Chemistry

Atomic Structure

⚛️ Chemistry · Atomic Structure
📅 Updated Jan 2026 ⏱️ 30 min read 🎯 High-yield

Atomic Structure — Quantum Numbers, Orbitals & Configuration

Complete MDCAT Chemistry chapter on atomic structure: proton discovery, photon energy, orbital shapes (s/p/d), quantum numbers, hydrogen spectrum (Lyman, Balmer, Paschen), and electronic configuration rules (Aufbau, Pauli, Hund). Includes 170 solved MCQs with explanations and year-wise past papers.

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⚡Quick Answers to Common Searches

Direct, snippet-optimized answers to the top questions students search about atomic structure.

What are canal rays?

Canal rays (positive rays) are beams of positive ions formed when cathode rays ionize residual gas in a discharge tube. Discovered by Goldstein (1886). e/m depends on the gas.

What is a photon?

A photon is a quantum of electromagnetic energy with no mass or charge. Energy: E = hν = hc/λ. h = 6.626 × 10⁻³⁴ J·s.

What are the 4 quantum numbers?

n (principal, shell), l (azimuthal, subshell), m_l (magnetic, orbital), m_s (spin, ±½).

Shapes of orbitals?

s = spherical, p = dumbbell (3 orbitals), d = cloverleaf (4) + d_z² (doughnut), f = complex.

Max electrons in a shell?

Max in shell n = 2n². n=1 → 2, n=2 → 8, n=3 → 18, n=4 → 32.

What is Aufbau principle?

Electrons fill lowest energy orbitals first. Order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s...

What is Pauli exclusion?

No two electrons in the same atom can have all 4 quantum numbers identical. Max 2 e⁻ per orbital with opposite spins.

What is Hund's rule?

In degenerate orbitals, electrons fill singly first with parallel spins before pairing. Minimizes repulsion.

Cr exceptional config?

Cr (Z=24): [Ar] 4s¹ 3d⁵ (not 4s² 3d⁴) — half-filled d stability.

Cu exceptional config?

Cu (Z=29): [Ar] 4s¹ 3d¹⁰ (not 4s² 3d⁹) — fully-filled d stability.

Lyman vs Balmer?

Lyman: n→1, UV. Balmer: n→2, visible (only visible series). Paschen: n→3, IR.

Bohr's angular momentum?

Electron's angular momentum quantized: mvr = nh/2π. Only works for 1-electron species (H, He⁺, Li²⁺).

Radius ∝ ?

Bohr radius: r ∝ n²/Z. 2nd orbit of H = 4× radius of 1st orbit.

Energy ∝ ?

For H-like atoms: E = -13.6 × Z²/n² eV. Ground state H = -13.6 eV.

Isoelectronic species sizes?

With same electrons, size decreases as nuclear charge increases. Cl⁻ (Z=17) > K⁺ (Z=19) > Ca²⁺ (Z=20).

❓People Also Ask — Atomic Structure

Questions students most frequently search about this chapter. Answers are direct and exam-focused.

QWhat are canal rays?
Canal rays (positive rays) are streams of positively charged ions produced in a discharge tube when cathode rays ionize residual gas molecules. They were discovered by Eugen Goldstein in 1886 using a perforated cathode. Their e/m ratio varies with the gas used, unlike electrons.
QWhat is a photon?
A photon is a quantum (discrete packet) of electromagnetic radiation energy with no mass and no charge. Energy: E = hν = hc/λ, where h is Planck's constant (6.626 × 10⁻³⁴ J·s).
QWhat are the four quantum numbers?
(1) Principal n — shell size/energy. (2) Azimuthal l — subshell shape (0=s, 1=p, 2=d, 3=f). (3) Magnetic m_l — orbital orientation (-l to +l). (4) Spin m_s — electron spin (+½ or -½).
QWhat are the shapes of s, p, and d orbitals?
s = spherical. p = dumbbell (3 orbitals: pₓ, pᵧ, p_z). d = cloverleaf for four orbitals + d_z² has two lobes with a doughnut ring. f = complex 8-lobed shapes.
QWhat is the Aufbau principle?
Electrons fill lowest-energy atomic orbitals first. Filling order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p. Note: 4s fills before 3d but loses electrons before 3d during ionization.
QWhat is the Pauli Exclusion Principle?
No two electrons in the same atom can have the same set of four quantum numbers. Result: each orbital holds a maximum of 2 electrons with opposite spins (+½ and -½).
QWhat is Hund's rule?
When electrons occupy degenerate orbitals (same energy), they fill singly first with parallel spins before any pairing occurs. This minimizes electron-electron repulsion.
QWhy does chromium have an exceptional configuration?
Chromium (Z=24) has [Ar] 4s¹ 3d⁵ instead of expected 4s² 3d⁴. A half-filled d subshell (d⁵) with parallel spins is more stable due to maximum exchange energy and symmetry.
QWhy does copper have an exceptional configuration?
Copper (Z=29) has [Ar] 4s¹ 3d¹⁰ instead of 4s² 3d⁹. A fully-filled d subshell (d¹⁰) offers maximum exchange energy and stability.
QDifference between Lyman and Balmer series?
Lyman: transitions to n=1, emits UV. Balmer: transitions to n=2, emits visible light (only visible series). Paschen: transitions to n=3, emits IR.
QMaximum electrons in a shell?
2n². n=1 → 2, n=2 → 8, n=3 → 18, n=4 → 32.
QWhat are degenerate orbitals?
Orbitals within the same subshell (e.g., pₓ, pᵧ, p_z) that have same energy but different orientations. They are filled singly first per Hund's rule.
QHow many electrons can a d subshell hold?
A d subshell (l=2) has 5 orbitals (m_l = -2 to +2). Each holds 2 electrons → max 10 electrons. s=2, p=6, d=10, f=14.
QRelationship between energy and wavelength?
E = hc/λ, so energy is inversely proportional to wavelength. Shorter wavelength → higher energy. Gamma rays highest, radio waves lowest.
QHow many MCQs from Atomic Structure in MDCAT?
Typically 2–3 MCQs. High-yield because quantum numbers, orbital filling, and electronic configuration are foundational to all later chemistry.
QWhat is the (n+l) rule?
Also called Madelung rule: orbitals fill in order of increasing (n+l). For equal (n+l), lower n fills first. E.g., 4s (n+l=4) before 3d (n+l=5).
QHow many radial nodes in an orbital?
Radial nodes = n - l - 1. Angular nodes = l. Total nodes = n - 1. Example: 3p has 1 radial + 1 angular = 2 total.
QWhat is the Bohr radius?
The radius of the first Bohr orbit of hydrogen is 0.529 Å (52.9 pm). For nth orbit: r_n = n² × 0.529 Å.
QParamagnetic vs diamagnetic?
Paramagnetic: has unpaired electrons, attracted by magnetic field. Diamagnetic: all paired, weakly repelled by magnetic field.
QWhat is the Heisenberg Uncertainty Principle?
It is impossible to determine both the exact position and momentum of an electron simultaneously. Δx · Δp ≥ h/4π. This is why orbitals are probability regions, not fixed paths.

📋Syllabus & Learning Outcomes

According to PMDC/PMC MDCAT syllabus, Atomic Structure covers:

📝Full Notes — Atomic Structure

1. Discovery of Proton (Positive Rays)

  • Goldstein (1886): used perforated cathode in discharge tube → observed rays behind cathode (positive rays or canal rays).
  • Positive rays consist of positively charged ions formed from residual gas in the tube.
  • e/m ratio of positive rays is smaller than that of electrons. The e/m depends on the gas used (unlike electrons).
  • The smallest positive ion (from hydrogen) is the proton (H⁺). Mass ≈ 1837 × mass of electron.
  • Later, Rutherford identified the proton as a fundamental particle in the nucleus (1919).

2. Photon as a Unit of Radiation Energy

  • Photon: A quantum of electromagnetic radiation; has energy E = hν = hc/λ, where h = Planck's constant (6.626 × 10⁻³⁴ J·s), ν = frequency, λ = wavelength, c = speed of light.
  • Photon has no mass and no charge; energy is directly proportional to frequency.
  • Introduced by Max Planck (1900) and Einstein (1905) to explain black-body radiation and photoelectric effect.
  • Key for atomic spectra: electrons absorb/emit photons of specific energies when jumping between energy levels.

3. Orbitals, Energy Levels, Sub-levels

  • Shell (principal energy level): designated by n = 1, 2, 3, … Maximum electrons = 2n².
  • Sub-shell (sub-level): s, p, d, f (based on azimuthal quantum number l = 0,1,2,3). Number of sub-levels in a shell = n.
  • Orbitals: regions in space where probability of finding an electron is maximum. Each orbital can hold max 2 electrons (Pauli).
  • Quantum numbers:
    • n – principal (size/energy)
    • l – azimuthal (shape) (0=s, 1=p, 2=d, 3=f)
    • m_l – magnetic (orientation of orbital in space, -l to +l)
    • m_s – spin (±½)
  • Sub-shell orbital count: s=1 orbital, p=3, d=5, f=7.

4. Shapes of s, p, d Orbitals

  • s orbital: spherical, symmetrical around nucleus; no nodal plane passing through nucleus (l=0).
  • p orbitals: dumbbell shape; three orbitals (pₓ, pᵧ, p_z) mutually perpendicular; each has a nodal plane passing through nucleus (l=1).
  • d orbitals: five orbitals (d_xy, d_xz, d_yz, d_x²−y², d_z²); cloverleaf shapes except d_z² which has a doughnut around the centre.

5. Hydrogen Atom — Quantum Theory

  • Bohr model: Energy of electron in nth orbit: Eₙ = −2.18 × 10⁻¹⁸ J × (Z²/n²). For hydrogen Z=1.
  • Emission/absorption: ΔE = hν = E_hi − E_lo. Balmer series (visible, n=2), Lyman (UV, n=1), Paschen (IR, n=3).
  • Quantum mechanical model: Electron behaviour described by wavefunction ψ; |ψ|² gives probability density. Orbitals are solutions to Schrödinger equation.
  • For hydrogen, energy depends only on n (degeneracy) in non-relativistic model.

6. Aufbau, Pauli, Hund's Rule

  • Aufbau Principle: Electrons fill lowest energy orbitals first. Order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, …
  • Pauli Exclusion Principle: No two electrons in an atom can have the same set of four quantum numbers. An orbital holds max 2 electrons with opposite spins.
  • Hund's Rule: Electrons fill degenerate orbitals singly with parallel spins before pairing. Minimises repulsion.

7. Writing Electronic Configuration

  • Use standard notation: e.g., C (Z=6): 1s² 2s² 2p².
  • Exceptions: Cr (Z=24): [Ar] 4s¹ 3d⁵ (half-filled stability); Cu (Z=29): [Ar] 4s¹ 3d¹⁰ (fully-filled stability).
  • For ions: remove electrons from the highest n first (e.g., Fe²⁺ from 4s then 3d).
  • Use noble gas shorthand: [He], [Ne], [Ar], etc.

High-Yield Points & Common Traps

  • Positive rays: e/m varies with gas; proton is H⁺.
  • Photon energy: E = hν = hc/λ; no mass.
  • Orbitals: s=sphere, p=dumbbell, d=cloverleaf.
  • Max electrons in a shell = 2n²; sub-levels s (2 e⁻), p (6 e⁻), d (10 e⁻), f (14 e⁻).
  • Aufbau order: 4s fills before 3d; but 4s loses electrons first when ionising.
  • Hund's rule: degenerate orbitals fill singly first.
  • Chromium (Cr): [Ar] 4s¹ 3d⁵; Copper (Cu): [Ar] 4s¹ 3d¹⁰ – due to half-filled/fully-filled d subshell stability.
  • Paramagnetic = unpaired electrons; diamagnetic = all paired.
  • Bohr model works only for one-electron species (H, He⁺, Li²⁺).
  • Traps: "4s is always lower energy than 3d for all atoms" → False (for multi-electron beyond K/Ca, 3d becomes lower). "Orbital is a fixed circular path" → False (probability region). "Photon has mass" → False. "Electron's spin means it rotates" → False (intrinsic quantum property).

✅Practice MCQs — Test 1

🧪 Basic Concepts

50 MCQs
1Canal rays were discovered by:
  • A. J.J. Thomson
  • B. William Crookes
  • C. Goldstein
  • D. James Chadwick
Answer: C. Goldstein
Eugen Goldstein discovered canal rays (positive rays) in 1886.
2The e/m ratio of positive rays is maximum for which gas?
  • A. Helium
  • B. Oxygen
  • C. Nitrogen
  • D. Hydrogen
Answer: D. Hydrogen
Hydrogen produces H⁺ ions with the smallest mass, giving the highest e/m ratio.
3The production of positive rays in a discharge tube is due to:
  • A. Ionization of gas molecules by cathode rays
  • B. Emission from the anode
  • C. Emission from the cathode
  • D. Radioactive decay of gas
Answer: A
Cathode rays (electrons) collide with gas molecules, ionizing them to produce positive ions.
4The mass of a proton is approximately how many times heavier than an electron?
  • A. 1836
  • B. 1840
  • C. 2000
  • D. 1000
Answer: A. 1836
Proton mass ≈ 1.6726×10⁻²⁷ kg, electron mass ≈ 9.109×10⁻³¹ kg, ratio ≈ 1836.
5Positive rays travel in a direction:
  • A. Parallel to cathode rays
  • B. Opposite to cathode rays
  • C. Perpendicular to cathode rays
  • D. Circular to cathode rays
Answer: B
Positive rays move from anode to cathode (opposite to electron flow).
6Which of the following is NOT a property of canal rays?
  • A. They carry a positive charge
  • B. They produce fluorescence
  • C. Their e/m ratio is independent of the gas used
  • D. They travel in straight lines
Answer: C
The e/m ratio of positive rays depends on the gas (mass and charge of ions).
7When the pressure in a discharge tube is reduced, the positive rays:
  • A. Disappear
  • B. Become more intense
  • C. Change their charge
  • D. Stop hitting the walls
Answer: B
At low pressure, more gas molecules are ionized, increasing positive ray intensity.
8The nature of positive rays depends upon:
  • A. The nature of the electrode
  • B. The nature of the residual gas
  • C. The voltage applied
  • D. The size of the discharge tube
Answer: B
Positive rays are ions of the residual gas; their properties vary with the gas.
9The charge on a proton is:
  • A. +1.602 × 10⁻¹⁹ C
  • B. -1.602 × 10⁻¹⁹ C
  • C. +1.602 × 10⁻²⁴ C
  • D. Zero
Answer: A
Proton charge is +1.602×10⁻¹⁹ C (elementary charge).
10Positive rays are also known as:
  • A. X-rays
  • B. Canal rays
  • C. Alpha rays
  • D. Beta rays
Answer: B. Canal rays
Canal rays is another name for positive rays.
11According to Planck's Quantum Theory, energy is emitted or absorbed in discrete packets called:
  • A. Electrons
  • B. Quanta
  • C. Positrons
  • D. Orbits
Answer: B. Quanta
Planck introduced the concept of quanta (photons) in 1900.
12The energy of a photon is inversely proportional to its:
  • A. Frequency
  • B. Velocity
  • C. Wavelength
  • D. Wave number
Answer: C. Wavelength
E = hc/λ, so energy ∝ 1/λ.
13The value of Planck's constant (h) is:
  • A. 6.626 × 10⁻³⁴ J.s
  • B. 6.626 × 10⁻²⁷ J.s
  • C. 3.0 × 10⁸ m/s
  • D. 1.6 × 10⁻¹⁹ J
Answer: A
Planck's constant h = 6.626×10⁻³⁴ J·s.
14Which of the following equations represents Planck's equation?
  • A. E = mc²
  • B. E = hν
  • C. E = h/λ
  • D. E = V/d
Answer: B. E = hν
E = hν is Planck's quantum energy equation.
15A unit of radiation energy is called a:
  • A. Joule
  • B. Photon
  • C. Hertz
  • D. Electron volt
Answer: B. Photon
A photon is a quantum (packet) of electromagnetic energy.
16The relationship between energy (E) and wave number (ν̄) is:
  • A. E = hcν̄
  • B. E = h/cν̄
  • C. E = c/hν̄
  • D. E = hc/ν̄
Answer: A
ν̄ = 1/λ, so E = hc/λ = hcν̄.
17If the frequency of radiation is doubled, its energy will:
  • A. Remain the same
  • B. Be halved
  • C. Be doubled
  • D. Be quadrupled
Answer: C. Be doubled
E ∝ ν, so doubling frequency doubles energy.
18The velocity of a photon is equal to:
  • A. Velocity of sound
  • B. Velocity of light
  • C. Velocity of an electron
  • D. Infinity
Answer: B. Velocity of light
Photons travel at the speed of light (c = 3×10⁸ m/s).
19Which region of the electromagnetic spectrum has the highest energy photons?
  • A. Radio waves
  • B. Infrared
  • C. Ultraviolet
  • D. Gamma rays
Answer: D. Gamma rays
Gamma rays have highest frequency, hence highest energy.
20The concept of 'Dual Nature of Matter' was inspired by the quantum nature of:
  • A. Protons
  • B. Light
  • C. Neutrons
  • D. Atoms
Answer: B. Light
De Broglie proposed matter waves after the dual nature of light was established.
21The quantum number that describes the orientation of an orbital in space is:
  • A. Principal Quantum Number
  • B. Azimuthal Quantum Number
  • C. Magnetic Quantum Number
  • D. Spin Quantum Number
Answer: C
Magnetic quantum number (m) specifies orbital orientation.
22For n=3, the possible values of l (Azimuthal Quantum Number) are:
  • A. 1, 2, 3
  • B. 0, 1, 2
  • C. 0, 1
  • D. 1, 2
Answer: B. 0, 1, 2
l ranges from 0 to n-1, so for n=3: l=0,1,2.
23The total number of orbitals in a shell with principal quantum number n is:
  • A. 2n
  • B. 2n²
  • C. n²
  • D. n-1
Answer: C. n²
Number of orbitals in shell = n².
24A sub-level with l = 2 is called:
  • A. s-subshell
  • B. p-subshell
  • C. d-subshell
  • D. f-subshell
Answer: C. d-subshell
l=0 → s, l=1 → p, l=2 → d, l=3 → f.
25The maximum number of electrons in a 4f subshell is:
  • A. 2
  • B. 6
  • C. 10
  • D. 14
Answer: D. 14
f subshell has 7 orbitals, each holds 2 electrons → 14.
26The Principal Quantum Number (n) represents:
  • A. Shape of orbital
  • B. Energy and size of shell
  • C. Spin of electron
  • D. Orientation of orbital
Answer: B
n indicates energy level and average distance from nucleus.
27An orbital can accommodate a maximum of how many electrons?
  • A. 2
  • B. 6
  • C. 10
  • D. 14
Answer: A. 2
Pauli exclusion principle: each orbital holds max 2 electrons with opposite spins.
28Which quantum number distinguishes between two electrons present in the same orbital?
  • A. n
  • B. l
  • C. m
  • D. s
Answer: D. s
Spin quantum number (s) differentiates electrons in the same orbital.
29If l = 1, the values of m (magnetic quantum number) are:
  • A. 0
  • B. +1, 0, -1
  • C. +2, +1, 0, -1, -2
  • D. +1/2, -1/2
Answer: B
For l=1, m = -1, 0, +1.
30The degenerate orbitals have:
  • A. Same shape but different energy
  • B. Same energy but different orientation
  • C. Different shape and different energy
  • D. Different energy but same orientation
Answer: B
Degenerate orbitals have the same energy but different spatial orientations.
31The shape of an 's' orbital is:
  • A. Dumb-bell
  • B. Spherical
  • C. Clover-leaf
  • D. Linear
Answer: B. Spherical
s orbitals are spherically symmetric.
32The number of nodes in a 2s orbital is:
  • A. 0
  • B. 1
  • C. 2
  • D. 3
Answer: B. 1
Radial nodes = n-l-1; for 2s (n=2,l=0) → 2-0-1=1.
33Which of the following 'p' orbitals is directed along the z-axis?
  • A. p_x
  • B. p_y
  • C. p_z
  • D. All of them
Answer: C. p_z
p_z orbital is oriented along the z-axis.
34The d-orbital that has a 'doughnut' or 'collar' shape is:
  • A. d_xy
  • B. d_{x²-y²}
  • C. d_{z²}
  • D. d_yz
Answer: C. d_{z²}
d_{z²} has a toroidal (doughnut) shape around the center.
35A p-orbital has how many lobes?
  • A. 1
  • B. 2
  • C. 3
  • D. 4
Answer: B. 2
Each p orbital has two lobes (dumbbell shape).
36The Lyman series of the hydrogen spectrum lies in which region?
  • A. Visible
  • B. Infrared
  • C. Ultraviolet
  • D. X-ray
Answer: C. Ultraviolet
Lyman series (n≥2 → n=1) emits UV radiation.
37When an electron jumps from n=3 to n=2, the spectral line belongs to:
  • A. Lyman series
  • B. Balmer series
  • C. Paschen series
  • D. Brackett series
Answer: B. Balmer series
Balmer series transitions end at n=2.
38The only series in the Hydrogen spectrum that lies in the visible region is:
  • A. Lyman
  • B. Balmer
  • C. Pfund
  • D. Paschen
Answer: B. Balmer
Balmer series (n≥3 → n=2) is visible.
39The wave number of the first line of the Lyman series is (where R_H is Rydberg constant):
  • A. 3/4 R_H
  • B. 5/36 R_H
  • C. 1/4 R_H
  • D. 8/9 R_H
Answer: A
First Lyman line: n=2→1, ν̄ = R_H(1/1² - 1/2²) = R_H(1 - 1/4) = 3/4 R_H.
40According to Bohr's theory, the angular momentum of an electron is quantized as:
  • A. nh/2π
  • B. h/2π
  • C. n²h/2π
  • D. nh/4π
Answer: A. nh/2π
Bohr's quantization: mvr = n(h/2π).
41The series obtained when an electron jumps to the 4th energy level (n=4) from higher levels is:
  • A. Paschen
  • B. Brackett
  • C. Pfund
  • D. Balmer
Answer: B. Brackett
Brackett series: transitions to n=4.
42Hydrogen spectrum is an example of:
  • A. Continuous spectrum
  • B. Line spectrum
  • C. Band spectrum
  • D. Absorption spectrum
Answer: B. Line spectrum
Hydrogen emits a line spectrum (discrete wavelengths).
43Which rule states that electrons enter the lowest energy orbital available?
  • A. Pauli Exclusion Principle
  • B. Hund's Rule
  • C. Aufbau Principle
  • D. Heisenberg Uncertainty Principle
Answer: C. Aufbau Principle
Aufbau principle: 'building up' from lowest energy.
44The electronic configuration of Copper (Z=29) is:
  • A. [Ar] 4s² 3d⁹
  • B. [Ar] 4s¹ 3d¹⁰
  • C. [Ar] 4s² 3d¹⁰
  • D. [Ar] 3d⁹ 4s¹
Answer: B. [Ar] 4s¹ 3d¹⁰
Copper exception: 4s¹ 3d¹⁰ (full d stability).
45Hund's rule is applicable when:
  • A. Electrons enter a new shell
  • B. Electrons are filled in degenerate orbitals
  • C. Electrons are removed from an atom
  • D. Orbitals are fully filled
Answer: B
Hund's rule: degenerate orbitals are singly occupied first with parallel spins.
46Which of the following sets of quantum numbers is NOT possible?
  • A. n=2, l=1, m=0
  • B. n=3, l=2, m=-2
  • C. n=1, l=1, m=0
  • D. n=4, l=0, m=0
Answer: C
For n=1, l can only be 0; l=1 is impossible.
47The electronic configuration of Cr³⁺ (Z=24) is:
  • A. [Ar] 4s¹ 3d²
  • B. [Ar] 3d³
  • C. [Ar] 4s² 3d¹
  • D. [Ar] 3d⁵
Answer: B. [Ar] 3d³
Cr ground state: [Ar] 4s¹ 3d⁵; remove 3 electrons → 3d³.
48The Pauli Exclusion Principle states that:
  • A. Electrons revolve in circular orbits
  • B. No two electrons in an atom can have the same four quantum numbers
  • C. Orbitals are filled in order of increasing (n+l)
  • D. Degenerate orbitals should be singly occupied first
Answer: B
Pauli: no two electrons can have identical n, l, m, s.
49The sequence of filling orbitals is 1s, 2s, 2p, 3s, ... Which orbital comes after 3p?
  • A. 3d
  • B. 4s
  • C. 4p
  • D. 4d
Answer: B. 4s
After 3p, the next is 4s (energy order: 4s < 3d).
50An atom with atomic number 15 has how many unpaired electrons?
  • A. 1
  • B. 2
  • C. 3
  • D. 0
Answer: C. 3
P (Z=15): [Ne] 3s² 3p³ → three unpaired electrons (Hund's rule).

✅Practice MCQs — Test 2

🧪 Advanced Concepts

50 MCQs
1The e/m ratio of canal rays for different gases varies because:
  • A. The charge on the ions is always the same, but masses differ
  • B. Both the mass and the net charge of the ions can vary depending on the gas
  • C. The velocity of the positive rays is constant for all gases
  • D. Protons are not actually present in all gases
Answer: B
Different gases produce ions with different mass-to-charge ratios.
2Which of the following ions would undergo the smallest deflection in a magnetic field when produced as positive rays?
  • A. H⁺
  • B. He²⁺
  • C. O₂⁺
  • D. Ne⁺
Answer: C. O₂⁺
Deflection ∝ e/m; O₂⁺ has largest mass among options (32 amu), smallest e/m.
3If the mass of an electron is m_e, the mass of a nucleus of Protium is approximately:
  • A. 1836 × m_e
  • B. 1 × m_e
  • C. 1840 × m_e
  • D. 3672 × m_e
Answer: A. 1836 × m_e
Proton mass ≈ 1836 m_e.
4The e/m ratio of a proton is 9.58 × 10⁷ C/kg. The e/m ratio of an α-particle is:
  • A. 9.58 × 10⁷ C/kg
  • B. 19.16 × 10⁷ C/kg
  • C. 4.79 × 10⁷ C/kg
  • D. 2.39 × 10⁷ C/kg
Answer: C
α-particle has charge +2e, mass 4 u → e/m = (2e)/(4m_p) = (1/2)(e/m_p) = 4.79×10⁷ C/kg.
5Positive rays were observed to pass through a perforated cathode. If the anode was perforated instead, we would observe:
  • A. Similar positive rays traveling toward the cathode
  • B. Cathode rays passing through the anode
  • C. No rays passing through the anode
  • D. Canal rays traveling in the same direction as cathode rays
Answer: C
Cathode rays (electrons) would pass through a perforated anode; positive rays would not be produced at the anode.
6The ratio of energies of two photons with wavelengths 2000 Å and 4000 Å is:
  • A. 1:2
  • B. 2:1
  • C. 4:1
  • D. 1:4
Answer: B. 2:1
E ∝ 1/λ; λ₁/λ₂ = 1/2 → E₁/E₂ = 2:1.
7According to Planck, a body can emit or absorb energy only in integral multiples of hν. This implies that energy is:
  • A. Continuous
  • B. Quantized
  • C. Infinite
  • D. Independent of frequency
Answer: B. Quantized
Energy is quantized, not continuous.
8The number of photons of light with a wavelength of 4000 Å that provide 1 Joule of energy is approximately:
  • A. 2 × 10¹⁸
  • B. 4 × 10¹⁸
  • C. 10²⁰
  • D. 3 × 10¹⁵
Answer: A
Energy per photon = hc/λ = (6.63×10⁻³⁴×3×10⁸)/(4000×10⁻¹⁰) = 4.97×10⁻¹⁹ J; Number = 1 / (4.97×10⁻¹⁹) ≈ 2×10¹⁸.
9Which of the following represents the correct slope of a graph plotted between Energy (E) and Frequency (ν)?
  • A. c
  • B. h
  • C. 1/h
  • D. λ
Answer: B. h
E = hν, slope = h.
10The work function of a metal is E₀. If light of energy 2E₀ falls on it, the maximum kinetic energy of the photoelectrons is:
  • A. E₀
  • B. 2E₀
  • C. Zero
  • D. 0.5E₀
Answer: A. E₀
K.E. = hν - φ = 2E₀ - E₀ = E₀.
11For a principal quantum number n=4, the total number of subshells and the maximum number of electrons are respectively:
  • A. 4 and 16
  • B. 4 and 32
  • C. 3 and 18
  • D. 16 and 32
Answer: B
Subshells: 4s,4p,4d,4f (4 subshells). Max electrons = 2n² = 2×16 = 32.
12The orbital angular momentum of an electron in a 2s orbital is:
  • A. h/2π
  • B. Zero
  • C. √2 h/2π
  • D. 0.5 h/2π
Answer: B. Zero
For s orbital (l=0), angular momentum = √[l(l+1)] (h/2π) = 0.
13Which of the following subshells is NOT allowed according to quantum mechanical rules?
  • A. 2p
  • B. 3f
  • C. 4d
  • D. 5s
Answer: B. 3f
For n=3, l max = 2 (d); f subshell (l=3) requires n≥4.
14If the magnetic quantum number m has a value of -3, the minimum possible value of the principal quantum number n is:
  • A. 3
  • B. 4
  • C. 5
  • D. 2
Answer: B. 4
m ranges from -l to +l, so l ≥ 3. Minimum n for l=3 is n=4 (4f).
15The number of radial nodes in a 3p orbital is:
  • A. 1
  • B. 2
  • C. 0
  • D. 3
Answer: A. 1
Radial nodes = n-l-1 = 3-1-1 = 1.
16The total number of values of the magnetic quantum number (m) for a given value of l is:
  • A. l+1
  • B. 2l+1
  • C. n²
  • D. 2l-1
Answer: B. 2l+1
m takes integer values from -l to +l → total 2l+1 values.
17Two electrons in the same orbital have the same n, l, and m values. They must have:
  • A. Same spin
  • B. Parallel spin
  • C. Opposite spin
  • D. Zero spin
Answer: C. Opposite spin
Pauli exclusion: electrons in same orbital have opposite spins.
18Which subshell has the lowest energy for an atom with more than one electron?
  • A. n=3, l=2
  • B. n=4, l=0
  • C. n=4, l=1
  • D. n=3, l=1
Answer: D
Lower (n+l) means lower energy. Among same (n+l), lower n has lower energy → 3p (n=3,l=1, n+l=4) has lowest.
19The azimuthal quantum number for the last electron of Sodium (Z=11) is:
  • A. 0
  • B. 1
  • C. 2
  • D. 3
Answer: A. 0
Na: [Ne] 3s¹; last electron in s (l=0).
20The 'Spin' quantum number is required to explain:
  • A. The shape of the orbital
  • B. The Zeeman effect
  • C. The fine structure of spectral lines/Doublet formation
  • D. The energy of the shell
Answer: C
Spin explains fine splitting of spectral lines (e.g., sodium D doublet).
21The probability of finding an electron in the xy plane for a p_z orbital is:
  • A. Maximum
  • B. Zero
  • C. 50%
  • D. Constant
Answer: B. Zero
p_z orbital has nodal plane in xy plane; probability is zero.
22How many d-orbitals have lobes lying directly along the axes?
  • A. 2
  • B. 3
  • C. 5
  • D. 1
Answer: A. 2
d_{x²-y²} and d_{z²} have lobes along axes; others are between axes.
23The nodal plane of a p_y orbital is the:
  • A. xy plane
  • B. yz plane
  • C. xz plane
  • D. z axis
Answer: C. xz plane
p_y orbital has nodal plane xz (where y=0).
24Which d-orbital does not have four lobes?
  • A. d_xy
  • B. d_{x²-y²}
  • C. d_{z²}
  • D. d_xz
Answer: C. d_{z²}
d_{z²} has two lobes along z-axis plus a doughnut ring.
25The region of zero electron density between two lobes of a p-orbital is called:
  • A. Antinode
  • B. Nodal plane
  • C. Core
  • D. Orbit
Answer: B. Nodal plane
Nodal plane passes through nucleus with zero electron density.
26The ratio of the frequency of the first line of Lyman series to the first line of Balmer series is:
  • A. 27:5
  • B. 5:27
  • C. 4:1
  • D. 1:4
Answer: A
First Lyman: n=2→1, ν̄ = R(1 - 1/4)=3R/4. First Balmer: n=3→2, ν̄ = R(1/4 - 1/9)=5R/36. Ratio = 27:5.
27The ionization energy of Hydrogen is 13.6 eV. The energy required to excite an electron from n=1 to n=2 is:
  • A. 13.6 eV
  • B. 3.4 eV
  • C. 10.2 eV
  • D. 1.85 eV
Answer: C. 10.2 eV
E₂ = -3.4 eV, E₁ = -13.6 eV, ΔE = 10.2 eV.
28The wavelength of the limiting line (series limit) of the Balmer series is:
  • A. R_H
  • B. 4/R_H
  • C. R_H/4
  • D. 9/R_H
Answer: B. 4/R_H
Limit: n₂→∞ to n=2, ν̄ = R_H(1/4 - 0) = R_H/4, so λ = 4/R_H.
29As the value of 'n' increases, the energy difference between adjacent shells:
  • A. Increases
  • B. Decreases
  • C. Remains constant
  • D. Becomes zero
Answer: B. Decreases
ΔE ∝ (1/n₁² - 1/n₂²); as n increases, difference decreases.
30The radius of the 2nd orbit of H atom is how many times the radius of the 1st orbit?
  • A. 2
  • B. 4
  • C. 8
  • D. 1.5
Answer: B. 4
r ∝ n²; r₂/r₁ = 2²/1² = 4.
31The ground state electronic configuration of Nitrogen (Z=7) follows Hund's Rule. If Hund's rule were violated, the atom would have:
  • A. Higher energy
  • B. Lower energy
  • C. Same energy
  • D. No electrons
Answer: A. Higher energy
Hund's rule minimizes energy; violation leads to higher energy.
32Which of the following ions is isoelectronic with CO?
  • A. N₂⁺
  • B. CN⁻
  • C. O₂⁺
  • D. F₂
Answer: B. CN⁻
CO has 14 electrons. CN⁻ has 6+7+1=14 electrons.
33The maximum number of electrons that can have n=3 and s = +1/2 is:
  • A. 18
  • B. 9
  • C. 5
  • D. 3
Answer: B. 9
For n=3, max electrons = 18. Half have s=+1/2 → 9.
34The (n+l) value for 4f and 5p orbitals are respectively:
  • A. 7 and 6
  • B. 7 and 7
  • C. 6 and 6
  • D. 5 and 6
Answer: A. 7 and 6
4f: n+l = 4+3=7; 5p: 5+1=6.
35If the (n+l) value is the same for two orbitals, the electron enters the orbital with:
  • A. Higher n value
  • B. Lower n value
  • C. Higher l value
  • D. Lower m value
Answer: B. Lower n value
For same (n+l), orbital with lower n is filled first (Aufbau rule).
36The number of unpaired electrons in Fe²⁺ (Z=26) is:
  • A. 4
  • B. 6
  • C. 0
  • D. 2
Answer: A. 4
Fe: [Ar] 4s² 3d⁶; Fe²⁺: [Ar] 3d⁶ → 4 unpaired electrons.
37Which element has the electronic configuration [Ar] 3d¹⁰ 4s¹?
  • A. Zinc
  • B. Copper
  • C. Chromium
  • D. Nickel
Answer: B. Copper
Cu (Z=29) has [Ar] 3d¹⁰ 4s¹.
38The transition element which has only one electron in its 4s orbital in ground state is:
  • A. Cr
  • B. Cu
  • C. Both A and B
  • D. Fe
Answer: C. Both A and B
Cr: [Ar] 3d⁵ 4s¹; Cu: [Ar] 3d¹⁰ 4s¹.
39For the configuration 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁵, 4s¹, the element is in its:
  • A. Excited state
  • B. Ground state
  • C. Anionic state
  • D. Impossible state
Answer: B. Ground state
This is ground state of Cr (Z=24).
40An element has 2 electrons in K-shell, 8 in L-shell, and 5 in M-shell. Its atomic number is:
  • A. 10
  • B. 12
  • C. 15
  • D. 17
Answer: C. 15
K:2, L:8, M:5 → total 15 electrons, Z=15.
41The wave number of a spectral line is 2 × 10⁶ m⁻¹. Its wavelength is:
  • A. 5 × 10⁻⁷ m
  • B. 2 × 10⁻⁶ m
  • C. 5 × 10⁷ m
  • D. 500 nm
Answer: A. 5 × 10⁻⁷ m
λ = 1/ν̄ = 1/(2×10⁶) = 5×10⁻⁷ m = 500 nm.
42The number of p-electrons in Bromine (Z=35) is:
  • A. 15
  • B. 17
  • C. 18
  • D. 5
Answer: B. 17
Br: [Ar] 4s² 3d¹⁰ 4p⁵; p electrons: 2p⁶ + 3p⁶ + 4p⁵ = 6+6+5 = 17.
43Which of the following has the maximum number of unpaired electrons?
  • A. Mg²⁺
  • B. Ti³⁺
  • C. V³⁺
  • D. Fe³⁺
Answer: D. Fe³⁺
Mg²⁺: 0; Ti³⁺: 1; V³⁺: 2; Fe³⁺: [Ar] 3d⁵ → 5 unpaired.
44The energy of an electron in the 3rd shell of Hydrogen is -1.51 eV. The energy in the 2nd shell is:
  • A. -3.4 eV
  • B. -0.85 eV
  • C. -13.6 eV
  • D. -4.53 eV
Answer: A. -3.4 eV
E ∝ 1/n²; E₂/E₃ = (3²/2²)=9/4; E₂ = (9/4)×(-1.51) = -3.4 eV.
45Which electronic jump in H-atom yields a photon of the shortest wavelength?
  • A. n=2 to n=1
  • B. n=3 to n=2
  • C. n=4 to n=3
  • D. n=∞ to n=1
Answer: D
Shortest wavelength (highest energy) corresponds to largest ΔE → n=∞→1 (ionization).
46A p-subshell can accommodate a maximum of 6 electrons because it has:
  • A. 3 orbitals
  • B. 6 orbitals
  • C. 1 orbital
  • D. 2 orbitals
Answer: A. 3 orbitals
p subshell has 3 orbitals, each holds 2 electrons → 6.
47According to Heisenberg, if the position of an electron is known with 100% accuracy, the uncertainty in its momentum is:
  • A. Zero
  • B. Infinite
  • C. h/4π
  • D. Constant
Answer: B. Infinite
Δx·Δp ≥ h/4π; if Δx=0, Δp → ∞.
48In the 3d subshell, the electron with m=+2 is in the:
  • A. d_xy orbital
  • B. d_{z²} orbital
  • C. Any one of the five d-orbitals (by convention)
  • D. First orbital only
Answer: C
m values are assigned arbitrarily to specific orbitals; m=+2 can be any d orbital with that m value.
49The species Cl⁻, K⁺, and Ca²⁺ are isoelectronic. The correct order of their sizes is:
  • A. Cl⁻ > K⁺ > Ca²⁺
  • B. Ca²⁺ > K⁺ > Cl⁻
  • C. K⁺ > Cl⁻ > Ca²⁺
  • D. Cl⁻ = K⁺ = Ca²⁺
Answer: A
All have 18 electrons. Size decreases with increasing nuclear charge: Cl⁻ (17), K⁺ (19), Ca²⁺ (20) → Cl⁻ largest, Ca²⁺ smallest.
50An electron is in one of the 3d orbitals. What are the possible values of n, l, and m?
  • A. n=3, l=2, m=0
  • B. n=3, l=3, m=1
  • C. n=3, l=1, m=0
  • D. n=2, l=3, m=-1
Answer: A
3d: n=3, l=2, m can be -2,-1,0,1,2.

✅Practice MCQs — Test 3

🧪 Exam-Level Questions

50 MCQs
1If a discharge tube contains a mixture of H₂, He, and Ne gases, which positive ion experiences the greatest deflection in a magnetic field?
  • A. He⁺
  • B. H⁺
  • C. He²⁺
  • D. Ne⁺
Answer: B. H⁺
Deflection ∝ e/m; H⁺ has smallest mass, highest e/m.
2The e/m value of an electron is 1.75 × 10¹¹ C/kg. What is the approximate e/m value of a deuteron (²H⁺)?
  • A. 9.58 × 10⁷ C/kg
  • B. 4.79 × 10⁷ C/kg
  • C. 1.75 × 10¹¹ C/kg
  • D. 8.95 × 10⁷ C/kg
Answer: B
Deuteron mass ≈ 2×1.67×10⁻²⁷ kg; e/m = 1.6×10⁻¹⁹/(2×1.67×10⁻²⁷) = 4.79×10⁷ C/kg.
3In a mass spectrometer, the radius of the circular path of an ion is proportional to:
  • A. √(m/e)
  • B. m/e
  • C. e/m
  • D. √(e/m)
Answer: A
r = (mv)/(qB) and v = √(2qV/m) → r ∝ √(m/q).
4If the velocity of cathode rays is 1/10th the velocity of light, and they strike a heavy metal target, they produce:
  • A. Canal rays
  • B. Alpha particles
  • C. Continuous X-rays
  • D. Gamma rays
Answer: C
High-energy electrons produce continuous X-rays (bremsstrahlung) when decelerated by metal target.
5Assuming the mass of a neutron equals the mass of a proton, the total mass of neutrons in 18 g of heavy water (D₂O) is:
  • A. 8 g
  • B. 10 g
  • C. 9 g
  • D. 20 g
Answer: C. 9 g
D₂O molar mass = 20 g/mol; 18 g = 0.9 mol. Each D₂O has 2+8 = 10 neutrons. Neutron mass ≈ 1 amu, so 0.9 × 10 = 9 g.
6Which graph correctly represents the relationship between e/m ratio of canal rays and atomic mass (A) of residual gas?
  • A. Straight line through origin
  • B. Rectangular hyperbola
  • C. Parabola
  • D. Straight line parallel to x-axis
Answer: B
e/m = e/(A×1.66×10⁻²⁷) ∝ 1/A → rectangular hyperbola.
7A proton and an alpha particle are accelerated through the same potential difference. The ratio of their de Broglie wavelengths (λ_p/λ_α) is:
  • A. 1:2
  • B. 2:1
  • C. 2√2:1
  • D. √2:1
Answer: C
λ = h/√(2mqV). λ_p/λ_α = √(m_α q_α / m_p q_p) = √(4×2/1×1) = √8 = 2√2.
8The charge on 1 gram of protons is approximately:
  • A. 96500 C
  • B. 1 C
  • C. 1.6×10⁻¹⁹ C
  • D. 6.02×10²³ C
Answer: A
1 g H⁺ = 1 mol protons. Charge = N_A × e = 6.022×10²³ × 1.6×10⁻¹⁹ = 9.65×10⁴ C ≈ 96500 C.
9A 100 W bulb emits monochromatic light of wavelength 400 nm. The number of photons emitted per second is approximately:
  • A. 2×10²⁰
  • B. 3×10¹⁸
  • C. 2×10¹⁹
  • D. 6×10²³
Answer: A
Energy per photon = hc/λ = 4.97×10⁻¹⁹ J. Photons/sec = 100 / 4.97×10⁻¹⁹ ≈ 2×10²⁰.
10The work function of a metal is 4.0 eV. To emit a photoelectron with zero kinetic energy, the wavelength of incident light must be approximately:
  • A. 310 nm
  • B. 400 nm
  • C. 620 nm
  • D. 155 nm
Answer: A
λ = hc/φ = (6.63×10⁻³⁴×3×10⁸)/(4×1.6×10⁻¹⁹) = 3.11×10⁻⁷ m = 311 nm.
11Two electromagnetic radiations have wave numbers in ratio 2:3. Their energies are in the ratio:
  • A. 3:2
  • B. 2:3
  • C. 4:9
  • D. 9:4
Answer: B. 2:3
E ∝ ν̄, so ratio E₁:E₂ = 2:3.
12According to de Broglie, a macroscopic object of mass 100 g moving at 100 m/s has a wavelength of:
  • A. 6.63×10⁻³⁴ m
  • B. 6.63×10⁻³⁵ m
  • C. 6.63×10⁻³³ m
  • D. 6.63×10⁻³² m
Answer: B
λ = h/mv = 6.63×10⁻³⁴ / (0.1×100) = 6.63×10⁻³⁵ m.
13If the kinetic energy of a free electron doubles, its de Broglie wavelength changes by the factor:
  • A. 1/2
  • B. 2
  • C. 1/√2
  • D. √2
Answer: C
λ ∝ 1/√K, so if K doubles, λ' = λ/√2.
14A photon of energy E ejects a photoelectron from a metal surface whose work function is W₀. If this electron enters a uniform magnetic field B perpendicularly and moves in a circular path of radius r, then r is:
  • A. √(2m(E-W₀))/(eB)
  • B. 2m(E-W₀)/(eB)
  • C. √(m(E-W₀))/(eB)
  • D. √(2mE)/(eB)
Answer: A
K.E. = E - W₀ = ½mv² → v = √(2(E-W₀)/m). r = mv/(eB) = √(2m(E-W₀))/(eB).
15If the uncertainty in position and momentum are equal, the uncertainty in velocity is:
  • A. (1/m)√(h/π)
  • B. (1/(2m))√(h/π)
  • C. √(h/(2π))
  • D. (1/m)√(h/(4π))
Answer: D
Δx·Δp = h/4π; given Δx = Δp, so (Δp)² = h/4π → Δp = √(h/(4π)). Δv = Δp/m = (1/m)√(h/(4π)).
16The total number of nodes (radial + angular) in a 4d orbital is:
  • A. 3
  • B. 4
  • C. 2
  • D. 1
Answer: A. 3
Total nodes = n-1 = 3.
17The orbital angular momentum of an electron in a 4f orbital is:
  • A. √12 (h/2π)
  • B. √6 (h/2π)
  • C. √2 (h/2π)
  • D. √20 (h/2π)
Answer: A
For f orbital, l=3; angular momentum = √[l(l+1)] h/2π = √12 h/2π.
18The number of radial nodes for a 5p orbital is:
  • A. 1
  • B. 2
  • C. 3
  • D. 4
Answer: C. 3
Radial nodes = n-l-1 = 5-1-1 = 3.
19Which of the following represents the correct order of penetration power of subshells for a given n?
  • A. s > p > d > f
  • B. f > d > p > s
  • C. p > s > d > f
  • D. s = p = d = f
Answer: A. s > p > d > f
Penetration power: s > p > d > f.
20The maximum number of electrons in an atom that can have n=4, m_l=0 is:
  • A. 2
  • B. 6
  • C. 8
  • D. 10
Answer: C. 8
m_l=0 can be in s, p, d, f subshells. For n=4, orbitals with m_l=0: 4s, 4p, 4d, 4f (4 orbitals), each holds 2 → 8.
21The angular nodes (nodal cones) for the d_{z²} orbital are:
  • A. 0
  • B. 1
  • C. 2
  • D. 3
Answer: C. 2
d_{z²} has two nodal cones.
22If a hypothetical atom had an orbital with quantum numbers n=5 and l=4, what would be the letter designation of this subshell?
  • A. 5f
  • B. 5g
  • C. 5h
  • D. 5d
Answer: B. 5g
l=0:s, 1:p, 2:d, 3:f, 4:g → 5g.
23Which set of quantum numbers represents the highest energy electron in a ground-state Gallium (Z=31) atom?
  • A. n=3, l=2, m_l=0, m_s=+1/2
  • B. n=4, l=0, m_l=0, m_s=-1/2
  • C. n=4, l=1, m_l=-1, m_s=+1/2
  • D. n=4, l=1, m_l=2, m_s=+1/2
Answer: C
Ga: [Ar] 4s² 3d¹⁰ 4p¹; highest energy electron is in 4p (n=4,l=1).
24The spin-only magnetic moment of a divalent ion in aqueous solution is 3.87 BM. The atomic number of the parent metal is likely:
  • A. 25 (Mn)
  • B. 26 (Fe)
  • C. 27 (Co)
  • D. 28 (Ni)
Answer: B. Fe (Z=26)
μ = √(n(n+2)) = 3.87 → n ≈ 4. Fe²⁺ (Z=26) has 4 unpaired electrons.
25The ratio of the energy of the electron in the ground state of Hydrogen to the electron in the first excited state of He⁺ is:
  • A. 1:1
  • B. 1:4
  • C. 4:1
  • D. 1:2
Answer: A. 1:1
E ∝ Z²/n². H (Z=1,n=1): -13.6 eV. He⁺ (Z=2,n=2): -13.6×4/4 = -13.6 eV. Ratio = 1:1.
26The wavelength of the spectral line emitted when an electron in H-atom falls from infinity to n=1 is (R = 1.097×10⁷ m⁻¹):
  • A. 91 nm
  • B. 121 nm
  • C. 400 nm
  • D. 364 nm
Answer: A. 91 nm
1/λ = R(1/1² - 0) = R; λ = 1/R = 9.12×10⁻⁸ m = 91.2 nm.
27If an electron makes a transition from the 5th orbit to the 2nd orbit in a Hydrogen atom, the total number of spectral lines produced is:
  • A. 3
  • B. 6
  • C. 10
  • D. 15
Answer: C. 10
When electrons are excited to n=5, total possible transitions = n(n-1)/2 = 10.
28What is the ratio of the time period of revolution of an electron in n=2 orbit of Hydrogen to n=1 orbit?
  • A. 2:1
  • B. 4:1
  • C. 8:1
  • D. 16:1
Answer: C. 8:1
T ∝ n³/Z²; T₂/T₁ = 2³/1³ = 8:1.
29The ionization energy of a H-like species is 54.4 eV. Which species is this?
  • A. H
  • B. He⁺
  • C. Li²⁺
  • D. Be³⁺
Answer: B. He⁺
IE = 13.6 × Z² eV. 13.6×Z² = 54.4 → Z² = 4 → Z=2 → He⁺.
30Which transition in He⁺ will have the same wavelength as n=4 to n=2 in Hydrogen?
  • A. n=8 to n=4
  • B. n=4 to n=2
  • C. n=6 to n=3
  • D. n=2 to n=1
Answer: A
1/λ ∝ Z²(1/n₁² - 1/n₂²). For H: (1/4-1/16)=3/16. For He⁺ (Z=2): need 3/64. n₁=4, n₂=8: (1/16-1/64)=3/64. So n=8→4.
31The velocity of an electron in the first Bohr orbit of Hydrogen is v. The velocity in the third orbit of Li²⁺ is:
  • A. v
  • B. v/3
  • C. 3v
  • D. v/9
Answer: A. v
v ∝ Z/n. For Li²⁺, Z=3, n=3 → v ∝ 3/3 = 1 → same v.
32The energy difference between the 3rd and 4th orbits of Hydrogen is ΔE₁, and between 4th and 5th is ΔE₂. Which is true?
  • A. ΔE₁ = ΔE₂
  • B. ΔE₁ > ΔE₂
  • C. ΔE₁ < ΔE₂
  • D. ΔE₁ = 2ΔE₂
Answer: B. ΔE₁ > ΔE₂
ΔE decreases as n increases.
33The ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series is:
  • A. 4:1
  • B. 1:4
  • C. 9:4
  • D. 27:5
Answer: A. 4:1
Series limit: Balmer λ_B = 4/R, Lyman λ_L = 1/R; ratio = 4:1.
34According to Bohr's model, the potential energy of an electron in the ground state of H is:
  • A. -13.6 eV
  • B. -27.2 eV
  • C. +13.6 eV
  • D. Zero
Answer: B. -27.2 eV
PE = -2 × KE; KE = +13.6 eV, so PE = -27.2 eV.
35If the radius of the first Bohr orbit is a₀, the de Broglie wavelength of an electron in the 3rd orbit is:
  • A. 2πa₀
  • B. 6πa₀
  • C. 9πa₀
  • D. 3πa₀
Answer: B. 6πa₀
rₙ = n²a₀, circumference = 2πrₙ = nλ → λ = 2π(9a₀)/3 = 6πa₀.
36The exchange energy is maximum for which configuration?
  • A. d⁴
  • B. d⁵
  • C. d⁹
  • D. p³
Answer: B. d⁵
Exchange energy is maximum for half-filled (d⁵) and fully-filled (d¹⁰) due to maximum parallel spins.
37Chromium (Z=24) has [Ar] 3d⁵ 4s¹. Total electrons with l=0 in Cr atom:
  • A. 6
  • B. 7
  • C. 8
  • D. 1
Answer: B. 7
l=0 for s orbitals: 1s², 2s², 3s², 4s¹ = 2+2+2+1 = 7.
38Which of the following ions is diamagnetic?
  • A. Cu²⁺
  • B. Zn²⁺
  • C. Ti³⁺
  • D. Ni²⁺
Answer: B. Zn²⁺
Zn²⁺ has [Ar] 3d¹⁰ (all paired, diamagnetic).
39For Palladium (Z=46), the ground state configuration is an exception. It is:
  • A. [Kr] 4d¹⁰ 5s⁰
  • B. [Kr] 4d⁸ 5s²
  • C. [Kr] 4d⁹ 5s¹
  • D. [Kr] 4d⁵ 5s² 5p³
Answer: A
Pd: [Kr] 4d¹⁰ 5s⁰ (fully filled d subshell stability).
40An element has successive ionization energies of 738, 1450, 7730, 10500 kJ/mol. The element belongs to:
  • A. Group 1
  • B. Group 2
  • C. Group 13
  • D. Group 14
Answer: B. Group 2
Large jump after 2nd IE → 2 valence electrons → Group 2.
41The number of electrons that can be accommodated in all orbitals having n+l = 5 is:
  • A. 18
  • B. 14
  • C. 10
  • D. 8
Answer: A. 18
n+l=5: 5s(l=0), 4p(l=1), 3d(l=2). Orbitals: 1+3+5=9, each holds 2 → 18 electrons.
42Correct sequence of sizes for isoelectronic series S²⁻, Cl⁻, K⁺, Ca²⁺:
  • A. S²⁻ > Cl⁻ > K⁺ > Ca²⁺
  • B. Ca²⁺ > K⁺ > Cl⁻ > S²⁻
  • C. K⁺ > Ca²⁺ > S²⁻ > Cl⁻
  • D. All equal
Answer: A
All have 18 electrons; size decreases with increasing nuclear charge.
43How many electrons in Iron (Z=26) have m_l = 0?
  • A. 14
  • B. 13
  • C. 7
  • D. 10
Answer: A. 14
Fe: [Ar] 4s² 3d⁶. m_l=0 orbitals: 1s,2s,3s,4s (8 e⁻), 2p_z,3p_z (4 e⁻), 3d_{z²} (2 e⁻) → 14.
44The shielding effect (screening constant σ) is maximum for an electron in which subshell?
  • A. s
  • B. p
  • C. d
  • D. f
Answer: D. f
f electrons have highest shielding effect due to diffuse shape and many inner electrons.
45If an electron is in a d_{x²-y²} orbital, the probability of finding it along the z-axis is:
  • A. Maximum
  • B. 50%
  • C. Zero
  • D. Constant but non-zero
Answer: C. Zero
d_{x²-y²} has lobes along x and y; along z, probability is zero.
46Copper (Z=29) has [Ar] 3d¹⁰ 4s¹. How many full-filled orbitals are present in Copper atom?
  • A. 15
  • B. 14
  • C. 10
  • D. 18
Answer: B. 14
Full-filled orbitals: 1s,2s,3s (3), 2p,3p (6), 3d (5). 4s is half-filled. Total = 3+6+5 = 14.
47According to (n+l) rule, after filling 6s orbital, the next electron enters the:
  • A. 6p orbital
  • B. 4f orbital
  • C. 5d orbital
  • D. 7s orbital
Answer: B. 4f orbital
Order: 6s, then 4f, then 5d, then 6p.
48Gadolinium (Z=64) belongs to f-block. Its correct outer config is:
  • A. 4f⁸ 5d⁰ 6s²
  • B. 4f⁷ 5d¹ 6s²
  • C. 4f⁶ 5d² 6s²
  • D. 4f⁹ 5d⁰ 6s¹
Answer: B
Gd: [Xe] 4f⁷ 5d¹ 6s² (half-filled f stability).
49An ion M³⁺ has magnetic moment √35 BM. Number of unpaired electrons and possible Z of element:
  • A. 5 and 26 (Fe)
  • B. 4 and 25 (Mn)
  • C. 5 and 25 (Mn)
  • D. 6 and 24 (Cr)
Answer: A
√(n(n+2)) = √35 → n=5. Fe³⁺ (Z=26) has [Ar] 3d⁵ → 5 unpaired.
50The exact mass of an atom is always slightly less than the sum of masses of its constituent nucleons. This mass defect is accounted for by:
  • A. Heisenberg's Uncertainty Principle
  • B. Binding Energy
  • C. Shielding Effect
  • D. Exchange Energy
Answer: B. Binding Energy
Mass defect is converted into binding energy (E=mc²).

📄Past Paper MCQs

📅 Historical/Conceptual Past Papers

20 MCQs
1Canal rays were discovered by Goldstein in which year?
PTB/UHS 2012
  • A. 1886
  • B. 1897
  • C. 1911
  • D. 1932
Answer: A. 1886
Goldstein discovered canal rays in 1886.
2The e/m ratio of positive rays is maximum for Hydrogen gas. This was confirmed by J.J. Thomson around:
UHS 2014
  • A. 1897
  • B. 1907
  • C. 1886
  • D. 1895
Answer: B. 1907
Thomson's work on positive rays extended into early 1900s.
3In the production of positive rays, the ionization of gas by cathode rays was a concept solidified in:
Federal Board 2015
  • A. 1886
  • B. 1895
  • C. 1900
  • D. 1913
Answer: B. 1895
By 1895, the mechanism of ionization was understood.
4The mass of a proton (1.67 × 10⁻²⁷ kg) was accurately determined following the work of Millikan and Rutherford in:
UHS 2011
  • A. 1909
  • B. 1911
  • C. 1920
  • D. 1897
Answer: C. 1920
Rutherford's gold foil (1911) and Millikan's oil drop (1909) helped determine proton mass.
5Max Planck presented his Quantum Theory, proposing that energy is quantized, in:
PTB/UHS 2013
  • A. 1900
  • B. 1905
  • C. 1913
  • D. 1924
Answer: A. 1900
Planck introduced quantum theory in 1900.
6The 'Photon' was named and treated as a discrete unit of radiation energy by Einstein in:
Federal Board 2017
  • A. 1900
  • B. 1905
  • C. 1921
  • D. 1911
Answer: B. 1905
Einstein's photoelectric effect paper (1905) introduced light quanta (photons).
7The value of Planck's constant (6.626 × 10⁻³⁴ J.s) is critical for MDCAT calculations. This constant was introduced in:
UHS 2016
  • A. 1900
  • B. 1913
  • C. 1905
  • D. 1895
Answer: A. 1900
Planck's constant was introduced in his 1900 quantum theory.
8Bohr's model of the Hydrogen atom was proposed in:
UHS 2012/2015
  • A. 1911
  • B. 1913
  • C. 1920
  • D. 1905
Answer: B. 1913
Niels Bohr proposed his model in 1913.
9The Lyman series, which lies in the Ultraviolet region, was discovered by Theodore Lyman in:
Federal Board 2018
  • A. 1885
  • B. 1906
  • C. 1914
  • D. 1924
Answer: B. 1906
Lyman discovered the UV series in 1906.
10The Balmer series (Visible region) was discovered in:
PTB 2014
  • A. 1885
  • B. 1900
  • C. 1913
  • D. 1897
Answer: A. 1885
Balmer discovered the visible series in 1885.
11The concept of 'Orbitals' as regions of 90% probability replaced the 'fixed orbit' concept following Schrödinger's work in:
UHS 2017
  • A. 1913
  • B. 1926
  • C. 1932
  • D. 1924
Answer: B. 1926
Schrödinger's wave equation (1926) introduced the orbital concept.
12The Principal Quantum Number (n) was first introduced in Bohr's theory in:
Federal Board 2013
  • A. 1913
  • B. 1900
  • C. 1926
  • D. 1911
Answer: A. 1913
Bohr introduced n in 1913.
13The Azimuthal Quantum Number (l) was introduced by Sommerfeld in:
UHS 2014
  • A. 1913
  • B. 1916
  • C. 1921
  • D. 1926
Answer: B. 1916
Sommerfeld introduced l in 1916 to explain fine structure.
14The Magnetic Quantum Number (m) was developed to explain the Zeeman effect, discovered in:
PTB 2015
  • A. 1896
  • B. 1913
  • C. 1925
  • D. 1900
Answer: A. 1896
Zeeman effect discovered in 1896; m was introduced later.
15The Spin Quantum Number (s) was proposed by Goudsmit and Uhlenbeck to explain the doublet structure of spectra in:
UHS 2011
  • A. 1925
  • B. 1913
  • C. 1932
  • D. 1905
Answer: A. 1925
Spin quantum number was proposed in 1925.
16The Aufbau Principle became a standard rule in:
Federal Board 2016
  • A. 1920s
  • B. 1900s
  • C. 1940s
  • D. 1890s
Answer: A. 1920s
Aufbau principle was established in the 1920s.
17Pauli's Exclusion Principle was formulated in:
UHS 2013
  • A. 1925
  • B. 1913
  • C. 1930
  • D. 1921
Answer: A. 1925
Pauli exclusion principle proposed in 1925.
18Hund's Rule was established in:
PTB 2017
  • A. 1927
  • B. 1913
  • C. 1935
  • D. 1905
Answer: A. 1927
Hund's rule was formulated in 1927.
19The Paschen series (Infrared) was discovered by Friedrich Paschen in:
Federal Board 2014
  • A. 1908
  • B. 1885
  • C. 1922
  • D. 1913
Answer: A. 1908
Paschen discovered the IR series in 1908.
20James Chadwick discovered the neutron, completing the basic atomic structure, in:
UHS 2012
  • A. 1911
  • B. 1920
  • C. 1932
  • D. 1935
Answer: C. 1932
Chadwick discovered the neutron in 1932.

🔍Direct Search Queries — Atomic Structure

Exact search terms students use on Google for this chapter. Click any to jump to the relevant section:

atomic structure MDCAT notes discovery of proton Goldstein canal rays properties photon energy E=hv Planck's quantum theory orbitals vs orbits difference quantum numbers n l m s principal vs azimuthal quantum number magnetic quantum number m values spin quantum number definition s p d orbital shapes p orbital dumbbell shape d orbital cloverleaf d z2 orbital doughnut hydrogen spectrum series Lyman Balmer Paschen series Balmer series visible region Bohr model hydrogen atom Bohr radius formula ionization energy hydrogen Aufbau principle order Pauli exclusion principle Hund's rule degenerate orbitals chromium exception configuration copper exception configuration what are canal rays in chemistry what is a photon in physics what are the 4 quantum numbers difference between orbital and orbit why do orbitals have different shapes atomic structure MCQs with answers atomic structure MCQs pdf quantum numbers MCQs MDCAT MDCAT past papers atomic structure UHS MDCAT chemistry MCQs NUMS atomic structure past papers atomic structure class 11 notes atomic structure class 12 notes electronic configuration rules

❔Frequently Asked Questions

What are canal rays?
Canal rays (positive rays) are beams of positively charged ions produced when cathode rays ionize residual gas in a discharge tube. Discovered by Goldstein (1886) using a perforated cathode. Their e/m ratio varies with the gas.
What is a photon?
A photon is a quantum of electromagnetic energy with no mass or charge. Energy: E = hν = hc/λ. Planck's constant h = 6.626 × 10⁻³⁴ J·s.
What are the four quantum numbers?
n (principal, shell), l (azimuthal, subshell), m_l (magnetic, orbital), m_s (spin, ±½).
What are the shapes of s, p, and d orbitals?
s = spherical. p = dumbbell (3 orbitals: pₓ, pᵧ, p_z). d = cloverleaf for four orbitals + d_z² has two lobes with a doughnut ring. f = complex 8-lobed shapes.
What is the Aufbau principle?
Electrons fill lowest-energy atomic orbitals first. Filling order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s...
What is the Pauli Exclusion Principle?
No two electrons can have the same 4 quantum numbers. Each orbital holds max 2 electrons with opposite spins.
What is Hund's rule?
Electrons fill degenerate orbitals singly first with parallel spins before pairing. Minimizes repulsion.
Why does chromium have an exceptional configuration?
Cr (Z=24): [Ar] 4s¹ 3d⁵ (not 4s² 3d⁴) — half-filled d stability.
Why does copper have an exceptional configuration?
Cu (Z=29): [Ar] 4s¹ 3d¹⁰ (not 4s² 3d⁹) — fully-filled d stability.
Difference between Lyman and Balmer series?
Lyman: n→1, UV. Balmer: n→2, visible. Paschen: n→3, IR.
Maximum electrons in a shell?
2n². n=1 → 2, n=2 → 8, n=3 → 18, n=4 → 32.
What are degenerate orbitals?
Orbitals with same energy but different orientations. Filled singly first per Hund's rule.
How many electrons can a d subshell hold?
5 orbitals → max 10 electrons. s=2, p=6, d=10, f=14.
Relationship between energy and wavelength?
E = hc/λ. Shorter wavelength → higher energy. Gamma rays highest, radio waves lowest.
How many MCQs from Atomic Structure in MDCAT?
Typically 2–3 MCQs. High-yield because quantum numbers and configuration are foundational.
What is the (n+l) rule?
Orbitals fill in order of increasing (n+l). For equal (n+l), lower n fills first. E.g., 4s (4) before 3d (5).
How many radial nodes in an orbital?
Radial nodes = n - l - 1. Angular nodes = l. Total = n - 1.
What is the Bohr radius?
Radius of first Bohr orbit of H = 0.529 Å (52.9 pm). For nth orbit: r_n = n² × 0.529 Å.
Paramagnetic vs diamagnetic?
Paramagnetic: unpaired electrons, attracted by magnet. Diamagnetic: all paired, repelled.
What is the Heisenberg Uncertainty Principle?
Impossible to determine both position and momentum exactly. Δx · Δp ≥ h/4π. Orbitals are probability regions.

📚Other Chemistry Chapters

2
Chemical Bonding
3
Gases
4
Hydrocarbons
5
Alkyl Halides
6
Alcohols & Phenols
7
Aldehydes & Ketones
8
Carboxylic Acids
9
Macromolecules

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