This chemistry category collects practice questions on the core areas of the subject, including atomic structure, chemical bonding, reactions, and the properties of elements and compounds. Questions are grouped so you can work through one topic at a time and review the answers as you go. Useful for building recall before a test and for spotting the concepts that still need work.
Reduction of 1-nitro-2-methylpropane ( (CH₃)2CHCH₂NO₂ ) with H₂/Ni yields (CH₃)2CHCH₂NH₂ (2-methylpropan-1-amine), a primary amine. This follows from NCERT principle where relation explains outcome clearly for students.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Hemoglobin contains an Fe coordination compound that acts as an oxygen carrier, unlike the others which are synthetic. This follows from NCERT principle where relation explains outcome clearly for students.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Vitamin B₁₂ (cyanocobalamin) is a Co coordination compound, unlike the others which are synthetic. This follows from NCERT principle where relation explains outcome clearly for students.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: The oxidation state of phosphorus in P_4O_{10 is: These values define the system as per NCERT data. Formula: In P_4O_{10, O is -2, so 4P + 10(-2) = 0, 4P - 20 = 0, 4P = 20, P = +5, as P shares electrons with oxygen.. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
S (0 in S to -2 in Na₂S ) is reduced by Na (0 to +1), so final state is -2. This follows from NCERT principle where relation explains outcome clearly for students.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: What is the mole fraction of glucose in a solution containing 18 g of glucose (C₆H₁₂O₆) and 72 g of water? (Molar masses: glucose = 180 g/mol, H₂O = 18 g/mol) These values define the system as per NCERT data. Formula: Moles of glucose = 18 / 180 = 0.1 mol. This is standard NCERT relation. Substitution & Calculation: Moles of H₂O = 72 / 18 = 4 mol. Total moles = 0.1 + 4 = 4.1. Mole fraction = 0.1 / 4.1 ≈ 0.0244. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: For PCl₅(g) PCl₃(g) + Cl₂(g), K_c = 0.04 at 250°C. If 1 mole of PCl₅ is placed in a 10 L vessel, what is the equilibrium concentration of Cl₂ ? These values define the system as per NCERT data. Formula: Initial [PCl₅] = 1 / 10 = 0.1 M. This is standard NCERT relation. Substitution & Calculation: Let x be the concentration of Cl₂ at equilibrium. Then, [PCl₅] = 0.1 - x, [PCl₃] = x, [Cl₂] = x . K_c = frac[PCl₃][Cl₂][PCl₅] = x · x/0.1 - x = 0.04 . x² = 0.04 (0.1 - x), x² + 0.04x - 0.004 = 0 . Solving: x = frac-0.04 pm sqrt(0.04)² + 4 · 0.0042 = 0.0336 M (positive root). Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: A solution contains 15 g of glycerol ( C₃H₈O₃ ) in 285 g of water. What is the molality of the solution? (Molar mass: C₃H₈O₃ = 92 g/mol ) These values define the system as per NCERT data. Formula: Moles of glycerol = 15/92 = 0.163 mol. This is standard NCERT relation. Substitution & Calculation: Mass of solvent in kg = 285/1000 = 0.285 kg . Molality = 0.163/0.285 = 0.572 mol/kg . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: What is the standard electrode potential of a galvanic ll where zinc is oxidized and copper is reduced, given their standard reduction potentials are -0.76 V and 0.34 V respectively? These values define the system as per NCERT data. Formula: Standard ll potential, E_{ll⁰ = E_{cathode⁰ - E_{anode⁰. This is standard NCERT relation. Substitution & Calculation: For Zn(s) → Zn²⁺(aq) + 2e⁻ (anode) and Cu²⁺(aq) + 2e⁻ → Cu(s) (cathode): E_{ll⁰ = 0.34 - (-0.76) = 1.10 V . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: What is the ll potential at 298 K for Ni(s) + 2Ag⁺(0.002 M) → Ni²⁺(0.160 M) + 2Ag(s) if E_{ll⁰ = 1.05 V ? These values define the system as per NCERT data. Formula: E_{ll = E_{ll⁰ - 0.059/2 log frac[Ni^{2+][Ag^{+]². This is standard NCERT relation. Substitution & Calculation: Q = 0.160/(0.002)² = 40000, log Q = 4.602 . E_{ll = 1.05 - 0.059/2 × 4.602 = 1.05 - 0.136 = 0.914 V approx 0.91 V . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: Calculate the power of a laser emitting 5.0 × __10POW₂₄__photons per second at 589 nm. (h = 6.626 × 10⁻³⁴J s, c = 3.0 × __10POW₈__m s^{-1) These values define the system as per NCERT data. Formula: Energy per photon E = hc/lambda = frac6.626 × 10⁻³⁴ × 3.0 × __10POW₈₅₈₉__ × 10⁻⁹= 3.375 × 10⁻¹⁹J. This is standard NCERT relation. Substitution & Calculation: Power = 5.0 × __10POW₂₄__ × 3.375 × 10⁻¹⁹= 1.6875 × __10POW₆__W. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
PH₃ has 3 bonding pairs and 1 lone pair on phosphorus, reducing the tetrahedral angle to approximately 93° due to lone pair repulsion. This follows from NCERT principle where relation explains outcome clearly for students.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
