Chemistry Question Papers Set (Class XI)

sets of Class XI Chemistry Question Papers (50 Marks Each) from the chapters:

📘 “Some Basic Concepts of Chemistry”
📗 “Atomic Structure”

🧪 SET – 1

Section A: MCQs [10 × 1 = 10 Marks]

  1. Which of the following has the highest number of atoms?
    • A. 1 mol H₂O
    • B. 18 g H₂O
    • C. 1 mol O₂
    • D. 1 mol CO₂
  2. Which law is obeyed in the formation of CO and CO₂ from C and O?
    • A. Law of Constant Proportion
    • B. Law of Multiple Proportions
    • C. Law of Conservation of Mass
    • D. Avogadro’s Law
  3. Which quantum number distinguishes the orientation of orbital?
    • A. Principal
    • B. Azimuthal
    • C. Magnetic
    • D. Spin
  4. The maximum number of electrons in a sublevel with l = 2 is:
    • A. 2
    • B. 6
    • C. 10
    • D. 14
  5. 22.4 L of a gas at STP weighs 44 g. Its molar mass is:
    • A. 22.4 g/mol
    • B. 44 g/mol
    • C. 11 g/mol
    • D. 88 g/mol
  6. Which of the following represents a correct set?
    • A. n = 2, l = 2
    • B. n = 3, l = 2
    • C. n = 1, l = 1
    • D. n = 2, l = 0
  7. A mole of NO₂ contains:
    • A. 6.022 × 10²³ atoms
    • B. 6.022 × 10²³ molecules
    • C. 1.8066 × 10²⁴ atoms
    • D. None of these
  8. What is the shape of the s-orbital?
    • A. Dumbbell
    • B. Spherical
    • C. Elliptical
    • D. Clover
  9. Who proposed the dual nature of matter?
    • A. Planck
    • B. Einstein
    • C. de Broglie
    • D. Heisenberg
  10. Ionization energy is highest for:
  • A. H
  • B. Li
  • C. He
  • D. Be

Section B: Assertion and Reason (1 × 1 = 1 mark)

11.
Assertion (A): Electrons in an atom are arranged in shells.
Reason (R): Shells are defined by principal quantum number ‘n’.

  • A. Both A and R are true, and R is the correct explanation.
  • B. Both A and R are true, but R is not the correct explanation.
  • C. A is true, R is false.
  • D. A is false, R is true.

Section C: Case-Based [3 × 1 = 3 Marks]

12. Read the case carefully and answer:

A student burns 12 g of carbon in excess oxygen to form carbon dioxide. He then collects 44 g of CO₂.

(a) Which law of chemical combination is illustrated here?
(b) Calculate the molar mass of CO₂.
(c) If 24 g of carbon is used, what will be the amount of CO₂ formed?

Section D: Short Answers [5 × 2 = 10 Marks]

  1. State the limitations of Dalton’s atomic theory.
  2. Explain Hund’s rule with an example.
  3. How many moles of nitrogen are present in 28 g of N₂?
  4. Define empirical and molecular formula.
  5. Differentiate between orbit and orbital.

Section E: Numerical Problems [3 × 3 = 9 Marks]

  1. A solution contains 10% KCl by mass. Its density is 1.06 g/mL. Calculate its molality.
  2. Calculate the wavelength (in nm) of the radiation emitted when an electron transitions from n = 5 to n = 2 in hydrogen atom. (Rydberg constant = 1.097 × 10⁷ m⁻¹)
  3. Write the four quantum numbers of the 10th electron of nitrogen (Z = 7).

Section F: Long Answer [2 × 5 = 10 Marks]

  1. Derive the relation between the radius and principal quantum number (n) for Bohr’s atom.
  2. An organic compound contains:
  • C = 40%, H = 6.7%, O = 53.3%
  • Molar mass = 60
    Find its empirical and molecular formula.

🧪 Sets 2 and 3 with improvements are coming up next…

You said:

set 2 & 3 please

ChatGPT said:

Here are the improved SET – 2 and SET – 3 Chemistry Question Papers (Class XI, 50 Marks Each), refined as requested:

🧪 SET – 2

Section A: MCQs [10 × 1 = 10 Marks]

  1. The number of moles in 8 g of O₂ is:
    • A. 0.25
    • B. 0.5
    • C. 1
    • D. 2
  2. Which law states that “A sample of water always has hydrogen and oxygen in the ratio 2:16 by mass”?
    • A. Avogadro’s Law
    • B. Law of Definite Proportion
    • C. Law of Multiple Proportions
    • D. Law of Conservation of Mass
  3. Which set of quantum numbers is not possible?
    • A. n = 2, l = 0, m = 0
    • B. n = 3, l = 2, m = -2
    • C. n = 2, l = 2, m = 1
    • D. n = 4, l = 3, m = 2
  4. Which orbital has zero angular momentum?
    • A. s
    • B. p
    • C. d
    • D. f
  5. Which of the following has the same number of protons, neutrons, and electrons?
    • A. ¹H
    • B. ⁴₂He
    • C. ¹⁴₇N
    • D. ¹²₆C
  6. Maximum number of electrons in a shell is given by:
    • A. 2n
    • B. 2n²
    • C. n²
    • D. n(n + 1)
  7. The shape of p-orbital is:
    • A. Spherical
    • B. Dumbbell
    • C. Double dumbbell
    • D. Circular
  8. The principal quantum number of 3d orbital is:
    • A. 2
    • B. 3
    • C. 4
    • D. 5
  9. Which of the following is correct about Planck’s quantum theory?
    • A. Energy is continuous
    • B. Energy is quantized
    • C. Only particles have energy
    • D. Radiation has no energy
  10. The quantum number that represents spin is:
  • A. n
  • B. l
  • C. m
  • D. s

Section B: Assertion and Reason [1 × 1 = 1 Mark]

  1.  

Assertion (A): Electrons revolve in definite orbits around the nucleus.
Reason (R): Bohr explained the stability of the atom using classical mechanics.

  • A. Both A and R are true, and R is the correct explanation.
  • B. Both A and R are true, but R is not the correct explanation.
  • C. A is true, R is false.
  • D. A is false, R is true.

Section C: Case-Based [3 × 1 = 3 Marks]

12. A chemist dissolves 36.5 g of HCl gas in 1.0 L water. He observes an exothermic reaction forming hydrochloric acid.

(a) What is the molar mass of HCl?
(b) What is the number of moles of HCl dissolved?
(c) Which law explains why the mass of reactants equals products here?

Section D: Short Answers [5 × 2 = 10 Marks]

  1. Define atomic mass unit and relate it to mass of a proton.
  2. Differentiate between isotopes and isobars with examples.
  3. State Heisenberg’s uncertainty principle.
  4. Calculate number of molecules in 18 g of water.
  5. Explain Aufbau principle with an example.

Section E: Numerical Problems [3 × 3 = 9 Marks]

  1. A 20% aqueous solution of NaCl has a density of 1.12 g/mL. Calculate its molality.
  2. Calculate the energy of a photon emitted when an electron drops from n = 4 to n = 1 in hydrogen atom.
  3. Give the quantum numbers of the last electron of Al (Z = 13).

Section F: Long Answer [2 × 5 = 10 Marks]

  1. Explain the postulates of Bohr’s model and derive the expression for energy levels.
  2. An organic compound has the following composition:
  • C = 54.5%, H = 9.1%, O = 36.4%
  • Molar mass = 88
    Find empirical and molecular formula.

🧪 SET – 3

Section A: MCQs [10 × 1 = 10 Marks]

  1. What is the volume occupied by 2 moles of gas at STP?
    • A. 11.2 L
    • B. 22.4 L
    • C. 33.6 L
    • D. 44.8 L
  2. Which of the following is not a postulate of Dalton’s theory?
    • A. Atoms are indivisible
    • B. Atoms of an element are identical
    • C. Atoms can be created or destroyed
    • D. Atoms combine in fixed ratios
  3. An orbital can hold a maximum of:
    • A. 1 electron
    • B. 2 electrons
    • C. 4 electrons
    • D. 8 electrons
  4. The value of Planck’s constant is:
    • A. 6.626 × 10⁻³⁴ J s
    • B. 3.00 × 10⁸ m/s
    • C. 1.602 × 10⁻¹⁹ C
    • D. 9.11 × 10⁻³¹ kg
  5. Which of the following represents an isotope?
    • A. ¹²₆C and ¹⁴₆C
    • B. ¹²₆C and ¹²₇N
    • C. ¹²₆C and ¹⁴₇N
    • D. ¹²₆C and ⁶₃Li
  6. The angular momentum of an electron in nth orbit (Bohr) is:
    • A. n × h
    • B. nh/2π
    • C. n²h
    • D. h/n
  7. Magnetic quantum number defines:
    • A. Energy level
    • B. Orbital shape
    • C. Orientation
    • D. Spin
  8. Energy of radiation is:
    • A. Directly proportional to frequency
    • B. Inversely proportional to frequency
    • C. Directly proportional to wavelength
    • D. Inversely proportional to energy
  9. Mass of 1 atom of oxygen is:
    • A. 16 g
    • B. 32 g
    • C. 2.66 × 10⁻²³ g
    • D. 1.66 × 10⁻²⁴ g
  10. Number of molecules in 1 L of gas at STP:
  • A. 6.022 × 10²³
  • B. 2.69 × 10²²
  • C. 3.01 × 10²³
  • D. 1.81 × 10²²

Section B: Assertion and Reason [1 × 1 = 1 Mark]

  1.  

Assertion (A): 2p orbitals are degenerate.
Reason (R): Degenerate orbitals have same energy in absence of external field.

  • A. Both A and R are true and R is correct explanation
  • B. Both A and R are true but R is not correct explanation
  • C. A is true but R is false
  • D. A is false but R is true

Section C: Case-Based [3 × 1 = 3 Marks]

12. A balloon is filled with 2.24 L of oxygen at STP.

(a) Calculate the number of moles of oxygen.
(b) How many molecules are present in this amount?
(c) Which law is applied here?

Section D: Short Answers [5 × 2 = 10 Marks]

  1. Explain Pauli’s exclusion principle with an example.
  2. Define wave number and give its SI unit.
  3. Calculate the number of atoms in 1.2 g of carbon (C-12).
  4. Write difference between mass number and atomic number.
  5. State two limitations of Rutherford’s atomic model.

Section E: Numerical Problems [3 × 3 = 9 Marks]

  1. A solution contains 12% glucose (C₆H₁₂O₆) by mass. Its density is 1.10 g/mL. Calculate its molality.
  2. An electron jumps from n = 6 to n = 2 in H atom. Identify the series and calculate the energy of emitted photon.
  3. What are the quantum numbers for the last electron of Cl (Z = 17)?

Section F: Long Answer [2 × 5 = 10 Marks]

  1. State the postulates of Dalton’s atomic theory. Which of these were later found to be incorrect?
  2. A compound has:
  • C = 85.7%, H = 14.3%, Molar Mass = 28
    Find the empirical and molecular formula.

Would you like the Answer Key for Sets 2 & 3 now?

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also give answer key of all 3 set

ChatGPT said:

Here’s the Answer Key for all 3 sets of the question papers:

Answer Key — SET 1

Section A: MCQs

  1. B
  2. B
  3. C
  4. C
  5. B
  6. B
  7. B
  8. B
  9. C
  10. C

Section B: Assertion & Reason

  1. A (Both true, R explains A)

Section C: Case-Based

  1.  

(a) Law of Conservation of Mass
(b) Molar mass of CO₂ = 44 g/mol
(c) 88 g CO₂ (if 24 g C used, CO₂ formed doubles)

Section D: Short Answers

  1. Dalton’s theory limitations: atoms divisible, isotopes exist, atoms can gain/lose electrons, etc.
  2. Hund’s rule: electrons fill orbitals singly first with parallel spins.
  3. Moles of N₂ in 28 g = 1 mole
  4. Empirical formula = simplest ratio; molecular formula = actual number of atoms in molecule.
  5. Orbit = fixed path (old); orbital = region of probable electron presence (modern).

Section E: Numerical Problems

  1. Molality = 1.00 m (Detailed: 10 g KCl in 90 g solvent → molality = moles/kg solvent)
  2. λ = 434 nm (using Rydberg formula)
  3. For 10th electron of N: n=2, l=1, m=0, s=+½

Section F: Long Answer

  1. Radius relation: r ∝ n²
  2. Empirical formula = CH₂O, Molecular formula = C₂H₄O₂

Answer Key — SET 2

Section A: MCQs

  1. B
  2. B
  3. C
  4. A
  5. A
  6. B
  7. B
  8. B
  9. B
  10. D

Section B: Assertion & Reason

  1. C (A true, R false)

Section C: Case-Based

  1.  

(a) Molar mass HCl = 36.5 g/mol
(b) Moles = 1 mol
(c) Law of Conservation of Mass

Section D: Short Answers

  1. 1 amu = 1/12 mass of ¹²C; mass proton ≈ 1 amu
  2. Isotopes: same Z, different A (e.g., ¹²C and ¹³C); Isobars: same A, different Z (e.g., ⁴⁰Ca and ⁴⁰Ar)
  3. Δx·Δp ≥ h/4π
  4. Molecules in 18 g water = 6.022×10²³
  5. Aufbau: electrons fill lower energy orbitals first

Section E: Numerical Problems

  1. Molality = 3.1 m
  2. Energy = 12.09 × 10⁻¹⁹ J (calculate with E = hc/λ or Rydberg)
  3. Al last electron: n=3, l=1, m=+1, s=+½

Section F: Long Answer

  1. Bohr postulates include quantized orbits, angular momentum quantization, energy levels.
  2. Empirical formula = C₂H₄O, Molecular formula = C₄H₈O₂

Answer Key — SET 3

Section A: MCQs

  1. D
  2. C
  3. B
  4. A
  5. A
  6. B
  7. C
  8. A
  9. C
  10. B

Section B: Assertion & Reason

  1. A (Both true and R explains A)

Section C: Case-Based

  1.  

(a) Moles = 0.1 mol
(b) Molecules = 6.022 × 10²²
(c) Avogadro’s Law

Section D: Short Answers

  1. No two electrons in atom have same 4 quantum numbers (Pauli exclusion).
  2. Wave number = 1/λ (cm⁻¹)
  3. Atoms in 1.2 g C = 6.022 × 10²²
  4. Mass number = protons + neutrons; Atomic number = protons only
  5. Rutherford model could not explain stability, discrete spectra

Section E: Numerical Problems

  1. Molality = 1.82 m
  2. Series: Balmer; Energy ≈ 3.03 × 10⁻¹⁹ J
  3. Cl last electron: n=3, l=1, m=0, s=–½

Section F: Long Answer

  1. Dalton’s postulates: atoms indivisible, identical, combine in ratios; isotopes disproved identical atom idea.
  2. Empirical formula = CH₂, Molecular formula = C₂H₄

If you want, I can format these as a neat PDF or Word file. Just let me know!

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with details solution

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Sure! Below is a detailed Answer Key with solutions for all 3 sets including explanations and calculations where applicable.

DETAILED ANSWER KEY — SET 1

Section A: MCQs

  1. B
    Molar mass of O₂ = 32 g/mol
    Moles = mass / molar mass = 8/32 = 0.25 moles → (Check options, 0.25 is option A? Correct option is A)
    Correction: The correct answer is A (0.25) not B.
  2. B — Law of Definite Proportion states a compound always contains elements in fixed mass ratio.
  3. C — n=2, l=2 is invalid because l ranges from 0 to n-1.
  4. C — d orbitals have angular momentum √(l(l+1))ħ with l=2, not zero.
  5. B — ⁴₂He has 2 protons, 2 neutrons, and 2 electrons.
  6. B — Maximum electrons in nth shell = 2n².
  7. B — p orbital shape is dumbbell.
  8. B — Principal quantum number for 3d is 3.
  9. C — Planck’s quantum theory states energy is quantized (E=nhv).
  10. C — Magnetic quantum number (m) represents orientation.

Section B: Assertion & Reason

  1. A
  • Assertion: Electrons revolve in definite orbits (Bohr’s postulate).
  • Reason: Bohr used classical mechanics + quantization to explain stability. Both true and Reason explains Assertion.

Section C: Case-Based

  1.  

(a) Law of Conservation of Mass — total mass of reactants = total mass of products.
(b) Molar mass CO₂ = 12 + 2×16 = 44 g/mol.
(c) If 24 g C reacts, produces 88 g CO₂ (based on reaction mass).

Section D: Short Answers

  1. Dalton’s theory limitations:
  • Atoms are divisible into protons, neutrons, electrons.
  • Isotopes exist (same Z, different A).
  • Atoms can gain or lose electrons (ions).
  1. Hund’s Rule:
    Electrons occupy orbitals singly with parallel spins before pairing.
  2. Moles N₂:
    Molar mass N₂ = 28 g/mol
    Moles = 28/28 = 1 mole.
  3. Empirical vs Molecular Formula:
  • Empirical: simplest whole number ratio of atoms.
  • Molecular: actual number of atoms in a molecule.
  1. Orbit vs Orbital:
  • Orbit: fixed circular path (Bohr).
  • Orbital: 3D region of probable electron location.

Section E: Numerical Problems

  1. Molality Calculation:
    Given: 10 g KCl in 90 g water.
    Moles KCl = 10/74.5 ≈ 0.134 mol
    Mass solvent in kg = 0.090 kg
    Molality = moles solute / kg solvent = 0.134 / 0.090 ≈ 1.49 m
  2. Wavelength λ for transition n=5→2:
    Using Rydberg:

1λ=R(122−152)=1.097×107(14−125)\frac{1}{\lambda} = R \left(\frac{1}{2^2} – \frac{1}{5^2}\right) = 1.097 \times 10^7 \left(\frac{1}{4} – \frac{1}{25}\right)λ1​=R(221​−521​)=1.097×107(41​−251​) =1.097×107×21100=2.3037×106 m−1= 1.097 \times 10^7 \times \frac{21}{100} = 2.3037 \times 10^6 \, m^{-1}=1.097×107×10021​=2.3037×106m−1 λ=12.3037×106=4.34×10−7m=434nm\lambda = \frac{1}{2.3037 \times 10^6} = 4.34 \times 10^{-7} m = 434 nmλ=2.3037×1061​=4.34×10−7m=434nm

  1. Quantum numbers for 10th electron of N:
    Electron configuration N (Z=7): 1s² 2s² 2p³
    The 10th electron is 2p with n=2, l=1, m=0, spin=+½

Section F: Long Answer

  1. Bohr’s Model Postulates:
  • Electrons revolve in fixed orbits without radiating energy.
  • Angular momentum is quantized: mvr=nℏmvr = n \hbarmvr=nℏ
  • Energy levels given by En=−13.6n2E_n = -\frac{13.6}{n^2}En​=−n213.6​ eV
  1. Empirical & Molecular Formula:
  • Calculate moles for each:
    C = 54.5 g → moles = 54.5/12 = 4.54
    H = 9.1 g → moles = 9.1/1 = 9.1
    O = 36.4 g → moles = 36.4/16 = 2.275
  • Simplify ratio dividing by smallest (2.275):
    C: 4.54/2.275 = 2, H: 9.1/2.275 = 4, O: 2.275/2.275 =1
    Empirical formula = C₂H₄O
    Molar mass given corresponds to molecular formula C₄H₈O₂ (double empirical).

DETAILED ANSWER KEY — SET 2

Section A: MCQs

  1. B
    Molar mass HCl = 36.5 g/mol.
  2. B — Law of Definite Proportion.
  3. C — n=2, l=2 invalid.
  4. A — s orbitals have zero angular momentum (l=0).
  5. A — Isotopes have same Z, different A (¹²C and ¹³C).
  6. B — Maximum electrons in nth shell = 2n².
  7. B — p orbital shape is dumbbell.
  8. B — Principal quantum number for 3d is 3.
  9. B — Planck’s theory: energy is quantized.
  10. D — Spin quantum number.

Section B: Assertion & Reason

  1. C
    A true (electrons revolve in orbits), R false (Bohr used quantization, not classical mechanics only).

Section C: Case-Based

  1.  

(a) Molar mass HCl = 36.5 g/mol
(b) Moles = 36.5/36.5 = 1 mole
(c) Law of Conservation of Mass.

Section D: Short Answers

  1. 1 amu = 1/12 mass of ¹²C; mass proton ≈ 1 amu.
  2. Isotopes same Z different A, isobars same A different Z.
  3. Uncertainty Principle: Δx⋅Δp≥h4π\Delta x \cdot \Delta p \geq \frac{h}{4\pi}Δx⋅Δp≥4πh​
  4. Molecules in 18 g water = 1818×6.022×1023=6.022×1023\frac{18}{18} \times 6.022 \times 10^{23} = 6.022 \times 10^{23}1818​×6.022×1023=6.022×1023
  5. Aufbau: Electrons fill lower energy orbitals first.

Section E: Numerical Problems

  1. Molality calculation:
  • Mass % = 20% NaCl, density = 1.12 g/mL
  • Mass of solution = 1000 mL × 1.12 g/mL = 1120 g
  • Mass NaCl = 20% of 1120 = 224 g
  • Mass water = 1120 – 224 = 896 g = 0.896 kg
  • Moles NaCl = 224/58.44 = 3.83 mol
  • Molality = 3.83 / 0.896 = 4.27 m
  1. Energy of photon:
    Transition n=4 → n=1 for H atom,

E=13.6(112−142)=13.6(1−116)=13.6×1516=12.75 eVE = 13.6 \left(\frac{1}{1^2} – \frac{1}{4^2}\right) = 13.6 \left(1 – \frac{1}{16}\right) = 13.6 \times \frac{15}{16} = 12.75 \text{ eV}E=13.6(121​−421​)=13.6(1−161​)=13.6×1615​=12.75 eV

Convert to Joules: 12.75 × 1.602 × 10⁻¹⁹ = 2.04×10−182.04 \times 10^{-18}2.04×10−18 J

  1. Quantum numbers for Al last electron:
    Al (Z=13) configuration ends at 3p¹:
    n=3, l=1, m=0 or ±1 (for first p electron, m=0), spin=+½

Section F: Long Answer

  1. Postulates: electrons in circular orbits, quantized angular momentum mvr=nℏmvr = n\hbarmvr=nℏ, energy quantized.
  2.  

Calculate empirical formula:

  • C: 54.5/12 = 4.54
  • H: 9.1/1 = 9.1
  • O: 36.4/16 = 2.275
    Ratio: C:H:O = 2:4:1
    Empirical formula = C₂H₄O
    Molecular mass is twice empirical → C₄H₈O₂

DETAILED ANSWER KEY — SET 3

Section A: MCQs

  1. D
    2 moles × 22.4 L/mol = 44.8 L
  2. C — Dalton said atoms can’t be created/destroyed (opposite to option C).
  3. B — Orbital max 2 electrons.
  4. A — Planck’s constant.
  5. A — Isotopes ¹²₆C and ¹⁴₆C.
  6. B — Angular momentum = nh/2π.
  7. C — Magnetic quantum number = orbital orientation.
  8. A — Energy ∝ frequency.
  9. C — Mass of one oxygen atom ≈ 16 amu / Avogadro’s number ≈ 2.66 × 10⁻²³ g.
  10. B — 1 mole of gas at STP occupies 22.4 L, so 1 L contains 1/22.4 × 6.022×10²³ ≈ 2.69×10²² molecules.

Section B: Assertion & Reason

  1. A Both true, R explains A.

Section C: Case-Based

  1.  

(a) Moles = Volume / molar volume = 2.24/22.4 = 0.1 mol
(b) Molecules = 0.1 × 6.022 × 10²³ = 6.022 × 10²²
(c) Avogadro’s Law.

Section D: Short Answers

  1. Pauli’s Principle: no two electrons in an atom can have the same set of four quantum numbers.
  2. Wave number = 1/λ (in cm⁻¹)
  3. Atoms in 1.2 g C = (1.2/12) × 6.022 × 10²³ = 6.022 × 10²²
  4. Atomic number = number of protons; mass number = protons + neutrons.
  5. Rutherford model failed to explain atom stability & discrete line spectra.

Section E: Numerical Problems

  1. Molality:
    Given 12% glucose, density = 1.10 g/mL
    Mass of solution = 1000 × 1.10 = 1100 g
    Glucose mass = 12% × 1100 = 132 g
    Water mass = 1100 – 132 = 968 g = 0.968 kg
    Moles glucose = 132 / 180 = 0.733 mol
    Molality = 0.733 / 0.968 = 0.757 m
  2. Electron jump n=6 → n=2 (Balmer series)
    Energy difference:

E=13.6(122−162)=13.6(14−136)=13.6×836=3.02 eVE = 13.6 \left( \frac{1}{2^2} – \frac{1}{6^2} \right) = 13.6 \left(\frac{1}{4} – \frac{1}{36}\right) = 13.6 \times \frac{8}{36} = 3.02 \text{ eV}E=13.6(221​−621​)=13.6(41​−361​)=13.6×368​=3.02 eV

Convert to Joules = 3.02 × 1.602 × 10⁻¹⁹ = 4.83 × 10⁻¹⁹ J

  1. Quantum numbers for Cl last electron:
    Configuration ends at 3p⁵:
    n=3, l=1, m=0 or ±1, spin = −½ (last electron spin opposite to paired ones)

Section F: Long Answer

  1. Dalton’s postulates: atoms indivisible, identical atoms for an element, atoms combine in fixed ratios; isotopes disprove identical atoms.
  2. Composition: C = 85.7%, H = 14.3%
    Calculate moles:
    C = 85.7/12 = 7.14
    H = 14.3/1 = 14.3
    Ratio: C:H = 1:2
    Empirical formula = CH₂
    Molecular mass = 28 → molecular formula = C₂H₄

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