Structure of Atom

What to remember

• Electrons, protons and neutrons are the subatomic particles; atomic number (Z) = number of protons, mass number (A) = protons + neutrons.
• Bohr's model gives Eₙ = −2.18 × 10⁻¹⁸ (Z²/n²) J and rₙ = 52.9 (n²/Z) pm; it works only for hydrogen-like one-electron species.
• Each orbital is defined by n, l and mₗ; an electron also has mₛ. Radial nodes = n − l − 1, angular nodes = l, total nodes = n − 1.
• Electrons fill orbitals by the aufbau principle (lowest (n + l) first), Pauli exclusion principle and Hund's rule; Cr is [Ar] 3d⁵ 4s¹ and Cu is [Ar] 3d¹⁰ 4s¹.

Subatomic particles and early atomic models

Cathode rays are produced in a discharge tube at very low pressure and high voltage. They travel in straight lines from the cathode, are deflected by electric and magnetic fields towards the positive plate, and are independent of the gas used. They are streams of negatively charged particles, the electrons. Thomson measured the charge-to-mass ratio: e/m = 1.758820 × 10¹¹ C kg⁻¹. Millikan's oil drop experiment gave the charge on the electron: 1.602 × 10⁻¹⁹ C. From these, the mass of the electron is 9.109 × 10⁻³¹ kg.

Canal rays (anode rays) led to the discovery of the proton. The neutron was discovered by Chadwick in 1932.

Thomson's model (plum pudding): the atom is a positively charged sphere with electrons embedded in it. It could not explain the results of the scattering experiment.

Rutherford's α-particle scattering experiment used a thin gold foil. Observations: most α-particles passed straight through; a small fraction were deflected through small angles; very few (about 1 in 20,000) bounced back. Conclusions: most of the atom is empty space; the positive charge and almost all the mass are concentrated in a tiny nucleus (radius about 10⁻¹⁵ m, compared with about 10⁻¹⁰ m for the atom); electrons move around the nucleus.

Limitations of Rutherford's model: it could not explain the stability of the atom (an orbiting electron should lose energy and spiral into the nucleus) and gave no information about the arrangement or energies of electrons.

Atomic number, mass number, isotopes, isobars and isotones

Atomic number Z = number of protons = number of electrons in a neutral atom. Mass number A = protons + neutrons. Number of neutrons = A − Z.

Isotopes have the same chemical properties and different physical properties such as mass and density.

Electromagnetic radiation, Planck's theory and photoelectric effect

Electromagnetic radiation consists of oscillating electric and magnetic fields. All such waves travel at the speed of light in vacuum, c = 3.0 × 10⁸ m s⁻¹.

• Wavelength (λ): distance between two successive crests, in m, nm or pm.
• Frequency (ν): number of waves passing a point per second, in Hz (s⁻¹).
• Wavenumber (ν̄): number of wavelengths per unit length, ν̄ = 1/λ, in m⁻¹ (or cm⁻¹).
• Relation: c = νλ.

Order of the spectrum, from short to long wavelength: γ-rays, X-rays, ultraviolet, visible (violet to red), infrared, microwaves, radio waves.

Planck's quantum theory: energy is emitted or absorbed in packets called quanta. Energy of one quantum E = hν = hc/λ, where h = 6.626 × 10⁻³⁴ J s. This explained black-body radiation.

Photoelectric effect: when light of suitable frequency falls on a metal surface, electrons are ejected.

• No electron is ejected unless the frequency is above a minimum, the threshold frequency ν₀.
• Number of electrons ejected depends on the intensity of light.
• Kinetic energy of an ejected electron depends on the frequency, not on the intensity: KE = hν − hν₀ (work function W₀ = hν₀).
• Ejection is instantaneous.

Worked example 1. Radiation of wavelength 600 nm: ν = c/λ = 3 × 10⁸ / 6 × 10⁻⁷ = 5 × 10¹⁴ Hz. Energy of one photon E = hν = 6.626 × 10⁻³⁴ × 5 × 10¹⁴ = 3.313 × 10⁻¹⁹ J.

Worked example 2 (photoelectric effect). A metal has threshold frequency 4 × 10¹⁴ Hz. Light of frequency 6 × 10¹⁴ Hz falls on it. KE = h(ν − ν₀) = 6.626 × 10⁻³⁴ × 2 × 10¹⁴ = 1.325 × 10⁻¹⁹ J.

Hydrogen spectrum and Bohr's model

An atom of hydrogen gives a line spectrum (discrete lines), not a continuous spectrum. Line spectra are characteristic of each element. The wavenumber of the lines is given by the Rydberg formula:

ν̄ = 1.097 × 10⁷ m⁻¹ × (1/n₁² − 1/n₂²), with n₂ > n₁.

Bohr's postulates for hydrogen-like species (H, He⁺, Li²⁺):

• 1. Electrons move in fixed circular paths (orbits or stationary states) without radiating energy.
• 2. The angular momentum of an electron in an orbit is an integral multiple of h/2π: mvr = nh/2π.
• 3. Energy is emitted or absorbed only when an electron jumps between orbits: ΔE = E₂ − E₁ = hν.

Formulae:

• Energy: Eₙ = −2.18 × 10⁻¹⁸ (Z²/n²) J per atom. In electron volts, E₁ for hydrogen = −13.6 eV (1 eV = 1.602 × 10⁻¹⁹ J).
• Radius: rₙ = 52.9 × (n²/Z) pm.
• Energy is negative because the electron is bound; Eₙ = 0 at n = ∞. Energy increases (becomes less negative) as n increases.
• Energy change in a transition: ΔE = 2.18 × 10⁻¹⁸ × Z² (1/n₁² − 1/n₂²) J.

Worked example 3. Hydrogen electron in n = 2: E₂ = −2.18 × 10⁻¹⁸/4 = −5.45 × 10⁻¹⁹ J. A jump from n = 2 to n = 1 releases ΔE = 2.18 × 10⁻¹⁸ (1 − 1/4) = 1.635 × 10⁻¹⁸ J.

Radius of the third orbit of Li²⁺ (Z = 3): r₃ = 52.9 × 9/3 = 158.7 pm.

Limitations of Bohr's model: it fails for atoms with more than one electron; it cannot explain the splitting of lines in a magnetic field (Zeeman effect) or an electric field (Stark effect); it does not explain the three-dimensional shape of atoms and violates the uncertainty principle by fixing exact orbits.

Dual nature of matter and uncertainty principle

de Broglie proposed that matter has wave properties: λ = h/mv = h/p. The wavelength is significant only for very small particles such as electrons. For large objects (a cricket ball) it is far too small to observe.

Worked example 4. A ball of mass 0.1 kg moving at 10 m s⁻¹ has λ = 6.626 × 10⁻³⁴ / (0.1 × 10) = 6.626 × 10⁻³⁴ m, which is undetectable. An electron (9.1 × 10⁻³¹ kg) moving at 7.28 × 10⁵ m s⁻¹ has λ ≈ 1 × 10⁻⁹ m.

Heisenberg's uncertainty principle: it is impossible to determine both the exact position and the exact momentum of a microscopic particle at the same time: Δx · Δp ≥ h/4π. It is important only for sub-atomic particles, not for everyday objects. It rules out the idea of definite orbits.

Quantum mechanical model, quantum numbers and orbitals

The quantum mechanical model is based on the Schrödinger equation. An orbital is a region of space where the probability of finding an electron is high. ψ² gives the probability density.

Counting rules: number of orbitals in a subshell = 2l + 1; in a shell = n²; maximum electrons in a shell = 2n²; each orbital holds 2 electrons.

Shapes: s orbitals are spherical; p orbitals are dumbbell-shaped, with three orientations (pₓ, p_y, p_z); four of the five d orbitals are double-dumbbell (cloverleaf) shaped, and d_z² has a dumbbell with a ring (collar) around the middle.

Nodes: a node is a place where the probability of finding the electron is zero.

• Radial nodes = n − l − 1.
• Angular nodes = l.
• Total nodes = n − 1.

Energies of orbitals and electronic configuration

In a hydrogen atom, the energy depends only on n, so 2s and 2p have the same energy. In multi-electron atoms, energy depends on n and l: the (n + l) rule says that the orbital with the lower (n + l) value has lower energy; if (n + l) is equal, the orbital with lower n has lower energy.

Order of filling: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f ... For example, 4s (n + l = 4) is lower than 3d (n + l = 5), and 3d is lower than 4p (n + l = 5, higher n).

Rules for filling:

• 1. Aufbau principle: orbitals are filled in order of increasing energy.
• 2. Pauli exclusion principle: no two electrons in an atom can have the same set of four quantum numbers. An orbital holds at most two electrons, with opposite spins.
• 3. Hund's rule: electrons occupy degenerate orbitals singly, with parallel spins, before any pairing begins.

Exceptions: Cr (Z = 24) is [Ar] 3d⁵ 4s¹, not 3d⁴ 4s². Cu (Z = 29) is [Ar] 3d¹⁰ 4s¹, not 3d⁹ 4s². Half-filled and completely filled subshells are extra stable because of symmetrical distribution and greater exchange energy.

When a transition metal forms a cation, electrons leave the 4s orbital first.

NEET traps

• 1. Isotopes: same Z; isobars: same A; isotones: same number of neutrons.
• 2. Energy of an orbit gets more negative as n decreases; n = 1 is the lowest energy state.
• 3. Bohr's radius formula has n² in the numerator and Z in the denominator; the energy formula has Z² in the numerator.
• 4. Number of radial nodes is n − l − 1, not n − l.
• 5. The maximum number of electrons in a shell is 2n²; the number of orbitals is n².
• 6. In a hydrogen atom 3s, 3p and 3d have equal energy; in multi-electron atoms they do not.
• 7. The 4s orbital is filled before 3d, but in ions 4s electrons are removed first.
• 8. The photoelectric effect: intensity changes the number of electrons, frequency changes their kinetic energy.
• 9. Hund's rule is about singly occupying degenerate orbitals; Pauli's principle is about opposite spins in one orbital.
• 10. The Lyman series is in the ultraviolet and Balmer is the only series in the visible region.

One-liners

• 1. e/m for the electron is 1.758820 × 10¹¹ C kg⁻¹.
• 2. Charge on the electron is 1.602 × 10⁻¹⁹ C (Millikan).
• 3. The neutron was discovered by Chadwick.
• 4. Number of neutrons = A − Z.
• 5. c = νλ; wavenumber = 1/λ.
• 6. E = hν, with h = 6.626 × 10⁻³⁴ J s.
• 7. Rydberg constant for hydrogen is 1.097 × 10⁷ m⁻¹.
• 8. Energy of hydrogen ground state is −2.18 × 10⁻¹⁸ J (−13.6 eV).
• 9. Radius of the first Bohr orbit of hydrogen is 52.9 pm.
• 10. de Broglie wavelength λ = h/mv.
• 11. Uncertainty principle: Δx · Δp ≥ h/4π.
• 12. For l = 2 the number of orbitals is 5; for l = 3 it is 7.
• 13. Total nodes in an orbital = n − 1.
• 14. Half-filled and full subshells are extra stable.
• 15. Fe³⁺ has five unpaired electrons.

Practice questions

1. Cathode rays are streams of

   1. neutral neutrons
   2. positively charged protons
   3. negatively charged electrons
   4. helium nuclei
   Answer

   C. negatively charged electrons

   Cathode rays are deflected towards the positive plate, so they are electrons.

2. The neutron was discovered by

   1. Chadwick
   2. Thomson
   3. Millikan
   4. Bohr
   Answer

   A. Chadwick

   Chadwick discovered the neutron in 1932.

3. The oil drop experiment was used to determine

   1. the size of the nucleus
   2. the charge on the electron
   3. the mass of the proton
   4. the charge-to-mass ratio of the electron
   Answer

   B. the charge on the electron

   Millikan's oil drop experiment gave e = 1.602 × 10⁻¹⁹ C.

4. The number of neutrons in an atom of chlorine-35 (atomic number 17) is

   1. 52
   2. 35
   3. 17
   4. 18
   Answer

   D. 18

   Neutrons = A − Z = 35 − 17 = 18.

5. Which pair of species are isotones?

   1. Hydrogen-1 and hydrogen-2
   2. Carbon-12 and carbon-14
   3. Carbon-14 and nitrogen-15
   4. Carbon-14 and nitrogen-14
   Answer

   C. Carbon-14 and nitrogen-15

   Carbon-14 has 14 − 6 = 8 neutrons and nitrogen-15 has 15 − 7 = 8 neutrons.

6. The wavenumber of radiation of wavelength 500 nm is

   1. 5 × 10⁶ m⁻¹
   2. 2 × 10⁷ m⁻¹
   3. 2 × 10⁵ m⁻¹
   4. 2 × 10⁶ m⁻¹
   Answer

   D. 2 × 10⁶ m⁻¹

   1/λ = 1/(5 × 10⁻⁷ m) = 2 × 10⁶ m⁻¹.

7. The frequency of light of wavelength 400 nm (c = 3 × 10⁸ m s⁻¹) is

   1. 7.5 × 10¹⁴ Hz
   2. 1.33 × 10¹⁵ Hz
   3. 7.5 × 10¹³ Hz
   4. 7.5 × 10¹⁵ Hz
   Answer

   A. 7.5 × 10¹⁴ Hz

   ν = c/λ = 3 × 10⁸/(4 × 10⁻⁷) = 7.5 × 10¹⁴ Hz.

8. The energy of the electron in the second orbit of He⁺ is

   1. −1.09 × 10⁻¹⁸ J
   2. −2.18 × 10⁻¹⁸ J
   3. −5.45 × 10⁻¹⁹ J
   4. −8.72 × 10⁻¹⁸ J
   Answer

   B. −2.18 × 10⁻¹⁸ J

   E = −2.18 × 10⁻¹⁸ × Z²/n² = −2.18 × 10⁻¹⁸ × 4/4 = −2.18 × 10⁻¹⁸ J.

9. The radius of the second Bohr orbit of hydrogen (r₁ = 52.9 pm) is

   1. 105.8 pm
   2. 476.1 pm
   3. 211.6 pm
   4. 52.9 pm
   Answer

   C. 211.6 pm

   r₂ = 52.9 × 2² = 211.6 pm.

10. The number of radial nodes in a 3p orbital is

    1. 0
    2. 2
    3. 3
    4. 1
    Answer

    D. 1

    Radial nodes = n − l − 1 = 3 − 1 − 1 = 1.

11. The energy of the electron in the third orbit of hydrogen is

    1. −2.42 × 10⁻¹⁹ J
    2. −6.05 × 10⁻²⁰ J
    3. −7.26 × 10⁻¹⁹ J
    4. −2.18 × 10⁻¹⁸ J
    Answer

    A. −2.42 × 10⁻¹⁹ J

    E₃ = −2.18 × 10⁻¹⁸/9 = −2.42 × 10⁻¹⁹ J.

12. Which of the following sets of quantum numbers (n, l, mₗ, mₛ) is not permitted?

    1. 3, 2, −2, +½
    2. 2, 2, 0, +½
    3. 4, 3, 3, +½
    4. 2, 1, 0, −½
    Answer

    B. 2, 2, 0, +½

    For n = 2, l can only be 0 or 1; l = 2 is not allowed.

13. The electronic configuration of chromium (Z = 24) is

    1. [Ar] 3d⁴ 4s¹ 4p¹
    2. [Ar] 3d⁵ 4s¹
    3. [Ar] 3d⁶ 4s⁰
    4. [Ar] 3d⁴ 4s²
    Answer

    B. [Ar] 3d⁵ 4s¹

    A half-filled 3d subshell with one 4s electron is extra stable.

14. The maximum number of electrons that can be accommodated in the shell with n = 3 is

    1. 18
    2. 32
    3. 8
    4. 9
    Answer

    A. 18

    2n² = 2 × 9 = 18.

15. Which is the correct order of increasing orbital energy in a multi-electron atom?

    1. 4s < 3d < 5s < 4p
    2. 4s < 4p < 3d < 5s
    3. 3d < 4s < 4p < 5s
    4. 4s < 3d < 4p < 5s
    Answer

    D. 4s < 3d < 4p < 5s

    (n + l): 4s = 4, 3d = 5, 4p = 5, 5s = 5; ties go to lower n, so 3d < 4p < 5s.

16. The de Broglie wavelength of a particle of mass m moving with velocity v is

    1. h/(m + v)
    2. hmv
    3. h/mv
    4. mv/h
    Answer

    C. h/mv

    de Broglie relation λ = h/mv.

17. Given below are two statements: Statement I: The Lyman series of the hydrogen spectrum lies in the ultraviolet region. Statement II: The Balmer series of hydrogen lies in the visible region. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Lyman (n₁ = 1) is ultraviolet; Balmer (n₁ = 2) is visible.

18. Given below are two statements: Statement I: Two electrons in the same orbital must have opposite spins. Statement II: Hund's rule states that pairing of electrons in a subshell begins before every orbital is singly occupied. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Pauli principle gives I; Hund's rule says singly occupy first, so II is wrong.

19. Given below are two statements: Statement I: The energy of an electron in a Bohr orbit becomes less negative as n decreases. Statement II: The energy of an electron at n = ∞ is zero. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Energy becomes more negative as n decreases; Eₙ = 0 at n = ∞.

20. Given below are two statements: Statement I: In Rutherford's experiment the positive charge was found to be spread uniformly through the atom. Statement II: In Rutherford's experiment most α-particles were deflected through large angles. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    Most α-particles passed undeflected; the positive charge is in a small nucleus.

21. Given below are two statements: Statement I: Photoelectrons are ejected only if the frequency of the light is above the threshold frequency. Statement II: The kinetic energy of ejected electrons increases with the frequency of the incident light. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    KE = hν − hν₀ rises with ν; no emission below ν₀.

22. Which of the following statements are correct? 1. A 2p orbital has no radial node. 2. A 3s orbital has two radial nodes. 3. A 2s orbital has one angular node.

    1. 1 and 3 only
    2. 1 and 2 only
    3. 2 and 3 only
    4. 1, 2 and 3
    Answer

    B. 1 and 2 only

    2p: 2 − 1 − 1 = 0; 3s: 3 − 0 − 1 = 2; 2s has l = 0 so no angular node.

23. Given below are two statements: Statement I: The number of orbitals in the nth shell is n². Statement II: The maximum number of electrons in the nth shell is n². Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Electrons in the shell are 2n².

24. Given below are two statements: Statement I: Heisenberg's uncertainty principle is significant for macroscopic objects. Statement II: The uncertainty principle is expressed as Δx · Δp ≥ h/4π. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    It matters only for microscopic particles; the expression is correct.

25. Which of the following statements are correct? 1. Copper has the configuration [Ar] 3d¹⁰ 4s¹. 2. Isotopes of an element differ in the number of protons. 3. Isobars have the same mass number.

    1. 1 and 2 only
    2. 2 and 3 only
    3. 1, 2 and 3
    4. 1 and 3 only
    Answer

    D. 1 and 3 only

    Isotopes differ in neutrons, so 2 is false.

26. Given below are two statements: Statement I: Thomson's model could not explain the results of the α-particle scattering experiment. Statement II: Rutherford's model could not explain the stability of the atom. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Both are recognised limitations.

27. Assertion (A): Bohr's model explains the line spectrum of hydrogen. Reason (R): Electrons in a hydrogen atom jump between discrete energy levels and emit radiation of definite frequency. Choose the correct option.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    Fixed energy gaps give lines of definite frequency, which explains A.

28. Assertion (A): The wave nature of a moving cricket ball is easily observed. Reason (R): The de Broglie wavelength is inversely proportional to momentum. Choose the correct option.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    The ball's wavelength is far too small to observe, so A is false; R is true.

29. Assertion (A): A p orbital has one angular node. Reason (R): The number of angular nodes equals n − l − 1. Choose the correct option.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    Angular nodes = l = 1 (A true); n − l − 1 is the number of radial nodes (R false).

30. Assertion (A): The Balmer series of hydrogen lies in the visible region. Reason (R): The Lyman series arises from transitions to the level n = 1. Choose the correct option.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    Both true, but R does not explain A.

31. Assertion (A): In a subshell, electrons occupy different orbitals singly before pairing (Hund's rule). Reason (R): Pairing of electrons in an orbital requires no energy. Choose the correct option.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    Hund's rule is true; pairing needs energy because of repulsion, so R is false.

32. Assertion (A): The kinetic energy of photoelectrons increases with the frequency of the light. Reason (R): The energy of a photon is directly proportional to its frequency. Choose the correct option.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    KE = hν − hν₀, and E = hν explains the rise.

33. Match List I with List II. List I: a. Lyman series b. Balmer series c. Paschen series d. Brackett series. List II (value of n₁): i. 3 ii. 1 iii. 4 iv. 2

    1. a-ii, b-iv, c-iii, d-i
    2. a-ii, b-iv, c-i, d-iii
    3. a-i, b-ii, c-iii, d-iv
    4. a-iv, b-ii, c-i, d-iii
    Answer

    B. a-ii, b-iv, c-i, d-iii

    Lyman 1, Balmer 2, Paschen 3, Brackett 4.

34. Match List I with List II. List I: a. n b. l c. mₗ d. mₛ. List II: i. orientation of the orbital ii. spin orientation of the electron iii. size and energy iv. shape of the orbital

    1. a-iv, b-iii, c-i, d-ii
    2. a-iii, b-iv, c-ii, d-i
    3. a-iii, b-iv, c-i, d-ii
    4. a-i, b-iv, c-iii, d-ii
    Answer

    C. a-iii, b-iv, c-i, d-ii

    n gives size and energy; l shape; mₗ orientation; mₛ spin.

35. Match List I (orbital) with List II (number of angular nodes). List I: a. 3s b. 3p c. 3d d. 4f. List II: i. 1 ii. 3 iii. 0 iv. 2

    1. a-iii, b-i, c-ii, d-iv
    2. a-iii, b-iv, c-i, d-ii
    3. a-i, b-iii, c-iv, d-ii
    4. a-iii, b-i, c-iv, d-ii
    Answer

    D. a-iii, b-i, c-iv, d-ii

    Angular nodes = l: 0, 1, 2, 3.

36. Match List I with List II. List I: a. Thomson b. Rutherford c. Millikan d. Chadwick. List II: i. nuclear model ii. e/m ratio of the electron iii. charge on the electron iv. discovery of the neutron

    1. a-ii, b-i, c-iv, d-iii
    2. a-ii, b-i, c-iii, d-iv
    3. a-iii, b-i, c-ii, d-iv
    4. a-i, b-ii, c-iii, d-iv
    Answer

    B. a-ii, b-i, c-iii, d-iv

    Standard contributions.

37. Match List I with List II. List I: a. Cu b. Cr c. Fe²⁺ d. Zn²⁺. List II (d-electron arrangement): i. 3d⁶ ii. 3d¹⁰ iii. 3d⁵ 4s¹ iv. 3d¹⁰ 4s¹

    1. a-iv, b-iii, c-i, d-ii
    2. a-iii, b-iv, c-i, d-ii
    3. a-ii, b-iii, c-i, d-iv
    4. a-iv, b-iii, c-ii, d-i
    Answer

    A. a-iv, b-iii, c-i, d-ii

    Cu [Ar]3d¹⁰4s¹; Cr [Ar]3d⁵4s¹; Fe²⁺ 3d⁶; Zn²⁺ 3d¹⁰.

38. Match List I with List II. List I: a. Aufbau principle b. Pauli exclusion principle c. Hund's rule d. Heisenberg principle. List II: i. no two electrons have the same four quantum numbers ii. orbitals are filled in order of increasing energy iii. pairing starts only after degenerate orbitals are singly occupied iv. position and momentum cannot both be known exactly

    1. a-iii, b-i, c-ii, d-iv
    2. a-ii, b-i, c-iv, d-iii
    3. a-i, b-ii, c-iii, d-iv
    4. a-ii, b-i, c-iii, d-iv
    Answer

    D. a-ii, b-i, c-iii, d-iv

    Each principle matched to its statement.

39. The energy of a photon of frequency 5 × 10¹⁴ Hz (h = 6.626 × 10⁻³⁴ J s) is

    1. 1.325 × 10⁻¹⁹ J
    2. 3.313 × 10⁻²⁰ J
    3. 3.313 × 10⁻¹⁹ J
    4. 6.626 × 10⁻¹⁹ J
    Answer

    C. 3.313 × 10⁻¹⁹ J

    E = hν = 6.626 × 10⁻³⁴ × 5 × 10¹⁴ = 3.313 × 10⁻¹⁹ J.

40. A metal has a threshold frequency of 4 × 10¹⁴ Hz. Light of frequency 6 × 10¹⁴ Hz falls on it (h = 6.626 × 10⁻³⁴ J s). The maximum kinetic energy of the ejected electrons is

    1. 2.651 × 10⁻¹⁹ J
    2. 1.325 × 10⁻¹⁹ J
    3. 6.626 × 10⁻²⁰ J
    4. 3.976 × 10⁻¹⁹ J
    Answer

    B. 1.325 × 10⁻¹⁹ J

    KE = h(ν − ν₀) = 6.626 × 10⁻³⁴ × 2 × 10¹⁴ = 1.325 × 10⁻¹⁹ J.

41. The energy released when the electron in a hydrogen atom falls from n = 2 to n = 1 is

    1. 2.18 × 10⁻¹⁸ J
    2. 5.45 × 10⁻¹⁹ J
    3. 1.635 × 10⁻¹⁸ J
    4. 1.09 × 10⁻¹⁸ J
    Answer

    C. 1.635 × 10⁻¹⁸ J

    ΔE = 2.18 × 10⁻¹⁸ (1 − 1/4) = 1.635 × 10⁻¹⁸ J.

42. The de Broglie wavelength of a ball of mass 0.1 kg moving at 10 m s⁻¹ (h = 6.626 × 10⁻³⁴ J s) is

    1. 6.626 × 10⁻³⁶ m
    2. 6.626 × 10⁻³³ m
    3. 6.626 × 10⁻³⁵ m
    4. 6.626 × 10⁻³⁴ m
    Answer

    D. 6.626 × 10⁻³⁴ m

    λ = h/mv = 6.626 × 10⁻³⁴/(0.1 × 10) = 6.626 × 10⁻³⁴ m.

43. The number of unpaired electrons in Fe³⁺ (atomic number 26) is

    1. 4
    2. 5
    3. 3
    4. 6
    Answer

    B. 5

    Fe³⁺ is [Ar] 3d⁵, and the five d orbitals are singly occupied.

44. Using R_H = 1.097 × 10⁷ m⁻¹, the wavenumber of the first line of the Lyman series of hydrogen is approximately

    1. 8.23 × 10⁶ m⁻¹
    2. 2.74 × 10⁶ m⁻¹
    3. 1.52 × 10⁶ m⁻¹
    4. 1.097 × 10⁷ m⁻¹
    Answer

    A. 8.23 × 10⁶ m⁻¹

    ν̄ = R(1/1² − 1/2²) = 0.75 × 1.097 × 10⁷ = 8.23 × 10⁶ m⁻¹.

45. The radius of the third Bohr orbit of Li²⁺ (r₁ of hydrogen = 52.9 pm) is

    1. 17.6 pm
    2. 52.9 pm
    3. 158.7 pm
    4. 476.1 pm
    Answer

    C. 158.7 pm

    r = 52.9 × n²/Z = 52.9 × 9/3 = 158.7 pm.