Chemistry Physical - How to Solve: Atomic Structure (Quantum Numbers, Orbital Shapes, Electronic Configuration, Bohr’s Model for H)

How to Solve: Atomic Structure (Quantum Numbers, Orbital Shapes, Electronic Configuration, Bohr’s Model for H)

For NEET UG (Physics & Chemistry) – Master This to Score 10+ Marks

Introduction

"If you can write the electronic configuration of any element in 10 seconds, predict orbital shapes, and solve Bohr’s model problems in under a minute, you’ll lock in 10+ marks in NEET—enough to push you into the top 10%."

WHAT YOU NEED TO KNOW FIRST

1. Basic atomic structure (protons, neutrons, electrons, atomic number, mass number).
2. Energy levels (shells) – Know that n=1 is the first shell, n=2 is the second, etc.
3. Electromagnetic spectrum – Understand that electrons absorb/emit energy as light (photons).

KEY TERMS & FORMULAS

1. Quantum Numbers (4 Types)

MEMORISE THIS: - l = 0 → s-orbital (spherical) - l = 1 → p-orbital (dumbbell) - l = 2 → d-orbital (cloverleaf) - l = 3 → f-orbital (complex)

2. Orbital Shapes & Nodes

MEMORISE THIS: - Radial nodes = n - l - 1 - Angular nodes = l

3. Electronic Configuration Rules

1. Aufbau Principle – Fill orbitals in order of increasing energy: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f… (Use the n + l rule: Lower n + l = lower energy. If same, lower n fills first.)

2. Pauli Exclusion Principle – No two electrons can have the same 4 quantum numbers. (Max 2 electrons per orbital, opposite spins.)

3. Hund’s Rule – Electrons fill degenerate orbitals (same energy) singly before pairing.

MEMORISE THIS: - Exceptions: Cr ([Ar] 4s¹ 3d⁵), Cu ([Ar] 4s¹ 3d¹⁰) – Half/full d-subshell is more stable.

4. Bohr’s Model for Hydrogen

Formulas:
1. Radius of nth orbit (rₙ): rₙ = (0.529 Å) × n² (MEMORISE: 0.529 Å = Bohr radius for n=1)

1. Velocity of electron (vₙ): vₙ = (2.18 × 10⁶ m/s) / n (MEMORISE: 2.18 × 10⁶ m/s = velocity in n=1)

2. Energy of nth orbit (Eₙ): Eₙ = -13.6 eV / n² (MEMORISE: -13.6 eV = ground state energy)

3. Wavelength of emitted/absorbed photon (Rydberg formula): 1/λ = R (1/n₁² - 1/n₂²) (R = 1.097 × 10⁷ m⁻¹ → Given on exam sheet)

STEP-BY-STEP METHOD

A. Quantum Numbers & Orbital Shapes

Step 1: Identify n (principal quantum number). Step 2: Find l (0 to n-1). Step 3: Find mₗ (-l to +l). Step 4: Assign mₛ (+½ or -½). Step 5: Sketch the orbital shape based on l.

B. Electronic Configuration

Step 1: Write the order of filling (1s, 2s, 2p, 3s, 3p, 4s, 3d…). Step 2: Count electrons (atomic number = total electrons). Step 3: Fill orbitals following Aufbau, Pauli, and Hund’s rules. Step 4: Check for exceptions (Cr, Cu, Mo, Ag, Au). Step 5: Write in noble gas notation (e.g., [Ar] 4s² 3d¹⁰).

C. Bohr’s Model Problems

Step 1: Identify n₁ (initial orbit) and n₂ (final orbit). Step 2: Use the correct formula based on what’s asked: - Radius? → rₙ = 0.529 × n² - Energy? → Eₙ = -13.6 / n² - Wavelength? → 1/λ = R (1/n₁² - 1/n₂²) Step 3: Plug in values and solve. Step 4: Check units (eV for energy, Å/nm for wavelength).

WORKED EXAMPLES

Example 1 – Basic: Quantum Numbers & Orbital Shape

Question: For n=3, list all possible quantum numbers and sketch the orbital for l=1.

Solution:
1. n = 3 → l can be 0, 1, 2.
2. For l=1 (p-orbital): - mₗ = -1, 0, +1 - mₛ = +½ or -½
3. Possible sets: (3, 1, -1, +½), (3, 1, -1, -½), (3, 1, 0, +½), (3, 1, 0, -½), (3, 1, +1, +½), (3, 1, +1, -½)
4. Orbital shape: Dumbbell (p-orbital).

What we did and why: - Used n to find l, then l to find mₗ. - Assigned spins to show all possible electrons in a p-orbital.

Example 2 – Medium: Electronic Configuration

Question: Write the electronic configuration of Fe (Z=26) and Fe²⁺.

Solution:
1. Fe (Z=26): - Order: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶ - Noble gas notation: [Ar] 4s² 3d⁶
2. Fe²⁺ (loses 2 electrons): - Remove from 4s first (higher energy when emptying). - [Ar] 3d⁶

What we did and why: - Followed Aufbau order but removed electrons from 4s first (exception to filling order). - Used noble gas notation for conciseness.

Example 3 – Exam-Style: Bohr’s Model (Energy & Wavelength)

Question: An electron jumps from n=4 to n=2 in a hydrogen atom. Calculate: a) Energy emitted (in eV). b) Wavelength of emitted light (in nm).

Solution: a) Energy emitted:
1. E₄ = -13.6 / 4² = -0.85 eV
2. E₂ = -13.6 / 2² = -3.4 eV
3. ΔE = E₂ - E₄ = -3.4 - (-0.85) = -2.55 eV (Negative sign means energy is emitted.)

b) Wavelength:
1. 1/λ = R (1/2² - 1/4²) = 1.097 × 10⁷ (1/4 - 1/16) = 2.057 × 10⁶ m⁻¹
2. λ = 1 / (2.057 × 10⁶) = 4.86 × 10⁻⁷ m = 486 nm

What we did and why: - Used Eₙ = -13.6 / n² for energy levels. - Applied Rydberg formula for wavelength, converting to nm for the answer.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 60-second atomic structure cheat sheet for NEET:

1. Quantum numbers:
2. n = shell (1, 2, 3…)
3. l = shape (0=s, 1=p, 2=d, 3=f)
4. mₗ = orientation (-l to +l)

5. mₛ = spin (+½ or -½)

6. Orbitals:

7. s = sphere, p = dumbbell, d = cloverleaf.

8. Nodes: radial = n - l - 1, angular = l.

9. Electronic configuration:

10. Fill 4s before 3d, but remove from 4s first.

11. Exceptions: Cr (4s¹ 3d⁵), Cu (4s¹ 3d¹⁰).

12. Bohr’s model:

13. rₙ = 0.529 n² Å
14. Eₙ = -13.6 / n² eV
15. 1/λ = R (1/n₁² - 1/n₂²)

Now go crush those 10 marks!