.

Step 3

Evaluate bridging Hydrogens. The two bridging Hydrogen atoms lie in a plane perpendicular to the horizontal plane (one above and one below). Hence, stating they lie in the same plane is INCORRECT.

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Question #2JEE Main 2023

Identify the correct order of oxidizing power for the tetravalent species of Group 14 elements: $\text{Ge}^{4+}$ , $\text{Sn}^{4+}$ , and $\text{Pb}^{4+}$ .

Solution Locked

Step-by-Step Solution & Concept

Step 1

Apply the inert pair effect down Group 14. Relativistic contraction and poor $d/f$ shielding reduce the stability of $+4$ oxidation state while increasing $+2$ stability down the group: $\text{Ge}^{4+} > \text{Sn}^{4+} > \text{Pb}^{4+}$ .

Step 2

Relate thermodynamic stability to redox behavior. A less stable high oxidation state species strongly prefers gaining electrons to reduce into the stable lower oxidation state.

Step 3

Because $\text{Pb}^{4+}$ is unstable and easily gains $2e^-$ to form stable $\text{Pb}^{2+}$ , $\text{Pb}^{4+}$ is the strongest oxidizing agent. Thus, oxidizing power increases as $\text{Ge}^{4+} < \text{Sn}^{4+} < \text{Pb}^{4+}$ .

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Mastering p-Block Elements (Group 13 & 14): Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Lewis Acidic Strength of Boron Halides

Inert Pair Effect & Oxidation State Stability (Group 13)

Inert Pair Effect & Oxidation State Stability (Group 14)

General Formula of Silicate Units

Hydrolysis of Group 14 Tetrahalides

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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