Atomic Orbitals

• Shapes of s, p, d orbitals and how l quantum number determines them.
• Node structures — radial and angular — and why 2s has a spherical node.
• Orbital capacity: s=2, p=6, d=10, f=14 electrons maximum.
• To relate orbital diagrams to bonding and magnetic properties (paired vs unpaired electrons).

Level 1 — Orbital shapes and quantum numbers

Verbal: An orbital is a region in space (probability ≥ ~90%) where an electron with specific n and l is found. Unlike Bohr orbits, shape depends on l.

Symbolic: s: l=0 (2 e⁻ max); p: l=1 (6 e⁻); d: l=2 (10 e⁻); total nodes = n − l − 1; 2n² electrons per shell n (rule of thumb).

Shapes:

• s (l=0): Spherical — 1s, 2s, 3s…
• p (l=1): Dumbbell along x, y, z — p_x, p_y, p_z (three orbitals)
• d (l=2): Cloverleaf + one unique (d_z²) — five d orbitals

Number of orbitals in subshell: 2l + 1 → p has 3, d has 5.

Level 2 — Nodes, energy order, and diagrams

Energy (one-electron): Depends mainly on n; for same n, E(s) < E(p) < E(d) slightly in multi-electron atoms due to penetration/shielding.

Orbital box diagram: □ each box = orbital; ↑↓ paired spins; Hund: spread ↑ first in degenerate set.

Paramagnetic: Unpaired electrons — attracted to magnetic field. Diamagnetic: All paired — weak repulsion.

NCERT spotlight — Quantum numbers and nodes

For 3p orbital: n=3, l=1, total nodes n-l-1 = 1 (one radial or angular depending on sublevel). Radial nodes increase with n for same l.

Shapes and exams: Draw boundary surface of px, py, pz. d orbitals: four clover plus dz squared — five orientations, max 10 electrons.

Magnetic behaviour: Diamagnetic if all paired; paramagnetic if unpaired electrons present — test with orbital box diagrams.

Draw orbital diagram for oxygen (Z = 8): 1s² 2s² 2p⁴. Count unpaired electrons and predict magnetic behaviour.

Step 1 — Fill: 1s ↓↑, 2s ↓↑, 2p: three boxes px py pz.
Step 2 — Hund's for 2p⁴: ↑↓ ↑ ↑ (two unpaired in p orbitals).
Step 3 — Unpaired count = 2 → paramagnetic O₂ (important for NEET: O₂ paramagnetic!).
Step 4 — Compare C (2p²): ↑ ↑ unpaired 2; N (2p³): ↑ ↑ ↑ three unpaired (Hund max).
Step 5 — All paired example: O²⁻ (2p⁶) diamagnetic.

Applications — bonding directionality

p orbital overlap end-on gives sigma bond; sideways overlap gives pi bond — explains why double bonds restrict rotation. sp3 hybridisation in carbon explains tetrahedral methane and diamond lattice. MRI uses nuclear spin states in magnetic field — quantum mechanical property of protons in water of tissues for medical imaging context.

Deep dive — radial probability and hybridisation intro

Radial probability density 4 pi r squared psi squared dr — 1s maximum near nucleus; 2s has radial node sphere where probability zero — explains why 2s higher energy than 1s. p orbitals dumbbell zero electron density at nucleus — penetrates less than s. d orbitals in transition metals colour from d-d transitions — crystal field splitting preview. Hybridisation sp3 one s + three p → four equivalent tetrahedral orbitals in CH4 — explains VSEPR geometry bond angles 109.5°. sp2 trigonal planar 120° BF3; sp linear 180° BeCl2. Paramagnetic O2 molecular orbital theory (NEET favourite): pi* antibonding unpaired electrons despite Lewis double bond appearance — orbital diagram trumps simple octet for diatomic homonuclear second period advanced questions. Nodal surfaces increase with n and l — higher nodes higher energy orbitals more diffuse.

Review and practice drill

Review checklist: (1) s spherical, p dumbbell, d clover. (2) Max electrons 2,6,10,14 for s,p,d,f. (3) Nodes n-l-1. (4) Orbital box diagrams for paramagnetism. Practice: O atom 2p4 — two unpaired electrons paramagnetic.

• How many orbitals in d subshell?
• Which orbital has a spherical node: 1s or 2s?
• Write orbital diagram for chromium (Z = 24) — note 4s¹ 3d⁵ exception.

Open the Practice tab for graded questions on Orbitals.