Atomic Structure and Chemical Bonding - Class Notes

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Student Quick Revision Focus High-yield exam revision guide for ICSE Class 9 Chemistry. Master subatomic particles ($e^-, p^+, n^0$), cathode & canal rays, Thomson, Rutherford & Bohr atomic models, Atomic Number ($Z$), Mass Number ($A$), Bohr-Bury electronic configurations ($2n^2$ rule), Isotopes, Isobars, and Electrovalent vs Covalent chemical bonding!

1. Fundamental Concepts & Dalton's Atomic Theory

1.1 Historical Perspective

Maharshi Kanada (6th Century B.C., India): Proposed that matter is composed of tiny indivisible particles called paramanu.
Democritus (Greek Philosopher): Named these indivisible particles "atomos" (meaning uncuttable / indivisible).
John Dalton (1808): Formulated the first scientific Atomic Theory.

Dalton's Postulates vs Modern Atomic Theory

Dalton's View: Atoms are indivisible, indestructible particles of an element.
Modern Atomic Theory: Atoms are divisible into subatomic particles (electrons, protons, neutrons). Atoms of the same element may differ in mass (Isotopes), and atoms of different elements may have identical mass (Isobars).

2. Discovery of Subatomic Particles

2.1 Discovery of Electrons (Cathode Rays)

Discovered by J.J. Thomson (1897) using a discharge tube at high voltage ($10,000\text{ V}$) and low pressure ($0.01\text{ mm Hg}$).

Fig 4.1: Discharge tube experiment for production of Cathode Rays

Deflection of Cathode Rays in Electric Field

Fig 4.2: Deflection of Cathode Rays toward positive plate in Electric Field

Properties: Travel in straight lines, carry negative charge (deflected toward positive plate), cause fluorescence, spin light paddle wheels (possess kinetic energy & momentum).
Electron Properties: Charge $= -1.602 \times 10^{-19}\text{ C}$, Mass $= 9.108 \times 10^{-31}\text{ kg} = \frac{1}{1837}\text{ mass of H-atom}$.

2.2 Discovery of Protons (Anode / Canal Rays)

Discovered by Eugen Goldstein using a discharge tube with a perforated cathode.

Fig 4.3: Production of Anode (Canal) Rays passing through perforated cathode

Properties: Positively charged particles originating from ionized gas molecules. Their $e/m$ ratio depends on the nature of gas in the tube (maximum for Hydrogen gas $\rightarrow$ bare proton $p^+$).
Proton Properties: Charge $= +1.602 \times 10^{-19}\text{ C}$, Mass $= 1.672 \times 10^{-27}\text{ kg} = 1\text{ amu}$.

2.3 Discovery of Neutrons

Discovered by James Chadwick (1932) by bombarding Beryllium sheet with $\alpha$-particles:

$${}_4^9\text{Be} + {}_2^4\text{He} \longrightarrow {}_6^{12}\text{C} + {}_0^1\text{n}$$

Neutral particle (charge $= 0$), Mass $= 1.676 \times 10^{-27}\text{ kg} \approx 1\text{ amu}$.

**Comparison of Fundamental Subatomic Particles**
Particle	Symbol	Relative Charge	Absolute Charge (C)	Mass (amu)	Mass (kg)	Discoverer
Electron	${}_{-1}^0e$ or $e^-$	$-1$	$-1.602 \times 10^{-19}$	$\frac{1}{1837} \approx 0.00055$	$9.108 \times 10^{-31}$	J.J. Thomson
Proton	${}_1^1p$ or $p^+$	$+1$	$+1.602 \times 10^{-19}$	$1.00727 \approx 1$	$1.672 \times 10^{-27}$	E. Goldstein
Neutron	${}_0^1n$ or $n^0$	$0$	$0$	$1.00866 \approx 1$	$1.676 \times 10^{-27}$	James Chadwick

3. Atomic Models & Structure of the Atom

3.1 Thomson's Plum-Pudding Model

Fig 4.4: J.J. Thomson's Plum-Pudding Model of the Atom

Atom is a positively charged sphere with electrons embedded like raisins in a pudding. Limitation: Failed to explain $\alpha$-scattering results.

3.2 Rutherford's $\alpha$-Particle Scattering Experiment

Fig 4.5: Deflection of $\alpha$-particles by gold foil in Rutherford's experiment

Fig 4.6: Rutherford's Nuclear Planetary Model of the Atom

Observations: Most $\alpha$-particles passed straight; a few deflected by small angles; $1$ in $10,000$ bounced back by $180^\circ$.
Nuclear Model: Extremely small, dense, positively charged Nucleus at center containing almost all mass; electrons revolve in orbits.

Drawback of Rutherford Model - Energy Radiation Collapse

Fig 4.7: Drawback - Accelerating electron radiates energy and spirals into nucleus

Drawback of Rutherford's Model According to Maxwell's classical electrodynamics, a revolving accelerated electron must continuously emit energy, spiral inwards, and collapse into the nucleus. Rutherford could not explain atomic stability!

3.3 Bohr's Atomic Model

Fig 4.8: Energy levels (K, L, M, N shells) around nucleus in Bohr's Model

Bohr's Postulates 1. Electrons revolve only in non-radiating, fixed circular paths called stationary orbits / energy levels (K, L, M, N or $n=1, 2, 3, 4$).
2. While revolving in a stationary orbit, the electron neither emits nor absorbs energy.
3. Energy is absorbed or emitted only when an electron jumps from one energy level to another.

Structure of an Atom - Nucleus vs Orbits

Fig 4.9: Structural components of an atom (Nucleus containing Nucleons & Outer Orbits)

4. Atomic Number, Mass Number & Electronic Configuration

Core Formulas $$\mathbf{Z = \text{Atomic Number} = \text{Protons } (p^+) = \text{Electrons } (e^-) \quad (\text{neutral atom})}$$ $$\mathbf{A = \text{Mass Number} = \text{Protons } (p^+) + \text{Neutrons } (n^0)}$$ $$\mathbf{\text{Number of Neutrons } (n^0) = A - Z}$$ Symbolic Representation: $\mathbf{{}_Z^A\text{X}}$ (e.g. ${}_{17}^{35}\text{Cl} \rightarrow Z=17, A=35, n=18$).

4.1 Bohr-Bury Rules for Electronic Configuration

Maximum Shell Capacity ($2n^2$):
• K Shell ($n=1$): Max $2 \times 1^2 = \mathbf{2 e^-}$
• L Shell ($n=2$): Max $2 \times 2^2 = \mathbf{8 e^-}$
• M Shell ($n=3$): Max $2 \times 3^2 = \mathbf{18 e^-}$
• N Shell ($n=4$): Max $2 \times 4^2 = \mathbf{32 e^-}$
Octet Rule: Outermost shell cannot hold more than 8 electrons (duplet rule: max 2 for K shell).
Stepwise Filling: Inner shells must be completely filled before outer shells begin filling.

Fig 4.10: Atomic orbit diagram of Magnesium atom ($Z=12$, Config: 2, 8, 2)

**Electronic Configurations of First 20 Elements (H to Ca)**
Element	Symbol	Atomic Number ($Z$)	Protons ($p^+$)	Neutrons ($n^0$)	Configuration (K, L, M, N)	Valency
Hydrogen	H	1	1	0	1	1
Helium	He	2	2	2	2	0
Lithium	Li	3	3	4	2, 1	1
Beryllium	Be	4	4	5	2, 2	2
Boron	B	5	5	6	2, 3	3
Carbon	C	6	6	6	2, 4	4
Nitrogen	N	7	7	7	2, 5	3
Oxygen	O	8	8	8	2, 6	2
Fluorine	F	9	9	10	2, 7	1
Neon	Ne	10	10	10	2, 8	0
Sodium	Na	11	11	12	2, 8, 1	1
Magnesium	Mg	12	12	12	2, 8, 2	2
Aluminium	Al	13	13	14	2, 8, 3	3
Silicon	Si	14	14	14	2, 8, 4	4
Phosphorus	P	15	15	16	2, 8, 5	3, 5
Sulfur	S	16	16	16	2, 8, 6	2
Chlorine	Cl	17	17	18	2, 8, 7	1
Argon	Ar	18	18	22	2, 8, 8	0
Potassium	K	19	19	20	2, 8, 8, 1	1
Calcium	Ca	20	20	20	2, 8, 8, 2	2

5. Isotopes, Isobars & Chemical Bonding Overview

Definitions & Examples

Isotopes: Atoms of the same element having the same atomic number ($Z$) but different mass numbers ($A$) due to different number of neutrons.
• Hydrogen: Protium (${}_1^1\text{H}$), Deuterium (${}_1^2\text{H}$), Tritium (${}_1^3\text{H}$).
• Chlorine: ${}_{17}^{35}\text{Cl}$ and ${}_{17}^{37}\text{Cl}$ (ratio $3:1$, average atomic mass $= 35.5\text{ amu}$).
Isobars: Atoms of different elements having the same mass number ($A$) but different atomic numbers ($Z$).
• Example: ${}_{18}^{40}\text{Ar}$ and ${}_{20}^{40}\text{Ca}$.
Isotones: Atoms of different elements having the same number of neutrons ($n = A - Z$).
• Example: ${}_6^{14}\text{C}$ ($n=8$) and ${}_8^{16}\text{O}$ ($n=8$).

5.1 Fundamentals of Chemical Bonding

Atoms combine to achieve a stable octet (8 valence electrons) or duplet (2 valence electrons) inert gas configuration.

Types of Chemical Bonds

Electrovalent (Ionic) Bond: Formed by complete transfer of electrons from a metal to a non-metal cation and anion formation held by strong electrostatic forces (e.g. $\text{NaCl}, \text{MgCl}_2, \text{CaO}$).
Covalent Bond: Formed by mutual sharing of electron pairs between non-metal atoms (e.g. $\text{H}_2, \text{O}_2, \text{N}_2, \text{H}_2\text{O}, \text{CH}_4$).

Student ICSE Exam Practice Set Q1 (ICSE Board): Calculate the number of protons, neutrons, and electrons in ${}_{19}^{39}\text{K}^+$.

Solution:
• Protons $= Z = \mathbf{19}$.
• Neutrons $= A - Z = 39 - 19 = \mathbf{20}$.
• Electrons (since $+1$ charge) $= 19 - 1 = \mathbf{18}$.

Q2 (ICSE Board): State the maximum number of electrons that can be accommodated in the M-shell.

Solution: Formula $= 2n^2$. For M-shell ($n=3$), Max capacity $= 2 \times (3)^2 = \mathbf{18\text{ electrons}}$.