CUET UG Chemistry Physical Chemistry Atomic Structure Quantum Numbers Orbital Shapes Electronic Configuration

Must-Know (15–20 detailed bullets)

• The principal quantum number (n) determines the energy level and size of the orbital; values are positive integers (n = 1, 2, 3, ...); for n = 3, electrons are in the M shell.
• The azimuthal quantum number (l) defines the shape of the orbital and ranges from 0 to (n – 1); for n = 4, l can be 0 (s), 1 (p), 2 (d), or 3 (f).
• The magnetic quantum number (mₗ) specifies the orientation of an orbital in space and ranges from –l to +l, including zero; for p-orbital (l = 1), mₗ = –1, 0, +1 → three p-orbitals (pₓ, pᵧ, p_z).
• The spin quantum number (mₛ) has only two values: +½ (↑) or –½ (↓), representing electron spin direction.
• For a given n, total number of orbitals = n²; for n = 2, there are 4 orbitals (one 2s + three 2p).
• Maximum number of electrons in a shell = 2n²; for n = 3, maximum 18 electrons.
• s-orbitals are spherically symmetric; size increases with n (e.g., 3s > 2s > 1s).
• p-orbitals have dumbbell shape with directional lobes along axes; each p-subshell has three orbitals.
• d-orbitals have complex shapes; five orientations (dxy, dyz, dzx, dx²–y², dz²); dz² has unique shape with a doughnut around the middle.
• No two electrons in an atom can have the same set of all four quantum numbers — Pauli Exclusion Principle.
• In ground state, electrons fill orbitals of lowest energy first — Aufbau Principle; order: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p...
• Half-filled and fully-filled subshells are more stable; Cr (Z = 24) has configuration [Ar] 4s¹ 3d⁵, not [Ar] 4s² 3d⁴.
• Cu (Z = 29) has electronic configuration [Ar] 4s¹ 3d¹⁰ due to extra stability of fully-filled d-subshell.
• The energy of orbitals in multi-electron atoms depends on both n and l; 4s is lower than 3d due to penetration effect.
• For hydrogen and hydrogen-like species, energy depends only on n; E(3s) = E(3p) = E(3d).
• Nodes are regions with zero probability of finding an electron; total nodes in an orbital = (n – 1); radial nodes = (n – l – 1), angular nodes = l.
• 3d orbital has (3 – 2 – 1) = 0 radial nodes and 2 angular nodes.
• The order of filling follows (n + l) rule: lower (n + l) fills first; if same, lower n fills first (e.g., 4s (4+0=4) before 3d (3+2=5)).
• An orbital can hold maximum two electrons with opposite spins; He atom: 1s², both electrons with opposite spins.
• The shape of f-orbitals is more complex with seven orientations; not typically drawn in detail in NCERT but mentioned for l = 3.

Difficulty Level

Intermediate — requires understanding of quantum number rules, exceptions in configurations, and visualization of orbital shapes, but all concepts are directly from NCERT Class 11 Chapter 2.

Common CUET Traps (3 bullets)

• Trap: Assuming 3d fills before 4s in all cases. Avoid: Remember 4s fills before 3d in neutral atoms (e.g., K: 4s¹, Ca: 4s²), but 4s loses electrons before 3d during ionization (e.g., Fe²⁺ loses 4s electrons first).
• Trap: Thinking dz² orbital looks like others; it has two lobes along z-axis and a ring (doughnut) in xy-plane. Avoid: Recall dz² is one of five d-orbitals with distinct appearance.
• Trap: Believing all orbitals in a subshell have different energies in absence of magnetic field. Avoid: In free atom, all orbitals of a subshell (e.g., px, py, pz) are degenerate (same energy).

Practice MCQs (5 questions)

Q1. Which set of quantum numbers is possible for the last electron of a neutral fluorine atom (Z = 9)?
A) n = 2, l = 0, mₗ = 0, mₛ = +½
B) n = 2, l = 1, mₗ = 0, mₛ = –½
C) n = 2, l = 1, mₗ = 2, mₛ = +½
D) n = 3, l = 1, mₗ = 0, mₛ = –½

Answer: B
Explanation: F (1s² 2s² 2p⁵) — last electron enters 2p with l = 1; mₗ = –1, 0, or +1; option B is valid.
Why others fail: C has mₗ = 2 which is invalid for p-orbital (l = 1 → mₗ max = ±1).

Q2. What is the maximum number of electrons in the n = 3 shell?
A) 8
B) 10
C) 18
D) 32

Answer: C
Explanation: Maximum electrons = 2n² = 2×(3)² = 18.
Why others fail: A (8) is valence shell capacity for octet rule, not total shell capacity.

Q3. Which of the following represents the correct order of increasing energy of orbitals in a potassium atom (Z = 19)?
A) 3d < 4s < 4p
B) 4s < 3d < 4p
C) 3d < 4p < 4s
D) 4p < 4s < 3d

Answer: B
Explanation: In multi-electron atoms, energy order follows (n + l) rule: 4s (4+0=4) < 3d (3+2=5) < 4p (4+1=5, higher n).
Why others fail: A assumes hydrogen-like energy levels where 3d < 4s, but not true for potassium.

Q4. Which orbital has two angular nodes and one radial node?
A) 3p
B) 4d
C) 3d
D) 4p

Answer: D
Explanation: For 4p: n = 4, l = 1 → angular nodes = l = 1? Wait — correction: angular nodes = l, radial nodes = n – l – 1. For 4p: angular nodes = 1, radial = 4–1–1=2 → not matching. Try 4d: n=4, l=2 → angular=2, radial=4–2–1=1 → matches. So correct is B.
Answer: B
Explanation: 4d orbital has angular nodes = l = 2 and radial nodes = n – l – 1 = 1.
Why others fail: C (3d): radial nodes = 3–2–1 = 0 → zero radial nodes, so incorrect.

Q5. Which element has the electronic configuration [Ar] 4s¹ 3d⁵?
A) Manganese
B) Chromium
C) Iron
D) Copper

Answer: B
Explanation: Chromium (Z = 24) has exceptional configuration due to stability of half-filled d-subshell.
Why others fail: A (Mn) has [Ar] 4s² 3d⁵ — tempting because also half-filled, but normal filling.

Last‑Minute Revision (15–20 one‑liners)

• ⚠️ n = 1, l = 0 → 1s orbital; only one orientation (mₗ = 0).
• ⚠️ For d-orbital, l = 2 → mₗ = –2, –1, 0, +1, +2 → five orbitals.
• ⚠️ s-subshell holds 2e⁻, p → 6e⁻, d → 10e⁻, f → 14e⁻.
• ⚠️ Total electrons with n = 2: max 8 (2s² 2p⁶).
• ⚠️ Orbital angular momentum = √[l(l+1)] × h/(2π); not mₗ.
• ⚠️ 2p orbital has 0 radial nodes (n – l – 1 = 2–1–1 = 0).
• ⚠️ Shape of s-orbital: spherical; p: dumbbell; d: double dumbbell or cloverleaf (except dz²).
• ⚠️ Pauli exclusion: no two electrons have identical four quantum numbers.
• ⚠️ Aufbau order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s...
• ⚠️ Cr (24): [Ar] 4s¹ 3d⁵; Cu (29): [Ar] 4s¹ 3d¹⁰ — remember exceptions.
• ⚠️ Fe²⁺: loses 4s electrons first → [Ar] 3d⁶, not 4s² 3d⁴.
• ⚠️ Number of orbitals in d-subshell = 5; f = 7.
• ⚠️ For n = 4, possible l values: 0, 1, 2, 3 → s, p, d, f.
• ⚠️ Electron density in s-orbital is maximum at nucleus.
• ⚠️ Mnemonic: "silly professors don't fail" → s, p, d, f.
• ⚠️ Energy: 5s < 4f < 5p < 6s (use n + l rule to verify).
• ⚠️ In hydrogen, 3s = 3p = 3d in energy.
• ⚠️ Radial probability curve of 2s has one minimum (node).
• ⚠️ dz² orbital has no nodal plane but has conical node.
• ⚠️ Maximum electrons with n = 3, l = 2: 10 (five d-orbitals × 2e⁻).