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Indian Navy SSR (Sailor) Chemistry — practice by topic

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Chemical Equilibrium (58)Organic Chemistry (57)The p-Block Elements (56)Alcohols Phenols Ethers (55)Atoms and Molecules (55)Classification of Elements (55)p-Block Elements (55)Polymers (55)S Block Elements (55)Alkenes (54)Classification of Elements and Periodicity (54)p-Block Elements (Groups 15-18) (54)States of Matter (54)Surface Chemistry (51)Solid State (50)Organic Chemistry: Some Basic Principles and Techniques (34)Chemistry (32)JEE Main 2023 Mixed (32)Atomic combinations (31)Atoms and Elements (31)Class9 Atom Structure (31)Ideal and Non-Ideal Solutions (31)Ideal Solutions (31)Non-Ideal Solutions (31)Reactions in Aqueous Solution (31)Structure of Atom (31)Coordination Compounds (30)Aldehydes Ketones (29)Atomic Structure (29)Atomic Structure and Valency (29)Chemical Bonding (29)Chemical Bonding and Molecular Structure (29)Classification of Elements and Periodicity in Properties (29)Electrochemistry (29)Equilibrium (29)Equilibrium Constant (29)Hydrogen (29)Periodic Classification (29)Periodic Classification of Elements (29)Periodic Table (29)Periodic Trends (29)The Periodic Table (29)Electrochemical Series (28)Aldehydes, Ketones and Carboxylic Acids (27)Atoms (27)The d- and f-Block Elements (27)Amines (26)Chemical Kinetics (26)Alcohols, Phenols and Ethers (25)d- and f-Block Elements (25)Haloalkanes and Haloarenes (25)Hydrocarbons (25)Some Basic Concepts of Chemistry (25)Surface Tension (25)Equipotential Surfaces (24)Redox & Electrochemistry (24)Redox Reactions (24)Surface Areas and Volumes (24)The Atom (24)Solutions (19)

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Worked practice questions

Arrange the bond angles of $\mathrm{H_2O}$, $\mathrm{NH_3}$ and $\mathrm{CH_4}$ in decreasing order.

  1. $\mathrm{CH_4} > \mathrm{H_2O} > \mathrm{NH_3}$
  2. $\mathrm{NH_3} > \mathrm{CH_4} > \mathrm{H_2O}$
  3. $\mathrm{H_2O} > \mathrm{NH_3} > \mathrm{CH_4}$
  4. $\mathrm{CH_4} > \mathrm{NH_3} > \mathrm{H_2O}$
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Correct answer: D. $\mathrm{CH_4} > \mathrm{NH_3} > \mathrm{H_2O}$

All three central atoms are $sp^3$ hybridised so the base bond angle is $109.5°$. Lone pairs compress that angle (the lone-pair-to-bond-pair repulsion is stronger than bond-pair-to-bond-pair). Approximate bond angles: - $\mathrm{CH_4}$ (0 lone pairs): $109.5°$ - $\mathrm{NH_3}$ (1 lone pair): $107°$ - $\mathrm{H_2O}$ (2 lone pairs): $104.5°$ More lone pairs means more compression, so the order is $\mathrm{CH_4} > \mathrm{NH_3} > \mathrm{H_2O}$.

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The modern periodic law states that the physical and chemical properties of elements are periodic functions of:

  1. Atomic mass
  2. Atomic radius
  3. Atomic number
  4. Number of neutrons
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Correct answer: C. Atomic number

Modern (Moseley's) periodic law: properties of elements are periodic functions of their **atomic number** ($Z$). This replaced Mendeleev's original law which was based on atomic mass. Why atomic number wins: it explains chemistry better (electronic structure follows $Z$, not mass) and resolves Mendeleev's anomalies (the Ar/K and Te/I pairs whose mass-order disagrees with chemistry-order).

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For an isothermal reversible expansion of an ideal gas from volume $V_1$ to $V_2$, the work done **by** the gas is:

  1. $-2.303\,nRT\,\log\dfrac{V_2}{V_1}$
  2. $+2.303\,nRT\,\log\dfrac{V_2}{V_1}$
  3. $0$
  4. $-p_\text{ext}(V_2 - V_1)$
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Correct answer: B. $+2.303\,nRT\,\log\dfrac{V_2}{V_1}$

For an isothermal reversible expansion, $W_\text{by gas} = +nRT\ln(V_2/V_1) = +2.303\,nRT\log(V_2/V_1)$. In the IUPAC convention $W_\text{on system} = -nRT\ln(V_2/V_1)$ is negative, but the question asks for work done *by* the gas, which is the positive of this.

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