The Quantum Mechanical Model of the Atom

The failures of Bohr's model, combined with de Broglie's wave-particle duality and Heisenberg's uncertainty principle, led to a fundamentally new approach to atomic structure — quantum mechanics.

Schrödinger Wave Equation (1926)

Erwin Schrödinger developed a mathematical equation that describes the wave-like behaviour of the electron in an atom:

H^ψ=Eψ\hat{H}\psi = E\psi

where:

  • H^\hat{H} is the Hamiltonian operator (represents total energy)
  • ψ\psi (psi) is the wave function
  • EE is the energy of the electron

You don't need to solve this equation at this level, but you need to understand what it tells us.

What is ψ\psi (Wave Function)?

The wave function ψ\psi is a mathematical function that describes the quantum state of an electron. By itself, ψ\psi has no direct physical meaning.

What is ψ2|\psi|^2 (Probability Density)?

ψ2|\psi|^2 gives the probability density — the probability of finding the electron per unit volume at a given point in space.

Key Point: We can never say "the electron IS here." We can only say "the probability of finding the electron here is X%."

Orbital vs Orbit

Feature Orbit (Bohr) Orbital (QM)
Definition Circular path of electron 3D region of space with high probability of finding electron
Shape Circle s, p, d, f shapes
Electron position Well-defined path Probability distribution
Based on Classical mechanics Quantum mechanics
Max electrons 2n22n^2 per shell 2 per orbital

Quantum Numbers

The solution of the Schrödinger equation for hydrogen gives a set of functions (orbitals), each characterised by three quantum numbers: nn, ll, and mlm_l. A fourth quantum number (msm_s) describes the spin of the electron.

1. Principal Quantum Number (nn)

n=1,2,3,4,n = 1, 2, 3, 4, \ldots

What it determines:

  • The shell (energy level) of the electron: n=1n = 1 (K), n=2n = 2 (L), n=3n = 3 (M), n=4n = 4 (N)
  • The size of the orbital: larger nn → larger orbital → electron is farther from the nucleus
  • The energy of the electron (in hydrogen): En=13.6/n2E_n = -13.6/n^2 eV
  • The maximum number of orbitals in a shell: n2n^2
  • The maximum number of electrons in a shell: 2n22n^2
Shell nn Orbitals (n2n^2) Max electrons (2n22n^2)
K 1 1 2
L 2 4 8
M 3 9 18
N 4 16 32

2. Azimuthal Quantum Number (ll) — also called Angular Momentum or Orbital Quantum Number

l=0,1,2,,(n1)l = 0, 1, 2, \ldots, (n-1)

What it determines:

  • The subshell (shape of the orbital)
  • The orbital angular momentum: L=l(l+1)h2πL = \sqrt{l(l+1)} \cdot \frac{h}{2\pi}
  • The number of subshells in a shell = nn
ll value 0 1 2 3
Subshell s p d f
Shape Spherical Dumbbell Double dumbbell Complex
Orbitals in subshell 1 3 5 7

3. Magnetic Quantum Number (mlm_l)

ml=l,(l1),,0,,(l1),lm_l = -l, -(l-1), \ldots, 0, \ldots, (l-1), l

Total values: 2l+12l + 1

What it determines:

  • The orientation of the orbital in space
  • The number of orbitals in a subshell
Subshell ll mlm_l values Number of orbitals
s 0 0 1
p 1 1,0,+1-1, 0, +1 3 (px,py,pzp_x, p_y, p_z)
d 2 2,1,0,+1,+2-2, -1, 0, +1, +2 5
f 3 3,2,1,0,+1,+2,+3-3, -2, -1, 0, +1, +2, +3 7

4. Spin Quantum Number (msm_s)

ms=+12 or 12m_s = +\frac{1}{2} \text{ or } -\frac{1}{2}

What it determines:

  • The spin of the electron: clockwise (+1/2+1/2, ↑) or anticlockwise (1/2-1/2, ↓)
  • Each orbital can hold a maximum of 2 electrons with opposite spins

The spin quantum number was proposed by Uhlenbeck and Goudsmit (1925) to explain certain features of atomic spectra.

Key Point: The four quantum numbers (n,l,ml,msn, l, m_l, m_s) together completely describe the state of an electron in an atom. No two electrons in an atom can have the same set of all four quantum numbers — this is the Pauli Exclusion Principle (covered in Section 11).

Allowed Combinations of Quantum Numbers

Not every combination of quantum numbers is valid. The rules are:

  1. nn = any positive integer (1, 2, 3, …)
  2. ll = 0 to (n1)(n - 1)
  3. mlm_l = l-l to +l+l
  4. msm_s = +1/2+1/2 or 1/2-1/2

Example: All Orbitals for n=3n = 3

nn ll Subshell mlm_l values Orbitals
3 0 3s 0 1
3 1 3p 1,0,+1-1, 0, +1 3
3 2 3d 2,1,0,+1,+2-2, -1, 0, +1, +2 5

Total orbitals = 1+3+5=9=n21 + 3 + 5 = 9 = n^2 ✓ Total electrons = 9×2=18=2n29 \times 2 = 18 = 2n^2

Invalid Combinations

These are NOT allowed:

  • n=1,l=1n = 1, l = 1 (because ll must be <n< n; for n=1n = 1, only l=0l = 0)
  • n=2,l=2n = 2, l = 2 (because ll can only be 0 or 1 for n=2n = 2)
  • l=1,ml=2l = 1, m_l = 2 (because mlm_l must be between l-l and +l+l)

[JEE Tip] To quickly check validity: l<nl < n and mll|m_l| \leq l. If either condition fails, the combination is invalid.

Designation of Orbitals

Orbitals are written as nlnl notation:

  • 1s,2s,2p,3s,3p,3d,4s,4p,4d,4f,1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f, \ldots

The number represents nn and the letter represents ll.

Solved Examples

Example 1: Listing Quantum Numbers

List all possible values of ll and mlm_l for n=3n = 3.

Solution: For n=3n = 3: l=0,1,2l = 0, 1, 2

  • l=0l = 0: ml=0m_l = 0 (1 orbital → 3s)
  • l=1l = 1: ml=1,0,+1m_l = -1, 0, +1 (3 orbitals → 3p)
  • l=2l = 2: ml=2,1,0,+1,+2m_l = -2, -1, 0, +1, +2 (5 orbitals → 3d)

Total orbitals = 9.

Example 2: Checking Validity of Quantum Numbers

Which of these sets of quantum numbers is not allowed? (a) n=2,l=1,ml=0,ms=+1/2n = 2, l = 1, m_l = 0, m_s = +1/2 (b) n=1,l=1,ml=0,ms=1/2n = 1, l = 1, m_l = 0, m_s = -1/2 (c) n=3,l=2,ml=2,ms=+1/2n = 3, l = 2, m_l = -2, m_s = +1/2

Solution: (a) Valid: l<nl < n (1 < 2) ✓, mll|m_l| \leq l (0 ≤ 1) ✓ (b) Invalid: l=1l = 1 but n=1n = 1, so ll can only be 0. ll must be <n< n. (c) Valid: l<nl < n (2 < 3) ✓, mll|m_l| \leq l (2 ≤ 2) ✓

Answer: (b) is not allowed.

Example 3: Number of Orbitals in a Subshell

How many orbitals are in the 4d subshell?

Solution: For 4d: n=4n = 4, l=2l = 2. Number of orbitals =2l+1=2(2)+1=5= 2l + 1 = 2(2) + 1 = 5.

Answer: 5 orbitals.

Example 4: Maximum Electrons in a Shell

What is the maximum number of electrons in the M shell?

Solution: M shell: n=3n = 3. Maximum electrons =2n2=2(9)=18= 2n^2 = 2(9) = 18.

Breakdown: 3s (2) + 3p (6) + 3d (10) = 18. ✓

Answer: 18 electrons.

Example 5: Orbital Angular Momentum

Calculate the orbital angular momentum of an electron in a 3p orbital.

Solution: For 3p: l=1l = 1. L=l(l+1)h2π=1×2h2π=26.626×10342πL = \sqrt{l(l+1)} \cdot \frac{h}{2\pi} = \sqrt{1 \times 2} \cdot \frac{h}{2\pi} = \sqrt{2} \cdot \frac{6.626 \times 10^{-34}}{2\pi} =1.414×1.055×1034=1.49×1034 J s= 1.414 \times 1.055 \times 10^{-34} = 1.49 \times 10^{-34} \text{ J s}

Answer: L=2=1.49×1034L = \sqrt{2} \cdot \hbar = 1.49 \times 10^{-34} J s.

Example 6: Identifying the Orbital

An electron has quantum numbers n=4,l=2,ml=1,ms=+1/2n = 4, l = 2, m_l = -1, m_s = +1/2. Identify the subshell.

Solution: n=4n = 4 and l=2l = 2 → This is a 4d orbital. ml=1m_l = -1 indicates one of the five orientations of d orbital. ms=+1/2m_s = +1/2 means spin-up electron.

Answer: The electron is in a 4d orbital.

Example 7: Total Orbitals for a Given nn

How many orbitals are present in n=4n = 4?

Solution: Total orbitals =n2=42=16= n^2 = 4^2 = 16.

Subshells: 4s (1) + 4p (3) + 4d (5) + 4f (7) = 16 ✓

Answer: 16 orbitals.

Example 8: Which Subshells Exist?

Which of the following subshells exist: 1p, 2s, 2d, 3f, 3d?

Solution:

  • 1p: n=1,l=1n = 1, l = 1 → Invalid (ll must be <n< n, so l=0l = 0 only) → Does not exist
  • 2s: n=2,l=0n = 2, l = 0 → Valid → Exists
  • 2d: n=2,l=2n = 2, l = 2 → Invalid (ll must be 0 or 1 for n=2n = 2) → Does not exist
  • 3f: n=3,l=3n = 3, l = 3 → Invalid (ll must be 0, 1, or 2 for n=3n = 3) → Does not exist
  • 3d: n=3,l=2n = 3, l = 2 → Valid → Exists

Answer: 2s and 3d exist. 1p, 2d, and 3f do not.

Example 9: Quantum Numbers for the 2p Subshell

Write all possible sets of quantum numbers for electrons in the 2p subshell.

Solution: For 2p: n=2,l=1n = 2, l = 1. Possible mlm_l values: 1,0,+1-1, 0, +1. Each can have ms=+1/2m_s = +1/2 or 1/2-1/2.

nn ll mlm_l msm_s
2 1 1-1 +1/2+1/2
2 1 1-1 1/2-1/2
2 1 00 +1/2+1/2
2 1 00 1/2-1/2
2 1 +1+1 +1/2+1/2
2 1 +1+1 1/2-1/2

Total: 6 sets → maximum 6 electrons in 2p subshell.

Example 10: Relating ψ2|\psi|^2 to Probability

What is the physical significance of ψ2|\psi|^2?

Solution: ψ2|\psi|^2 at a point gives the probability density of finding the electron at that point. To find the probability of finding the electron in a small volume dVdV around a point, we calculate ψ2dV|\psi|^2 \cdot dV.

The total probability over all space must equal 1: ψ2dV=1\int |\psi|^2 \, dV = 1

This is called the normalisation condition.

Answer: ψ2|\psi|^2 is the probability density — it gives the probability of finding the electron per unit volume at a given point.