Why is tritium ($\text{}^3_1H$) radioactive while protium and deuterium are non-radioactive?

Tritium has $1$ proton and $2$ neutrons, yielding a high neutron-to-proton ratio ($n/p = 2$). This high ratio destabilizes the nucleus, causing tritium to decay via low-energy $\beta^-$ emission with a half-life of $t_{1/2} \approx 12.33\text{ years}$.

Why does heavy water ($D_2O$) react slower than ordinary water ($H_2O$)?

Due to the higher mass of deuterium ($D$), the zero-point vibrational energy of the $O-D$ bond is lower than that of the $O-H$ bond. Consequently, the activation energy for breaking an $O-D$ bond is higher, leading to a slower reaction rate (Kinetic Isotope Effect).

How do I instantly differentiate between electron-deficient, electron-precise, and electron-rich hydrides?

Examine the valence electron count of the central atom in covalent hydrides: Group 13 hydrides ($B_2H_6$) lack sufficient electrons for normal covalent bonds (electron-deficient); Group 14 hydrides ($CH_4$) have exact octet pairs (electron-precise); Group 15-17 hydrides ($NH_3, H_2O, HF$) possess lone pairs (electron-rich).

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Solved High-Yield Sample Questions

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Question #1NEET 2023

Calculate the volume strength of a $1.5\text{ M}$ aqueous solution of $H_2O_2$ at $STP$ .

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Step-by-Step Solution & Concept

Step 1

Use the direct formula relating volume strength to molarity ( $M$ ) of $H_2O_2$ : $\text{Volume Strength} = 11.2 \times M$ .

Step 2

Substitute the given molarity $M = 1.5\text{ M}$ into the expression: $\text{Volume Strength} = 11.2 \times 1.5$ .

Step 3

Perform multiplication: $11.2 \times 1.5 = 16.8\text{ V}$ . Thus, the correct option is $16.8\text{ V}$ .

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Question #2JEE Main 2024

A sample of hard water contains $0.001\text{ mole}$ of $CaCl_2$ dissolved in $1\text{ L}$ of water. The degree of hardness of this water sample in $\text{ppm}$ of $CaCO_3$ equivalent is (Molar mass of $CaCl_2 = 111\text{ g/mol}$ ):

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Step-by-Step Solution & Concept

Step 1

Calculate mass of $CaCO_3$ equivalent for $0.001\text{ mole}$ of $CaCl_2$ . Since $1\text{ mole } CaCl_2 \equiv 1\text{ mole } CaCO_3$ , $0.001\text{ mole } CaCl_2 = 0.001 \times 100\text{ g} = 0.1\text{ g}$ of $CaCO_3$ .

Step 2

Mass of $1\text{ L}$ of water is approximately $1000\text{ g}$ .

Step 3

Compute degree of hardness in $\text{ppm}$ : $\text{Hardness} = \frac{0.1\text{ g}}{1000\text{ g}} \times 10^6 = 100\text{ ppm}$ .

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Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Volume Strength to Molarity of $H_2O_2$

Volume Strength to Normality of $H_2O_2$

Degree of Hardness of Water (ppm)

$CaCO_3$ Equivalence Factor

Percentage Strength of $H_2O_2$ Solution

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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