Structure of the Atom: Models, Electron Configuration, Valency, and Nuclear Properties

At the end of the nineteenth century, one of the greatest challenges before scientists was to reveal the internal structure of the atom and to explain its fundamental properties. John Dalton’s atomic theory had previously proposed that the atom was an indivisible, ultimate particle of matter. However, the discovery of static electricity and subatomic particles—electrons, protons, and neutrons—shattered this view.

Understanding the structure of an atom is crucial because it provides the foundational framework for all of chemistry and modern physics. It explains why elements react the way they do, how chemical bonds form, why materials possess distinct physical and chemical characteristics, and how energy is released in nuclear processes. This study guide offers a comprehensive breakdown of atomic models, electronic configuration rules, valency, and nuclear properties as prescribed in the NCERT Class 9 Science curriculum.

1. In-Depth Conceptual Breakdown

1.1 Discovery of Subatomic Particles

Matter is electrically neutral under normal conditions, yet simple experiments show that rubbing two objects together (such as a glass rod with silk or a plastic comb with dry hair) gives them an electric charge. This phenomenon proved that atoms contain smaller, charged constituents.

                  ┌─────────────────────────────────────────┐
                  │               THE ATOM                  │
                  └────────────────────┬────────────────────┘
                                       │
         ┌─────────────────────────────┼─────────────────────────────┐
         ▼                             ▼                             ▼
   Electron (e⁻)                 Proton (p⁺)                   Neutron (n⁰)
  - Discovered by Thomson       - Discovered by Goldstein     - Discovered by Chadwick
  - Negative charge             - Positive charge             - Neutral (no charge)
  - Negligible mass             - Mass ≈ 1 u                  - Mass ≈ 1 u

The Electron ($e^-$)

• Discovery: J.J. Thomson (1897) identified negatively charged particles during experiments with cathode ray discharge tubes.
• Charge: $-\text{1.6} \times 10^{-19}\text{ Coulombs}$ (Relative charge = $-1$).
• Mass: $9.11 \times 10^{-31}\text{ kg}$, which is approximately $\frac{1}{1840}$ of the mass of a hydrogen atom (considered negligible for atomic mass calculations).

The Proton ($p^+$)

• Discovery: E. Goldstein (1886) discovered new radiations in a gas discharge tube called canal rays (anode rays), which led to the discovery of the positively charged proton.
• Charge: $+\text{1.6} \times 10^{-19}\text{ Coulombs}$ (Relative charge = $+1$).
• Mass: $1.672 \times 10^{-27}\text{ kg}$ (Taken as $1\text{ atomic mass unit (u)}$, approximately 2000 times that of an electron).

The Neutron ($n^0$)

• Discovery: James Chadwick (1932) discovered an uncharged subatomic particle by bombarding beryllium with alpha particles.
• Charge: Neutral ($0$ charge).
• Mass: $1.675 \times 10^{-27}\text{ kg}$ (Slightly greater than a proton, taken as $1\text{ u}$). Neutrons reside in the nucleus of all atoms except hydrogen (${}_{1}^{1}\text{H}$).

1.2 Evolution of Atomic Models

A. Thomson’s Model of the Atom (Plum Pudding Model)

J.J. Thomson proposed that an atom is structurally similar to a Christmas pudding or a watermelon.

• Postulates:
  1. An atom consists of a positively charged sphere with electrons embedded in it.
  2. The negative and positive charges are equal in magnitude; therefore, the atom as a whole is electrically neutral.

$\text{Total Positive Charge of Sphere} = \text{Total Negative Charge of Embedded Electrons}$

• Limitations: Although Thomson’s model explained electrical neutrality, it failed to explain the results of experiments carried out by Ernest Rutherford.

       Thomson's Model                    Rutherford's Model
     ┌─────────────────┐                 ┌─────────────────┐
     │  +  -  +  -  +  │                 │    e⁻    e⁻     │
     │ -  +  -  +  -  +│                 │     ┌───┐       │
     │  +  -  +  -  +  │                 │ e⁻  │p⁺n│  e⁻   │
     │ (Pos. Sphere with               │     └───┘       │
     │  embedded e⁻)   │                 │    e⁻    e⁻     │
     └─────────────────┘                 └─────────────────┘

B. Rutherford’s $\alpha$-Particle Scattering Experiment

Rutherford designed an experiment to probe the structure inside an atom by bombarding a very thin gold foil ($\approx 1000$ atoms thick) with fast-moving alpha ($\alpha$) particles (doubly charged helium ions, $\text{He}^{2+}$, mass $= 4\text{ u}$).

• Observations:

  1. Most of the fast-moving $\alpha$-particles passed straight through the gold foil without any deflection.
  2. A small fraction of $\alpha$-particles was deflected by small angles.
  3. A very tiny fraction (about 1 out of every 12,000 particles) rebounded back completely ($180^\circ$ deflection).

• Conclusions:

  1. Most of the space inside the atom is empty because most $\alpha$-particles passed straight through.
  2. The positive charge occupies a very tiny volume because only a few particles were deflected from their path.
  3. All the positive charge and mass of the atom are concentrated in a very small region called the nucleus.

• Features of Rutherford’s Nuclear Model:

  1. There is a positively charged, extremely dense center in an atom called the nucleus. Nearly all the mass of an atom resides in the nucleus.
  2. Electrons revolve around the nucleus in circular paths called orbits.
  3. The size of the nucleus ($10^{-15}\text{ m}$) is very small compared to the size of the atom ($10^{-10}\text{ m}$).

• Major Drawback: According to classical electromagnetic theory, any charged particle undergoing circular motion experiences acceleration. An accelerated electron must continuously radiate energy. As a result, the revolving electron would lose energy, slow down, and ultimately spiral into the nucleus. This would render the atom unstable, meaning matter could not exist in the stable form we observe.

C. Bohr’s Model of the Atom

To overcome the limitations of Rutherford’s model, Niels Bohr (1913) proposed revised postulates:

• Postulates:
  1. Only certain special orbits known as discrete orbits (or distinct energy levels) of electrons are allowed inside the atom.
  2. While revolving in discrete orbits, electrons do not radiate energy.
  3. These orbits or shells are called energy levels. They are represented by the letters $K, L, M, N...$ or by numbers $n = 1, 2, 3, 4...$.

                        Shell n=4 (N-shell)
                     Shell n=3 (M-shell)
                  Shell n=2 (L-shell)
               Shell n=1 (K-shell)
                   ┌─────────┐
                   │ Nucleus │
                   │ (p⁺, n⁰)│
                   └─────────┘

1.3 Distribution of Electrons in Shells (Bohr-Bury Scheme)

The distribution of electrons into various energy shells of an atom is governed by the Bohr-Bury Rules:

1. Maximum Capacity Rule ($2n^2$): The maximum number of electrons present in a shell is given by the formula $2n^2$, where $n$ is the orbit number (or energy level index).

   • For $K\text{-shell } (n = 1)$: Maximum electrons $= 2(1)^2 = 2$
   • For $L\text{-shell } (n = 2)$: Maximum electrons $= 2(2)^2 = 8$
   • For $M\text{-shell } (n = 3)$: Maximum electrons $= 2(3)^2 = 18$
   • For $N\text{-shell } (n = 4)$: Maximum electrons $= 2(4)^2 = 32$

2. Octet Rule for Valence Shell: The maximum number of electrons that can be accommodated in the outermost shell is 8 (even if the shell has capacity for more under the $2n^2$ formula).

3. Step-wise Filling Rule: Electrons are not accommodated in a given shell unless the inner shells are completely filled. Shells are filled in a step-by-step manner.

1.4 Valency and Electronic Configurations

• Valence Electrons: The electrons present in the outermost shell of an atom are known as its valence electrons.
• Valency: The combining capacity of an atom of an element to form chemical bonds and achieve a stable octet (8 electrons in the valence shell, or 2 for hydrogen/helium—a duplet) is called its valency.

Rules for Determining Valency:

1. If the number of valence electrons ($V$) is $1, 2, 3,$ or $4$: $\text{Valency} = V$

2. If the number of valence electrons ($V$) is $5, 6, 7,$ or $8$: $\text{Valency} = 8 - V$

Electronic Configurations and Valencies of First 18 Elements:

1.5 Atomic Number, Mass Number, and Notation

Atomic Number ($Z$)

The total number of protons present in the nucleus of an atom of an element is called its Atomic Number ($Z$).

• Every element has a unique atomic number.
• In a neutral atom:

$Z = \text{Number of protons} = \text{Number of electrons}$

Mass Number ($A$)

The total sum of the number of protons and neutrons present in the nucleus of an atom is called its Mass Number ($A$). Protons and neutrons together are referred to as nucleons.

$A = \text{Number of protons } (Z) + \text{Number of neutrons } (N)$ $N = A - Z$

Standard Atomic Notation

An element $\text{X}$ with atomic number $Z$ and mass number $A$ is written as:

${}_{Z}^{A}\text{X}$

Example: ${}_{11}^{23}\text{Na}$ indicates Sodium with $Z = 11$ (11 protons, 11 electrons) and mass number $A = 23$ ($N = 23 - 11 = 12$ neutrons).

1.6 Isotopes and Isobars

                             NUCLEAR VARIATIONS
                                     │
                 ┌───────────────────┴───────────────────┐
                 ▼                                       ▼
             ISOTOPES                                ISOBARS
  - Same Atomic Number (Z)                - Different Atomic Number (Z)
  - Different Mass Number (A)             - Same Mass Number (A)
  - Same Chemical Properties              - Different Chemical Properties
  - Example: ¹²₆C and ¹⁴₆C                - Example: ⁴⁰₁₈Ar and ⁴⁰₂₀Ca

Isotopes

Definition: Atoms of the same element having the same atomic number ($Z$) but different mass numbers ($A$).

• Examples:

  1. Hydrogen: Protium (${}_{1}^{1}\text{H}$), Deuterium (${}_{1}^{2}\text{H}$ or $\text{D}$), Tritium (${}_{1}^{3}\text{H}$ or $\text{T}$).
  2. Carbon: Carbon-12 (${}_{6}^{12}\text{C}$) and Carbon-14 (${}_{6}^{14}\text{C}$).
  3. Chlorine: Chlorine-35 (${}_{17}^{35}\text{Cl}$) and Chlorine-37 (${}_{17}^{37}\text{Cl}$).

• Properties:

  • Chemical Properties: Identical, because they have the same atomic number and same electron configuration.
  • Physical Properties: Different (such as mass, density, boiling point), because their mass numbers differ due to different neutron counts.

• Average Atomic Mass Formula: If an element exists in isotopic forms with fractional abundances:

$\text{Average Atomic Mass} = \left( \text{Mass}_1 \times \frac{\%_1}{100} \right) + \left( \text{Mass}_2 \times \frac{\%_2}{100} \right)$

• Applications of Isotopes:
  1. An isotope of Uranium (${}^{235}\text{U}$) is used as fuel in nuclear reactors.
  2. An isotope of Cobalt (${}^{60}\text{Co}$) is used in the treatment of cancer.
  3. An isotope of Iodine (${}^{131}\text{I}$) is used in the treatment of goitre.

Isobars

Definition: Atoms of different elements with different atomic numbers ($Z$), which possess the same mass number ($A$).

• Examples:
  1. Argon (${}_{18}^{40}\text{Ar}$) and Calcium (${}_{20}^{40}\text{Ca}$): Both have a mass number of $40$, but their atomic numbers are $18$ and $20$ respectively.
  2. Carbon-14 (${}_{6}^{14}\text{C}$) and Nitrogen-14 (${}_{7}^{14}\text{N}$).

2. Real-World Applications & Analogies

1. The Solar System Analogy (Bohr’s Planetary Model)

Visualize the atom like our solar system. The heavy nucleus acts like the Sun, sitting in the center. The electrons are like planets orbiting at distinct, fixed distances.

Just as a space rocket needs a precise burst of fuel (energy) to jump from an orbit closer to Earth to a higher orbit, an electron must absorb a specific quantum of light energy to jump to a higher shell. When it drops back down, it emits that exact amount of energy as light.

2. Medical Radiotherapy & Diagnostics

• Cobalt-60 (${}^{60}\text{Co}$): Cancer cells divide rapidly and are vulnerable to high-energy radiation. Radiotherapy machines utilize the gamma rays emitted by radioactive Cobalt-60 to target and destroy cancerous tumors without surgical incisions.
• Iodine-131 (${}^{131}\text{I}$): The thyroid gland uses iodine to make hormones. When a patient suffers from goitre or thyroid disorders, doctors administer tiny doses of radioactive Iodine-131, which selectively concentrates in the thyroid, helping to map or treat the diseased tissue.

3. Carbon Dating in Archaeology

Living plants and animals absorb both stable Carbon-12 and radioactive Carbon-14 (${}_{6}^{14}\text{C}$) in a fixed ratio. When the organism dies, it stops absorbing carbon.

Over thousands of years, the radioactive ${}_{6}^{14}\text{C}$ slowly decays while ${}_{6}^{12}\text{C}$ stays constant. By measuring the ratio of Carbon-14 to Carbon-12 in ancient fossils or wood samples, archaeologists can calculate the exact age of ancient artifacts.

3. Step-by-Step Solved Examples

Example 1: Calculating Average Atomic Mass

Question: Natural chlorine consists of two isotopes: ${}_{17}^{35}\text{Cl}$ with relative abundance $75\%$ and ${}_{17}^{37}\text{Cl}$ with relative abundance $25\%$. Calculate the average atomic mass of chlorine.

Solution:

• Step 1: Identify the given data.

  • Mass of Isotope 1 (${}^{35}\text{Cl}$) $= 35\text{ u}$, Abundance $\%_1 = 75\%$
  • Mass of Isotope 2 (${}^{37}\text{Cl}$) $= 37\text{ u}$, Abundance $\%_2 = 25\%$

• Step 2: Apply the Average Atomic Mass formula.

$\text{Average Atomic Mass} = \left( \text{Mass}_1 \times \frac{\%_1}{100} \right) + \left( \text{Mass}_2 \times \frac{\%_2}{100} \right)$

• Step 3: Substitute the numerical values.

$\text{Average Atomic Mass} = \left( 35 \times \frac{75}{100} \right) + \left( 37 \times \frac{25}{100} \right)$

$\text{Average Atomic Mass} = \left( 35 \times \frac{3}{4} \right) + \left( 37 \times \frac{1}{4} \right)$

$\text{Average Atomic Mass} = \frac{105}{4} + \frac{37}{4} = \frac{142}{4} = 35.5\text{ u}$

Final Answer: The average atomic mass of chlorine is $35.5\text{ u}$.

Example 2: Determining Subatomic Composition of Ions

Question: An ion $\text{M}^{3+}$ has $10$ electrons and $14$ neutrons. Find the atomic number ($Z$) and mass number ($A$) of the neutral element $\text{M}$. Identify the element.

Solution:

• Step 1: Determine the number of electrons in the neutral atom. The species is a tripositive ion ($\text{M}^{3+}$), meaning it has lost $3$ electrons.

$\text{Electrons in neutral atom } \text{M} = (\text{Electrons in } \text{M}^{3+}) + 3 = 10 + 3 = 13$

• Step 2: Determine Atomic Number ($Z$). In a neutral atom, number of protons = number of electrons.

$Z = \text{Number of protons} = 13$

• Step 3: Calculate Mass Number ($A$).

$A = \text{Protons } (Z) + \text{Neutrons } (N)$ $A = 13 + 14 = 27$

• Step 4: Identify the element. Element with atomic number $Z = 13$ is Aluminium ($\text{Al}$).

Final Answer: Atomic number $Z = 13$, Mass number $A = 27$. The element is Aluminium (${}_{13}^{27}\text{Al}$).

Example 3: Bohr-Bury Configuration & Valency Calculation

Question: The atomic number of an element $\text{X}$ is $15$.

1. Write its electronic configuration.
2. Calculate its valency.
3. Draw its atomic structure representation symbolically.

Solution:

• Step 1: Write electronic configuration. Total electrons = $15$.
  • $K\text{-shell} = 2$ (remains $13$)
  • $L\text{-shell} = 8$ (remains $5$)
  • $M\text{-shell} = 5$

$\text{Electronic Configuration} = 2, 8, 5$

• Step 2: Determine valency. Number of valence electrons ($V$) = $5$. Since $V > 4$:

$\text{Valency} = 8 - V = 8 - 5 = 3$

(Note: Phosphorus can also show a valency of 5 by sharing all 5 valence electrons, but its primary combining capacity for Class 9 syllabus is 3).

Final Answer: Electronic configuration is $2, 8, 5$ and its valency is $3$ (Element is Phosphorus).

4. Common Student Mistakes to Avoid

1. Confusing "Valence Electrons" with "Valency"

• The Error: Writing that Nitrogen (atomic number $7$, configuration $2, 5$) has a valency of $5$.
• The Correction: Valence electrons are the electrons in the outermost shell ($5$). Valency is the combining capacity required to complete the octet ($8 - 5 = 3$).

2. Misapplying the $2n^2$ Rule for Outermost Shells

• The Error: Filling the $M$-shell of Potassium ($Z = 19$) as $2, 8, 9$ because the $M$-shell can hold up to $18$ electrons.
• The Correction: The Octet Rule strictly limits the outermost shell of any neutral atom to a maximum of 8 electrons. Therefore, after filling 8 electrons in the $M$-shell, the 19th electron must enter the $N$-shell. The correct configuration for Potassium is $2, 8, 8, 1$.

3. Confusing Mass Number with Average Atomic Mass

• The Error: Assuming Mass Number ($A$) can be a decimal value like $35.5$.
• The Correction: Mass number ($A$) is always a whole number integer because it is the sum of whole subatomic particles (protons + neutrons). The fractional value $35.5\text{ u}$ is the Average Atomic Mass, which accounts for natural isotopic abundances.

4. Swapping Definitions of Isotopes and Isobars

• The Error: Stating that ${}_{18}^{40}\text{Ar}$ and ${}_{20}^{40}\text{Ca}$ are isotopes.
• The Correction: Remember the mnemonic:
  • Isotopes $\rightarrow$ Same Top-level chemistry / Same Atomic number ($Z$).
  • Isobars $\rightarrow$ Same Bulk mass / Same Mass number ($A$).

5. Practice Questions for Self-Assessment

Question 1

An atom of an element has $3$ electrons in its $M$-shell.

1. What is its atomic number?
2. What is its electronic configuration?
3. What is its valency?
4. Identify the element.

Question 2

The mass number of an element $\text{Y}$ is $31$, and its nucleus contains $16$ neutrons.

1. Find the atomic number of $\text{Y}$.
2. Write the electronic configuration of $\text{Y}$.
3. What will be the charge on the ion formed by $\text{Y}$ to achieve a stable octet?

Question 3

An element has two natural isotopes: ${}^{10}\text{B}$ ($20\%$ abundance) and ${}^{11}\text{B}$ ($80\%$ abundance). Calculate the average atomic mass of Boron.

Solutions to Practice Questions

Solution 1:

1. Since electrons are in the $M$-shell, the inner $K$ and $L$ shells must be completely filled.
   • $K = 2$, $L = 8$, $M = 3$.
   • Total electrons $= 2 + 8 + 3 = 13$.
   • Atomic Number ($Z$) = 13.
2. Electronic configuration = $2, 8, 3$.
3. Valence electrons = $3$. Since $V \le 4$, Valency = 3.
4. The element is Aluminium ($\text{Al}$).

Solution 2:

1. $A = 31$, $N = 16$.
   • $Z = A - N = 31 - 16 = 15$.
   • Atomic Number ($Z$) = 15.
2. Electronic configuration = $2, 8, 5$.
3. To achieve a stable octet ($8$ valence electrons), atom $\text{Y}$ needs to gain $3$ electrons. Gaining 3 negative charges forms an anion with a charge of $-3$ (Formula: $\text{Y}^{3-}$).

Solution 3:

• Apply the average atomic mass formula:

$\text{Average Atomic Mass} = \left( 10 \times \frac{20}{100} \right) + \left( 11 \times \frac{80}{100} \right)$

$\text{Average Atomic Mass} = \left( 10 \times 0.2 \right) + \left( 11 \times 0.8 \right) = 2.0 + 8.8 = 10.8\text{ u}$

• Final Answer: The average atomic mass of Boron is $10.8\text{ u}$.

6. Exam Revision & FAQs

Q1: Why did Rutherford select a gold foil for his $\alpha$-particle scattering experiment?

Answer: Rutherford selected gold foil because he needed a layer as thin as possible. Gold is the most malleable metal known; the foil used was only about $1000$ atoms thick, ensuring that $\alpha$-particles interacted with a minimal depth of atomic structures.

Q2: Why are noble gases chemically unreactive (inert)?

Answer: Noble gases (such as Helium, Neon, Argon) have completely filled outermost shells. Helium has a stable duplet ($2$ electrons in $K$-shell), while Neon ($2,8$) and Argon ($2,8,8$) have stable octets in their valence shells. Because their valence shells are full, their combining capacity (valency) is zero, making them chemically inert.

Q3: An element has $Z = 8$. Explain why its valency is $2$ and not $8$.

Answer: For $Z = 8$, the electronic configuration is $2, 6$. The number of valence electrons is $6$. Valency is the number of electrons gained, lost, or shared to achieve a stable octet. Since it is easier for an atom with $6$ valence electrons to gain $2$ electrons than to lose all $6$, its valency is calculated as:

$\text{Valency} = 8 - 6 = 2$

Q4: State two main differences between Isobars and Isotopes with one example each.

Answer:

1. Isotopes are atoms of the same element having the same atomic number ($Z$) but different mass numbers ($A$).
   • Example: ${}_{1}^{1}\text{H}$ and ${}_{1}^{2}\text{H}$.
2. Isobars are atoms of different elements having different atomic numbers ($Z$) but the same mass number ($A$).
   • Example: ${}_{18}^{40}\text{Ar}$ and ${}_{20}^{40}\text{Ca}$.