Chemistry Chapter 2: Atomic Structure | Interactive Learning

Atomic Structure

Explore the fundamental building blocks of matter. Learn about atomic number, quantum numbers, electronic configuration, and more with interactive animations and detailed explanations.

Key Notes – Atomic Structure

2.1 Atomic Number, Proton Number, Nucleon Number

Moseley’s Law (1913): √ν ∝ Z where ν = frequency of characteristic X-rays, Z = atomic number.

Atomic Number (Z): Number of protons in the nucleus. Determines element identity.

Proton Number: Same as atomic number.

Nucleon Number (A): Total protons + neutrons (mass number).

A = Z + N (where N = number of neutrons)

Memorization Tip

Moseley = √ν ∝ Z – Think “Moseley measures Z with X-rays”

A = Z + N – Remember “All Zealous Neutrons”

Example: \(^{27}_{13}Al\) → Z=13, A=27, N=14

For Ions: Electrons change, protons stay same.

  • Cations (positive): Lose electrons → fewer electrons than protons
  • Anions (negative): Gain electrons → more electrons than protons

2.2 Effect of Electric Field on Fundamental Particles

Particle Charge Deflection in Electric Field
Electron -1 (Negative) Toward positive plate (maximum deflection)
Proton +1 (Positive) Toward negative plate
Neutron 0 (Neutral) No deflection (straight path)

Memorization Tip

Opposites Attract: Negative electrons → positive plate. Positive protons → negative plate.

Neutrons are Neutral: No charge = no deflection.

Reason for electron’s greater deflection: Electron mass is ~1/1836 of proton mass (lighter = more deflection).

2.3 Experimental Evidences for Electronic Configuration

Two main sources:

  1. Atomic spectra
  2. Ionization energies
Ionization Energy (IE): Energy required to remove an electron from gaseous atom.

Successive IEs of same element: Shows electron shells (big jumps indicate new shell).

First IEs across periodic table:

  • Down a group: IE decreases (atomic size increases, shielding increases)
  • Across a period: IE increases (nuclear charge increases, same shell)

Memorization Tip

IE Trends: “Down decreases, across increases” (except anomalies at Be-B, N-O, Mg-Al).

Jumps in successive IE: Big jump = new shell closer to nucleus.

2.4 Quantum Numbers

Describe electron’s position and energy in atom (from Schrödinger equation).

Quantum Number Symbol Values Describes
Principal n 1, 2, 3, … Shell (size, energy)
Azimuthal l 0 to (n-1) Subshell shape (s, p, d, f)
Magnetic m -l to +l Orbital orientation
Spin s +½ or -½ Electron spin direction

Memorization Tip

Quantum Numbers Mnemonic: “Please (Principal) Attend (Azimuthal) My (Magnetic) Seminar (Spin)”

l values: 0=s, 1=p, 2=d, 3=f (remember: sharp, principal, diffuse, fundamental)

Maximum electrons in shell: 2n²

Orbitals in subshell: 2l+1

Degenerate orbitals: Same energy (e.g., all three p-orbitals in a subshell).

2.5 Shapes of Atomic Orbitals

Orbital: 3D region where probability of finding electron is maximum.

Orbital Shape No. of Orbitals Nodes
s Spherical 1 n-1
p Dumbbell 3 (px, py, pz) n-2
d Double dumbbell/cloverleaf 5 n-3
f Complex 7 n-4

Memorization Tip

Orbital Shapes: “s is Sphere, p is Peanut, d is Double peanut, f is Fancy/complex”

dz² looks like “dumbbell with collar” – remember the unique shape.

2.6 Electronic Configuration

Definition: Distribution of electrons among orbitals.

Aufbau Principle: Fill orbitals in increasing energy order.

(n + l) Rule:

  1. Lower (n + l) value → lower energy → filled first
  2. If (n + l) equal → lower n filled first
Energy order: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s
Pauli Exclusion Principle: No two electrons can have same set of four quantum numbers.

Hund’s Rule: Electrons fill degenerate orbitals singly first (with parallel spins).

Memorization Tip

Electron filling order: Remember “1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s”

Exceptions: Cr (Z=24): [Ar]4s¹3d⁵ and Cu (Z=29): [Ar]4s¹3d¹⁰ (half-filled/full d-subshell stability).

Solved Multiple Choice Questions

1. 2311Na atom loses one electron to form Na+. Which of the following is true about Na+?
A) It has 10 protons
B) It has 10 neutrons
C) It has 10 electrons
D) All of these
Solution: Sodium (Na) has atomic number 11 → 11 protons. When it loses 1 electron to form Na+, protons remain 11, electrons become 10. Neutrons = 23-11 = 12. So only statement C is correct.
2. The square root of the frequency of the X-rays is directly proportional to the atomic number of an element. This is according to:
A) Boyle’s law
B) Moseley’s law
C) Charles’s law
D) Avogadro’s law
Solution: In 1913, Henry Moseley discovered that √ν ∝ Z where ν = frequency of characteristic X-rays, Z = atomic number. This is Moseley’s law.
3. The e/m ratio for the positive rays is maximum for:
A) Hydrogen
B) Helium
C) Oxygen
D) Nitrogen
Solution: Positive rays are cations. e/m = charge/mass ratio. Hydrogen ion (H⁺) has smallest mass among all ions (1 amu), so highest e/m ratio.
4. Which group has lowest ionization energy in a given period of the periodic table?
A) VIII-A
B) I-A
C) III-A
D) IV-A
Solution: Group I-A (alkali metals) have lowest ionization energy in each period because they have single valence electron in s-orbital, largest atomic size in period, and effective nuclear charge is lowest.
5. All ionization energies are:
A) Strongly exothermic
B) Strongly endothermic
C) Weakly exothermic
D) None of these
Solution: Ionization energy is the energy required to remove an electron from gaseous atom. Energy is always absorbed (endothermic). Breaking attraction between electron and nucleus requires energy input.
6. Which formula helps us to determine the number of orbitals in a subshell?
A) 2n²
B) n²
C) n – l
D) 2l + 1
Solution: Number of orbitals in a subshell = 2l + 1 where l = azimuthal quantum number. For s (l=0): 1 orbital; p (l=1): 3 orbitals; d (l=2): 5 orbitals; f (l=3): 7 orbitals.
7. Magnetic quantum number explains:
A) Shape
B) Orientation
C) Size
D) Spin
Solution: Magnetic quantum number (m) describes the orientation of orbital in space. For p-orbitals (l=1), m = -1, 0, +1 correspond to px, py, pz orientations.
8. A node is space in which probability of finding the electron is:
A) Zero
B) More than 95%
C) 50%
D) Infinite
Solution: Node is region where probability density (ψ²) = 0, so probability of finding electron is zero. Number of radial nodes = n – l – 1, angular nodes = l.
9. The shape of dz² atomic orbital is:
A) Spherical
B) Dumb-bell
C) Dumb-bell with collar
D) Savage shaped
Solution: dz² orbital has unique shape: dumbbell along z-axis with doughnut-shaped collar in xy-plane. Other four d-orbitals have cloverleaf/double dumbbell shapes.
10. No two electrons in an atom can have the same values for all the four quantum numbers. This is:
A) Hund’s rule
B) Pauli’s exclusion principle
C) Aufbau principle
D) n + l rule
Solution: Pauli Exclusion Principle states no two electrons can have identical set of four quantum numbers (n, l, m, s). In an orbital, two electrons must have opposite spins.

Solved Exercise Questions

Short Question 1

Question: There are three orientations of p-orbital due to three values of magnetic quantum number. Justify it.

Solution:
For p-orbitals, azimuthal quantum number l = 1.
Magnetic quantum number m = -l to +l = -1, 0, +1 (three values).
These three m values correspond to three spatial orientations along x, y, and z axes:
• m = -1 → py orbital
• m = 0 → pz orbital
• m = +1 → px orbital
Each orientation has same shape (dumbbell) but different direction in space.

Short Question 2

Question: I₃ of Mg is much bigger than its I₂. Justify.

Solution:
Magnesium (Mg, Z=12) electronic configuration: 1s² 2s² 2p⁶ 3s²

• I₁: Removing 1st electron from 3s² → Mg → Mg⁺ + e⁻
• I₂: Removing 2nd electron from 3s¹ → Mg⁺ → Mg²⁺ + e⁻
• I₃: Removing 3rd electron from 2p⁶ (inner shell) → Mg²⁺ → Mg³⁺ + e⁻

Reason for large I₃:
1. After I₂, Mg²⁺ has noble gas configuration (Ne-like, stable)
2. 3rd electron comes from inner 2p subshell (closer to nucleus)
3. Much stronger nuclear attraction (less shielding, smaller size)
4. Requires significantly more energy → large jump in ionization energy

Short Question 3

Question: Among Li, K, Ca, S, Kr which has lowest first ionization energy? Which has highest?

Solution:
Lowest IE: Potassium (K)
Reason: Group 1 alkali metal, largest atomic size among these, single valence electron in 4s orbital, high shielding effect, lowest effective nuclear charge.

Highest IE: Krypton (Kr)
Reason: Noble gas (Group 18), completely filled stable electronic configuration (1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶), small atomic size for its period, high effective nuclear charge, extremely stable → hardest to remove electron.

Short Question 4

Question: Consider potassium (Z=19). (i) Write full electronic configuration. (ii) Explain why 4s is filled before 3d.

Solution:
(i) Electronic configuration of potassium (Z=19):
1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹

(ii) Why 4s fills before 3d:
According to (n + l) rule:
• 4s: n + l = 4 + 0 = 4
• 3d: n + l = 3 + 2 = 5
Lower (n + l) value → lower energy → filled first.

Additional reasons:
1. 4s orbital penetrates closer to nucleus than 3d
2. 4s has better shielding from nuclear charge
3. For multi-electron atoms, energy order: 4s < 3d (contrary to hydrogen-like atoms)

Short Question 5

Question: Element X: Z=17, A=35. (i) Find protons, neutrons, electrons. (ii) If forms X⁻ ion, find particles.

Solution:
(i) For neutral atom:
Atomic number Z = 17 → protons = 17
Mass number A = 35 → neutrons = A – Z = 35 – 17 = 18
Neutral atom → electrons = protons = 17

(ii) For X⁻ ion (charge = -1):
Gains 1 electron → electrons = 17 + 1 = 18
Protons unchanged = 17 (defines element)
Neutrons unchanged = 18
So X is Chlorine (Cl), and Cl⁻ has p=17, n=18, e=18

Short Question 6

Question: Draw orbital box diagram for valence electrons of phosphorus (Z=15) following Hund’s rule and Pauli principle.

Solution:
Phosphorus (Z=15) electronic configuration: 1s² 2s² 2p⁶ 3s² 3p³
Valence electrons: 3s² 3p³ (5 electrons)

Orbital box diagram:
3s: ↑↓ (2 electrons, opposite spins – Pauli principle)
3p: ↑ ↑ ↑ (three orbitals: px, py, pz)
• Each p-orbital gets 1 electron first (Hund’s rule: maximize parallel spins)
• All three electrons have same spin direction
• No pairing occurs in p-orbitals (only 3 electrons available)

This gives half-filled p-subshell (stable configuration).

Textbook Exercise MCQ Solutions

Q.No. Question Correct Answer Explanation
i Quantum number ‘m’ of free gaseous atom C Spatial orientation of orbital
ii After 3d filled, next electron goes to C 4p (energy order: 4s < 3d < 4p)
iii Quantum numbers for 2p orbitals A n=2, l=1 (p orbital has l=1)
iv n=4, l=0, m=0, s=+½ lies in C 4s orbital (n=4, l=0 → s-subshell)
v Quantum numbers for 4th electron of Be B Be (Z=4): 1s² 2s²; 4th electron: n=2, l=0, m=0
vi Correct order of first IEs B He > F > N > O > Mg (noble gas highest, anomalies N>O)
vii p orbital lobes B 2 lobes (dumbbell shape)
viii Three p orbitals orientation C Mutually perpendicular along x, y, z axes
ix d orbitals in energy level C 5 d-orbitals (m = -2, -1, 0, +1, +2)
x Principal quantum number gives A Size of orbital (and energy)
xi Unpaired electrons in cobalt B 3 unpaired electrons (Co: [Ar]4s²3d⁷ → 3 unpaired in d)

Memorization Tips & Tricks

1. Quantum Numbers Mnemonics

Remembering All Four Quantum Numbers

“Please Attend My Seminar”

  • Please = Principal (n)
  • Attend = Azimuthal (l)
  • My = Magnetic (m)
  • Seminar = Spin (s)

l-values for Subshells

“Some People Don’t Fly”

  • Some = Subshell (l=0)
  • People = P subshell (l=1)
  • Don’t = D subshell (l=2)
  • Fly = F subshell (l=3)

2. Electron Configuration Order

Filling Order Mnemonic

Remember this sequence:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s

Memory Aid: “1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s”
Note the pattern: after 3p comes the block 4s-3d-4p, then 5s-4d-5p, then 6s-4f-5d-6p

(n + l) Rule Shortcut

When (n + l) is equal, the orbital with lower n fills first.
Example: 3d vs 4p → both have (n+l)=5, but 3d fills first (n=3 < n=4).

3. Ionization Energy Trends

General Trend Memory Aid

“Down decreases, Across increases”

  • Down a group: IE decreases (easier to remove electron)
  • Across a period: IE increases (harder to remove electron)

Exceptions to Remember

Three important exceptions in period 2 & 3:

  1. Be (1.57) > B (1.36): Be has full 2s², B starts 2p¹
  2. N (1.40) > O (1.31): N has half-filled 2p³ (stable)
  3. Mg (1.38) > Al (1.42): Mg has full 3s², Al starts 3p¹

Memory: “BeB, NO, MgAl” – remember these pairs where first has higher IE.

4. Shapes of Orbitals

Shape Visualization

  • s-orbital: Spherical like a “ball” or “marble”
  • p-orbital: Dumbbell like “peanut” or “figure-8”
  • d-orbitals: Four are “cloverleaf” (like 4-leaf clover), dz² is “dumbbell with collar”
  • f-orbitals: Complex shapes – just remember there are 7 of them

Number of Orbitals Memory

Formula: Number of orbitals in subshell = 2l + 1

Quick recall: s:1, p:3, d:5, f:7 (odd numbers increasing by 2)

5. Important Exceptions in Electronic Configuration

Cr and Cu Family Exceptions

Remember: Half-filled and fully filled d-subshells are extra stable.

  • Cr (Z=24): Expected: [Ar]4s²3d⁴ → Actual: [Ar]4s¹3d⁵ (half-filled d⁵)
  • Cu (Z=29): Expected: [Ar]4s²3d⁹ → Actual: [Ar]4s¹3d¹⁰ (fully filled d¹⁰)

Memory: “Cr and Cu steal from 4s to complete 3d”

Writing Configuration Quickly

Steps:

  1. Find previous noble gas (shorthand notation)
  2. Fill s-orbital of next shell
  3. Fill d-orbitals of previous shell (if needed)
  4. Fill p-orbitals of current shell
  5. Check for exceptions (Cr, Cu, etc.)

Interactive Atomic Model

Bohr’s Model Visualization

This interactive model shows electrons orbiting the nucleus in different shells. Adjust the element to see different configurations.

How to Read the Model:

  • Red center: Nucleus (protons + neutrons)
  • Green circles: Electrons
  • Dashed circles: Electron shells/orbits (K, L, M, N…)
  • Shell capacities: K=2, L=8, M=18, N=32 electrons

Quantum Numbers Interactive Chart

Shell (n) Subshells (l) Orbitals (m) Max Electrons Example Element