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ChemistryStructure of Atom
Match List I (quantum numbers n, l, mₗ) with List II (orbital):

List I (n, l, mₗ)

A. 3, 0, 0
B. 4, 1, 0
C. 3, 2, 1
D. 4, 2, −2

List II (Orbital)

I. 4p
II. 4d
III. 3d
IV. 3s
Options
1
A–IV, B–I, C–III, D–II
2
A–III, B–I, C–IV, D–II
3
A–IV, B–II, C–III, D–I
4
A–II, B–I, C–III, D–IV
Correct Answer
A–IV, B–I, C–III, D–II
Solution
1

Quick Rules: n = shell, l→subshell (0=s,1=p,2=d,3=f), orbital name = n + subshell letter

2
n, l, mₗl→subshellmₗ valid?Orbital
A: 3,0,00→s0 to 0 ✓3s = IV ✓
B: 4,1,01→p−1 to +1 ✓4p = I ✓
C: 3,2,12→d−2 to +2 ✓3d = III ✓
D: 4,2,−22→d−2 to +2 ✓4d = II ✓
Theory: Quantum Numbers
1. The Four Quantum Numbers

Every electron in an atom is uniquely described by four quantum numbers. The Pauli Exclusion Principle states no two electrons can have the same set of all four quantum numbers. Together they define the energy, shape, orientation, and spin of each electron's orbital.

n → Principal quantum number → 1, 2, 3, 4... (shell)

l → Azimuthal quantum number → 0 to (n−1) (subshell shape)

mₗ → Magnetic quantum number → −l to +l (orbital orientation)

ms → Spin quantum number → +½ or −½ only

2. l Values and Subshell Names

📌 l = 0 → s subshell → spherical → 1 orbital (mₗ = 0 only)

📌 l = 1 → p subshell → dumbbell → 3 orbitals (mₗ = −1, 0, +1)

📌 l = 2 → d subshell → cloverleaf → 5 orbitals (mₗ = −2 to +2)

📌 l = 3 → f subshell → complex → 7 orbitals (mₗ = −3 to +3)

📌 Number of orbitals in subshell l = 2l + 1

📌 Max electrons in subshell = 2(2l + 1)

3. Constraints on Quantum Numbers

The quantum numbers are not independent — each must satisfy strict constraints. For a given n: l can be 0, 1, 2... up to (n−1). For a given l: mₗ can be −l, −l+1, ... 0, ... l−1, +l (total 2l+1 values). ms is always +½ or −½ regardless of n, l, or mₗ. A quantum number set is invalid if l ≥ n or |mₗ| > l. For example, (2,2,0) is invalid because for n=2, l can only be 0 or 1, not 2.

4. Shell-wise Orbital Count
n (shell)Subshells (l)Total orbitalsMax e⁻
1 (K)1s12
2 (L)2s, 2p48
3 (M)3s, 3p, 3d918
4 (N)4s, 4p, 4d, 4f1632
5. Aufbau Principle — Filling Order

Orbitals fill in order of increasing (n + l). When two subshells have the same (n + l), the one with lower n fills first. The correct filling order: 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d → 6p. This explains why 4s fills before 3d — 4s has n+l=4 while 3d has n+l=5.

6. Pauli Exclusion and Hund's Rules

Pauli Exclusion: maximum 2 electrons per orbital, with opposite spins. This limits s-subshell to 2 electrons, p to 6, d to 10, f to 14. Hund's Rule: electrons fill degenerate orbitals one each with parallel spins before pairing begins. This maximises total spin and gives the most stable configuration. For carbon (1s²2s²2p²): the two 2p electrons are in separate 2p orbitals with parallel spins, not both in the same orbital.

7. Exceptions to Aufbau — Cr and Cu

Chromium (Cr, Z=24): Expected [Ar]3d⁴4s², Actual [Ar]3d⁵4s¹. The half-filled 3d⁵ configuration (all five 3d orbitals singly occupied with parallel spins) has extra stability due to maximum exchange energy and spherical electron distribution. Copper (Cu, Z=29): Expected [Ar]3d⁹4s², Actual [Ar]3d¹⁰4s¹. The completely filled 3d¹⁰ subshell is extra stable. These are the most commonly tested exceptions in NEET.

8. Hydrogen Spectrum and Quantum Numbers

When electrons transition between energy levels, photons are emitted or absorbed. The energy of emitted photon: E = 13.6(1/n₁² − 1/n₂²) eV for hydrogen. Different series correspond to different final states: Lyman series (n₁=1, UV), Balmer (n₁=2, visible), Paschen (n₁=3, IR), Brackett (n₁=4, IR), Pfund (n₁=5, far IR). The Balmer series (H-alpha at 656 nm etc.) gives the visible colours of hydrogen discharge tubes.

Frequently Asked Questions
1. What orbital does (3, 2, −1) represent? ⌄
n=3 (3rd shell), l=2 (d subshell), mₗ=−1 (valid since −2 ≤ −1 ≤ +2). Orbital = 3d. Specifically one of the five 3d orbitals (mₗ=−1 corresponds to the 3dyz orbital in one convention). The electron is in the third shell, d subshell.
2. Is (3, 3, 0) a valid quantum number set? ⌄
No. For n=3, l can only be 0, 1, or 2 (l ranges from 0 to n−1 = 2). l=3 requires n ≥ 4. So (3,3,0) is invalid — you would need n=4 to have an f orbital (l=3). The first f orbitals appear in the 4th shell: 4f.
3. How many electrons can have n=3, l=1? ⌄
n=3, l=1 is the 3p subshell. mₗ = −1, 0, +1 → 3 orbitals. Each holds 2 electrons. Total = 6 electrons maximum in 3p. Similarly: 3s max 2e, 3p max 6e, 3d max 10e. Total for n=3 shell = 18 electrons = 2n² = 2×9 = 18 ✓.
4. Why does 4s fill before 3d? ⌄
Aufbau order is by increasing (n+l): 4s has 4+0=4, 3d has 3+2=5. Lower (n+l) fills first → 4s fills before 3d. However, once 3d is partially filled, its energy drops below 4s, which is why transition metal cations lose 4s electrons first when forming ions (e.g., Fe²⁺ is [Ar]3d⁶, not [Ar]3d⁴4s²).
5. What is the electronic configuration of Cu (Z=29)? ⌄
[Ar] 3d¹⁰ 4s¹ (NOT [Ar] 3d⁹ 4s²). The completely filled 3d¹⁰ has extra stability due to maximum exchange energy. One electron transfers from 4s to 3d to achieve this. This exception is very commonly tested in NEET — always remember Cr and Cu as the two main exceptions.
6. What is the maximum value of mₗ for f-orbital? ⌄
For f-orbital: l=3. mₗ ranges from −3 to +3: −3,−2,−1,0,+1,+2,+3. Maximum value = +3. There are 2l+1 = 7 f-orbitals total, each holding 2 electrons for a maximum of 14 electrons in any f-subshell.
7. What does the Heisenberg Uncertainty Principle state? ⌄
It is impossible to simultaneously determine both the exact position (Δx) and exact momentum (Δp) of an electron: Δx × Δp ≥ h/4π. This fundamental quantum uncertainty is why orbitals (probability distributions) are used instead of definite electron paths. The more precisely we know position, the more uncertain momentum is, and vice versa.
8. What is the de Broglie wavelength of an electron? ⌄
λ = h/mv = h/p. For an electron accelerated through voltage V: λ = h/√(2meV). This wave nature of electrons explains why they occupy orbitals (standing wave patterns around the nucleus) rather than circular orbits as Bohr proposed. The Bohr orbits only work for hydrogen because they happen to satisfy the de Broglie condition: nλ = 2πr.
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