Structure of the Atom — NCERT Solutions | Class 9 Science | Academia Aeternum
Ch 4  ·  Q–
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Class 9 Science Exercise NCERT Solutions Olympiad Board Exam
Chapter 4

Structure of the Atom

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

19 Questions
40–60 min Ideal time
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Q1
NUMERIC3 marks
Compare the properties of electrons, protons and neutrons.
📘 Concept & Theory Concept Builder ›

Every atom is made up of three fundamental subatomic particles—electrons, protons, and neutrons. Understanding their location, charge, and mass is essential for explaining atomic structure, chemical bonding, atomic number, mass number, isotopes, and the behavior of matter.

The discovery of these particles revolutionized our understanding of matter:

  • Electron was discovered by J.J. Thomson.
  • Proton was identified through canal rays by E. Goldstein.
  • Neutron was discovered by James Chadwick.

The nucleus contains protons and neutrons, while electrons revolve around the nucleus in definite energy levels or shells.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Identify the three fundamental particles of an atom.

  2. Compare their location inside the atom.

  3. Compare their electric charge.

  4. Compare their masses.

  5. State their importance in atomic structure.

  6. Summarize the comparison in tabular form.

📊 Graph / Figure Graph / Figure ›
STRUCTURE OF AN ATOM LITHIUM-7 ISOTOPE (³Li) ELEMENT INFO Name: Lithium-7 Protons (p⁺): 3 Neutrons (n⁰): 4 Electrons (e⁻): 3 + + + NUCLEUS Protons (p⁺) & Neutrons (n⁰) Dense positive center containing 99.9% mass ELECTRON (e⁻) Negatively charged particle Orbiting the nucleus in probability shells ORBITAL SHELL Energy levels / Electron cloud Paths defined by quantum mechanics + Proton (p⁺) Neutron (n⁰) Electron (e⁻)
Atomic Structure Visualization
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  4 steps
  1. Electron

  2. Electrons are negatively charged subatomic particles present outside the nucleus. They occupy different shells or energy levels around the nucleus. The mass of an electron is extremely small, approximately \[ \frac{1}{1836} \] times the mass of a proton.

    Electrons are responsible for chemical bonding and determine many physical and chemical properties of an element.

  3. Proton

  4. Protons are positively charged particles located inside the nucleus. The number of protons present in an atom determines the atomic number and identity of the element.

    A proton has a relative mass of approximately \[ 1\;amu \] and contributes significantly to the mass of an atom.

  5. Neutron

  6. Neutrons are electrically neutral particles present inside the nucleus. Their mass is nearly equal to that of a proton and is slightly greater than the proton's mass.

    Neutrons help stabilize the nucleus by reducing the repulsive force between positively charged protons.

  7. Comparison of the Three Particles
  8. Property Electron Proton Neutron
    Symbol e− p+ n0
    Location Outside the nucleus (shells) Inside the nucleus Inside the nucleus
    Charge −1 +1 0
    Actual Charge −1.602 × 10−19 C +1.602 × 10−19 C 0 C
    Mass (amu) 0.00055 amu 1.0073 amu 1.0087 amu
    Relative Mass \(\frac{1}{1836}\) 1 1
    Role in Atom Chemical bonding and reactivity Determines identity of element Provides nuclear stability
💡 Answer Final Answer ›

Electrons, protons, and neutrons differ in their location, charge, and mass. Electrons are negatively charged particles present outside the nucleus and have negligible mass. Protons are positively charged particles located in the nucleus and determine the identity of an element. Neutrons are neutral particles present in the nucleus and help stabilize it. Together, these three particles constitute the basic structure of an atom.

🎯 Exam Significance Exam Significance ›
  • This is one of the most frequently asked foundational questions in CBSE examinations.
  • Knowledge of subatomic particles is essential for understanding atomic number and mass number.
  • Forms the basis for concepts such as isotopes, isobars, ions, and electronic configuration.
  • Frequently appears in NTSE, Olympiads, Foundation courses, and other competitive entrance examinations.
  • Direct comparison-based MCQs are commonly framed from these properties.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  7 points

  1. Atoms are composed of electrons, protons, and neutrons.

  2. Electrons carry negative charge and occupy shells around the nucleus.

  3. Protons carry positive charge and determine the atomic number of an element.

  4. Neutrons are neutral particles that contribute to nuclear stability.

  5. Most of the mass of an atom is concentrated in the nucleus.

  6. The mass of an electron is approximately \(\frac{1}{1836}\) of a proton.

  7. Protons and neutrons together are called nucleons.
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NUMERIC3 marks
What are the limitations of J.J. Thomson’s model of the atom?
📘 Concept & Theory Concept Builder ›

J.J. Thomson proposed the "Plum Pudding Model" (also called the Watermelon Model) of the atom after the discovery of electrons. According to this model, an atom was imagined as a positively charged sphere with negatively charged electrons embedded throughout it, just as plums are embedded in a pudding.

Thomson's model was the first attempt to explain how electrons exist inside an atom. However, later experiments, especially Rutherford's alpha-particle scattering experiment, revealed several shortcomings in this model.

Scientific models remain valid only until new experimental evidence disproves them. Thomson's model was an important milestone but could not explain the actual structure of the atom.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Recall the main assumptions of Thomson's atomic model.

  2. Compare these assumptions with later experimental observations.

  3. Identify the features that the model failed to explain.

  4. List the major limitations systematically.

  5. Conclude why the model was eventually replaced.

📊 Graph / Figure Graph / Figure ›
J.J. THOMSON'S PLUM PUDDING MODEL HISTORICAL ATOMIC MODEL (1904) & ITS LIMITATION + + + + + + + - - - - - POSITIVE CHARGE CLOUD Electrons embedded like plums in a pudding RUTHERFORD'S DISCOVERY Gold Foil Scattering Results (1909) GOLD FOIL α α α Most α- particles passed straight Some were deflected Few bounced back Thomson's Model Could Not Explain Large-Angle Deflections
Thomson's Plum Pudding Model and Its Limitation
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  3 steps
  1. Understanding Thomson's Model
  2. According to Thomson, the atom consisted of a uniformly distributed positive charge throughout a spherical volume, with electrons embedded in it. The total positive charge balanced the total negative charge of electrons, making the atom electrically neutral.
  3. Examining the Experimental Evidence
  4. As scientific investigations progressed, several experimental observations could not be explained by Thomson's model. These observations highlighted weaknesses in the model.
  5. Identifying the Limitations
    • 1. Could not explain atomic stability
      The model did not clearly explain how negatively charged electrons remained embedded within the positively charged sphere without collapsing or escaping. Thus, the stability of the atom remained unexplained.
    • 2. Failed to predict the existence of a nucleus
      Thomson assumed that positive charge was spread uniformly throughout the atom. Later discoveries showed that almost the entire positive charge and mass are concentrated in a tiny central nucleus.
    • 3. Could not explain Rutherford's alpha-particle scattering results
      Rutherford observed that while most alpha particles passed straight through the gold foil, some were deflected through large angles and a few even bounced back. Thomson's model could not account for these observations because it lacked a dense central nucleus.
    • 4. Could not explain the arrangement of electrons
      The model gave no information about how electrons are arranged inside the atom or how they move around it.
    • 5. Failed to explain atomic spectra
      Atoms emit characteristic spectral lines when excited. Thomson's model could not explain the origin of these discrete line spectra because it did not describe electron energy levels.
    • 6. Lack of strong experimental support
      Subsequent experiments contradicted the assumptions of the plum pudding model, leading to its rejection and replacement by more accurate atomic models.
💡 Answer Final Answer ›

J.J. Thomson's model of the atom had several limitations. It could not explain the stability of atoms, failed to predict the existence of a central nucleus, could not account for Rutherford's alpha-particle scattering experiment, gave no satisfactory explanation for the arrangement of electrons, failed to explain atomic spectra, and lacked experimental support from later discoveries. Therefore, Thomson's model was eventually replaced by Rutherford's nuclear model of the atom.

🎯 Exam Significance Exam Significance ›
  • Frequently asked as a short-answer and long-answer question in CBSE examinations.
  • Important for understanding the evolution of atomic models.
  • Forms the conceptual bridge between Thomson's and Rutherford's models.
  • Helpful for objective questions in NTSE, Olympiads, Foundation courses, and other competitive examinations.
  • Often appears as a comparison-based question with Rutherford's model.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  8 points

  1. Thomson proposed the "Plum Pudding Model" of the atom.

  2. The model assumed positive charge was uniformly distributed throughout the atom.

  3. Electrons were considered embedded in the positive sphere.

  4. The model could not explain atomic stability.

  5. It failed to predict the existence of the nucleus.

  6. Rutherford's experiment disproved Thomson's assumptions.

  7. The model could not explain atomic spectra and electron arrangement.

  8. Scientific models evolve when new experimental evidence becomes available.
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NUMERIC3 marks
What are the limitations of Rutherford’s model of the atom?
📘 Concept & Theory Concept Builder ›

Rutherford's nuclear model of the atom was proposed in 1911 based on the results of the alpha-particle scattering experiment. The model successfully explained that:

  • Most of the atom is empty space.
  • A tiny, dense, positively charged nucleus exists at the center.
  • Electrons revolve around the nucleus.

Although Rutherford's model solved many problems associated with Thomson's model, it could not explain several important observations related to atomic stability and the behavior of electrons. These shortcomings led to the development of Bohr's atomic model.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Recall the main features of Rutherford's nuclear model.

  2. Apply the laws of classical physics to the revolving electrons.

  3. Compare the model's predictions with experimental observations.

  4. Identify the phenomena that Rutherford's model failed to explain.

  5. List the limitations systematically.

📊 Graph / Figure Graph / Figure ›
RUTHERFORD'S MODEL & LIMITATION THE CLASSICAL ELECTROMAGNETIC COLLAPSE + - e⁻ Energy Radiated Away SPIRAL COLLAPSE Continuous energy loss causes orbit decay into the nucleus UNANSWERED QUESTIONS Why are atoms stable? Accelerating charges should collapse. Why are spectra discrete? Classical physics predicts continuous emission. How are electrons arranged? Orbits and distribution remained unexplained. Rutherford Explained the Nucleus but Not Atomic Stability
Limitation of Rutherford's Model
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. Rutherford's Basic Idea

  2. According to Rutherford, electrons revolve around a small, dense, positively charged nucleus, much like planets revolve around the Sun.

    While this model explained the scattering of alpha particles successfully, it created new theoretical difficulties.

  3. Problem of Atomic Stability

  4. According to classical electromagnetic theory, any charged particle moving in a circular path continuously undergoes acceleration.

    Since electrons are negatively charged particles revolving around the nucleus, they should continuously radiate energy.

    As energy is lost, the electron should move closer and closer to the nucleus in a spiral path and finally collapse into it.

    This prediction implies that atoms should be unstable.

    However, atoms are actually stable. Therefore, Rutherford's model failed to explain atomic stability.

  5. Failure to Explain Atomic Spectra

  6. Experiments showed that elements emit or absorb light only at specific wavelengths, producing characteristic line spectra.

    Rutherford's model could not explain why only certain frequencies of light are emitted instead of a continuous spectrum.

    Thus, the model failed to explain the origin of atomic spectra.

  7. No Explanation for Electron Arrangement

  8. The model did not describe how electrons are distributed around the nucleus. It gave no information about electron shells, energy levels, or permitted orbits.

    Therefore, the arrangement of electrons inside the atom remained unexplained.

  9. Incomplete Description of the Nucleus

  10. Rutherford's model considered the nucleus to be positively charged but could not explain its internal composition.

    The existence and role of neutrons were not known at that time. Hence, the structure of the nucleus remained incomplete.

⚠️ Limitations of Rutherford's Model ›
  1. Could not explain atomic stability: Revolving electrons should continuously lose energy and eventually fall into the nucleus.
  2. Failed to explain line spectra: The model could not account for the discrete spectral lines emitted by atoms.
  3. No explanation of electron arrangement: The distribution of electrons around the nucleus was not described.
  4. No concept of energy levels: The model did not explain why electrons occupy specific regions around the nucleus.
  5. Incomplete description of the nucleus: It could not explain the existence and role of neutrons discovered later.
💡 Answer Final Answer ›

Rutherford's model of the atom had several limitations. It could not explain the stability of atoms because revolving electrons should continuously lose energy and fall into the nucleus. It failed to explain the discrete line spectra of elements, did not describe the arrangement of electrons or energy levels, and provided no explanation for the internal structure of the nucleus. These shortcomings led to the development of Bohr's atomic model.

🎯 Exam Significance Exam Significance ›
  • This is one of the most important conceptual questions from the chapter "Structure of the Atom".
  • Frequently asked in CBSE board examinations as a short-answer or long-answer question.
  • Helps students understand why Bohr's model was proposed.
  • Important for competitive examinations such as NTSE, Olympiads, Foundation courses, and scholarship tests.
  • Often appears as a comparison question between Rutherford's and Bohr's atomic models.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  7 points

  1. Rutherford discovered the nucleus through the alpha-particle scattering experiment.

  2. The model established that most of the atom is empty space.

  3. It could not explain why atoms are stable.

  4. Revolving electrons should radiate energy according to classical physics.

  5. The model failed to explain atomic line spectra.

  6. No concept of fixed energy levels or electron shells was provided.

  7. The model was later improved by Niels Bohr.
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NUMERIC3 marks
Describe Bohr’s model of the atom.
📘 Concept & Theory Concept Builder ›

Rutherford's model successfully established the existence of a small, dense nucleus, but it could not explain why electrons revolving around the nucleus do not lose energy and fall into it. It also failed to explain the line spectra of atoms.

To overcome these difficulties, the Danish physicist Niels Bohr proposed a new atomic model in 1913. Bohr combined Rutherford's nuclear model with ideas from quantum theory and suggested that electrons move around the nucleus only in certain permitted circular paths called shells or energy levels.

Bohr's model successfully explained the stability of atoms and the origin of atomic spectra, making it one of the most important developments in atomic theory.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Recall the shortcomings of Rutherford's model.

  2. Understand Bohr's assumptions about electron motion.

  3. Learn the concept of fixed energy levels or shells.

  4. Understand how electrons absorb and emit energy.

  5. Summarize the main postulates of Bohr's atomic model.

📊 Graph / Figure Graph / Figure ›
BOHR'S MODEL OF THE ATOM K Shell (n = 1) L Shell (n = 2) M Shell (n = 3) N Shell (n = 4) + + + + + Incoming Photon (hν) e⁻ Energy Absorption Electron jumps to a higher orbit Emitted Photon (hν) e⁻ Energy Emission Electron drops to a lower orbit e⁻ e⁻ e⁻ e⁻ e⁻ e⁻ e⁻ e⁻ e⁻ e⁻ LEGEND Proton (+) Neutron (0) Electron (e⁻) Photon (hν) ENERGY EQUATION ΔE = E₂ - E₁ = hν h = Planck's Constant ν = Photon Frequency Electrons occupy quantized, fixed energy levels (shells) around the nucleus.
Bohr's Atomic Model
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. Presence of a Central Nucleus
  2. According to Bohr's model, an atom consists of a small, dense, positively charged nucleus located at its center. Almost the entire mass of the atom is concentrated in this nucleus.
  3. Electrons Revolve in Fixed Orbits

  4. Electrons revolve around the nucleus only in certain fixed circular paths called orbits, shells, or energy levels.

    These shells are represented as:

    K, L, M, N, ...

    or by principal quantum numbers:

    \[ n = 1,\;2,\;3,\;4,\ldots \]

    where:

    • K-shell corresponds to \(n = 1\)
    • L-shell corresponds to \(n = 2\)
    • M-shell corresponds to \(n = 3\)
    • N-shell corresponds to \(n = 4\)
  5. Electrons Do Not Lose Energy in Stable Orbits
  6. While moving in these permitted orbits, electrons do not radiate energy. Therefore, they do not spiral into the nucleus
    This successfully explains the stability of atoms.
  7. Each Orbit Has Fixed Energ

  8. Every orbit possesses a definite amount of energy.

    The energy of the shells increases as the distance from the nucleus increases:

    \[ E_K < E_L < E_M < E_N \]

    Thus, the K-shell has the lowest energy and is closest to the nucleus.

  9. Absorption and Emission of Energy

  10. An electron can move from one energy level to another only by absorbing or releasing a fixed amount of energy.

    • When an electron absorbs energy, it jumps from a lower energy level to a higher energy level.
    • When an electron loses energy, it falls from a higher energy level to a lower energy level.

    The energy change is given by:

    \[ \Delta E = E_2 - E_1 \]

    This explains why atoms emit or absorb only specific wavelengths of light, producing characteristic line spectra.

⚖️ Main Postulates of Bohr's Model ›
  1. An atom consists of a small positively charged nucleus at the center.
  2. Electrons revolve around the nucleus in fixed circular orbits.
  3. Each orbit has a definite and fixed energy.
  4. Electrons do not radiate energy while moving in permitted orbits.
  5. Energy is emitted or absorbed only when electrons jump from one orbit to another.
  6. The shells are designated as K, L, M, N, etc.
💡 Answer Final Answer ›

According to Bohr's model, an atom consists of a small positively charged nucleus surrounded by electrons moving in fixed circular orbits called shells or energy levels. Each shell possesses a definite amount of energy. Electrons do not lose energy while revolving in these permitted orbits. They can move from one energy level to another only by absorbing or emitting a fixed amount of energy. This model successfully explains the stability of atoms and the origin of atomic line spectra.

🎯 Exam Significance Exam Significance ›
  • One of the most important theoretical questions from the chapter "Structure of the Atom".
  • Frequently asked in CBSE board examinations.
  • Forms the foundation for electronic configuration and valency.
  • Important for understanding atomic spectra and energy levels.
  • Frequently appears in NTSE, Olympiads, Foundation courses, and competitive entrance examinations.
  • Essential for higher studies in Chemistry and Physics.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  8 points

  1. Bohr improved Rutherford's atomic model.

  2. Electrons revolve in fixed circular orbits called shells.

  3. Shells are designated as K, L, M, N, etc.

  4. Each shell possesses a definite amount of energy.

  5. Electrons do not radiate energy in stable orbits.

  6. Atoms remain stable because electrons do not continuously lose energy.

  7. Energy is emitted or absorbed during transitions between energy levels.

  8. Bohr's model successfully explained atomic spectra.
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NUMERIC3 marks
Compare all the proposed models of an atom given in this chapter.
📘 Concept & Theory Concept Builder ›

Our understanding of the atom evolved gradually through the contributions of several scientists. Each atomic model was proposed to explain the experimental observations available at that time.

As new experiments revealed limitations in existing theories, improved atomic models were developed. The progression from Dalton's model to Bohr's model represents one of the most important scientific journeys in Chemistry.

The sequence of development is:

Dalton → Thomson → Rutherford → Bohr

Comparing these models helps us understand how scientific theories evolve and how each model contributed to modern atomic structure.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Identify the scientists who proposed the atomic models.

  2. Recall the main features of each model.

  3. Compare the structure of the atom according to each model.

  4. Examine how each model explained electrons and the nucleus.

  5. Analyze the strengths and limitations of each model.

  6. Summarize the comparison in tabular form.

📊 Graph / Figure Graph / Figure ›
EVOLUTION OF ATOMIC MODELS Dalton Billiard Ball Model (1803) + - - - - Thomson Plum Pudding Model (1904) + - - - Rutherford Nuclear Model (1911) + - - - Bohr Planetary Model (1913) + Schrödinger Electron Cloud Model (1926) Dalton (1803) ➔ Thomson (1904) ➔ Rutherford (1911) ➔ Bohr (1913) ➔ Schrödinger (1926)
Evolution of Atomic Models
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. Dalton's Atomic Model
  2. Dalton considered atoms to be tiny, indivisible, solid spheres. According to him, atoms could neither be created nor destroyed during a chemical reaction. His model introduced the concept of atoms but did not describe any internal structure.
  3. Thomson's Atomic Model
  4. After the discovery of electrons, Thomson proposed the Plum Pudding Model. He suggested that electrons are embedded in a sphere of positive charge. This was the first model to include subatomic particles.
  5. Rutherford's Nuclear Model
  6. Rutherford's alpha-particle scattering experiment revealed the existence of a small, dense nucleus. He proposed that most of the atom is empty space and electrons revolve around the nucleus.
  7. Bohr's Atomic Model
  8. Bohr improved Rutherford's model by introducing fixed energy levels or shells. Electrons revolve around the nucleus only in certain permitted orbits and do not lose energy while doing so.
  9. Property Dalton's Model Thomson's Model Rutherford's Model Bohr's Model
    Scientist John Dalton J.J. Thomson Ernest Rutherford Niels Bohr
    Year 1808 1904 1911 1913
    Internal Structure Solid, indivisible sphere Electrons embedded in positive sphere Dense nucleus with revolving electrons Electrons occupy fixed energy shells around nucleus
    Subatomic Particles Included None Electrons Protons (nucleus) and electrons Protons (nucleus) and electrons in fixed orbits
    Nucleus Present No No Yes Yes
    Arrangement of Electrons Not specified Embedded randomly in positive charge Orbit around nucleus Occupy fixed shells (K, L, M, N)
    Atomic Stability Explained No No No Yes
    Atomic Spectra Explained No No No Yes (especially hydrogen spectrum)
    Main Strength Introduced atomic theory Included electrons and charge neutrality Discovered the nucleus Explained stability and line spectra
    Main Limitation No internal structure No nucleus; failed to explain stability Failed to explain stability and spectra Applicable mainly to hydrogen-like atoms
💡 Answer Final Answer ›

Dalton's model considered atoms as indivisible solid spheres. Thomson proposed that electrons were embedded in a positively charged sphere. Rutherford introduced the concept of a dense central nucleus with electrons revolving around it. Bohr further refined Rutherford's model by proposing fixed energy levels for electrons. Among these models, Bohr's model most successfully explained atomic stability and atomic spectra, making it the most advanced model discussed in this chapter.

📖 Evolution of Atomic Models at a Glance ›
Model Major Contribution Why It Was Replaced
Dalton Introduced atomic theory Could not explain subatomic particles
Thomson Introduced electrons into atomic structure Could not explain nucleus and scattering results
Rutherford Discovered nucleus Could not explain atomic stability and spectra
Bohr Introduced quantized energy levels Could not fully explain multi-electron atoms
🎯 Exam Significance Exam Significance ›
  • One of the most important long-answer questions from the chapter.
  • Frequently asked in CBSE board examinations as a comparison-based question.
  • Helpful for understanding the historical development of atomic theory.
  • Important for NTSE, Olympiads, Foundation courses, and scholarship examinations.
  • Many MCQs are framed from the strengths and limitations of different atomic models.
  • Provides a complete conceptual overview of the chapter.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  7 points

  1. Scientific knowledge develops through continuous experimentation and improvement.

  2. Dalton introduced the concept of atoms.

  3. Thomson discovered electrons and proposed the Plum Pudding Model.

  4. Rutherford discovered the nucleus through alpha-particle scattering.

  5. Bohr introduced fixed energy levels for electrons.

  6. Each model corrected the shortcomings of the previous one.

  7. Modern atomic theory evolved from the combined contributions of all these scientists.
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Q6
NUMERIC3 marks
Summarise the rules for writing of distribution of electrons in various shells for the first eighteen elements.
📘 Concept & Theory Concept Builder ›

According to Bohr's model, electrons revolve around the nucleus in definite shells or energy levels. These shells are designated as K, L, M, N, and so on.

The arrangement of electrons in different shells of an atom is known as its electronic distribution or electronic configuration.

The distribution of electrons follows specific rules so that the atom remains stable and attains the lowest possible energy state.

For Class 9 NCERT, we study the distribution of electrons only for the first eighteen elements of the periodic table.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Understand the concept of electron shells.

  2. Learn the maximum electron capacity of each shell.

  3. Apply the formula for shell capacity.

  4. Study the order in which shells are filled.

  5. Summarize the rules used for the first eighteen elements.

📊 Graph / Figure Graph / Figure ›
BOHR-BURY ELECTRON DISTRIBUTION MODEL Visualization of shell occupancy and distribution rules for the first 18 elements SHELL OCCUPANCY K Shell (n=1) 2 / 2 e⁻ L Shell (n=2) 8 / 8 e⁻ M Shell (n=3) 8 / 18 e⁻ Valence Shell (Stable Octet) ARGON ATOM (₁₈Ar) Electron Config: 2, 8, 8 BOHR-BURY RULES 2n² Shell Capacity Max electrons = 2n² per shell K=2, L=8, M=18, N=32... In Stepwise Filling Inner shells fill completely before outer shells occupy. 8 Octet Limit Outermost shell cannot exceed 8 valence electrons. VALENCE STABILITY Argon (2,8,8) has a complete octet, making it inert. n=1 (K) n=2 (L) n=3 (M) Ar 18p⁺ 22n⁰ BOHR-BURY MODEL INFOGRAPHIC • ATOMIC NUMBER 18 (ARGON)
BOHR-BURY ELECTRON DISTRIBUTION MODEL
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  6 steps
  1. Electrons Occupy Shells Around the Nucleus

  2. Electrons are arranged in different shells around the nucleus. These shells are named:

    K, L, M, N, ...

    The shells correspond to principal quantum numbers:

    \[ n = 1,\;2,\;3,\;4,\ldots \]

    where:

    • K-shell → \(n = 1\)
    • L-shell → \(n = 2\)
    • M-shell → \(n = 3\)
    • N-shell → \(n = 4\)
  3. Maximum Number of Electrons in a Shell

  4. The maximum number of electrons that can be accommodated in a shell is given by:

    \[ 2n^2 \]

    where \(n\) is the shell number.

    Applying the formula:

    Shell Value of \(n\) Maximum Electrons
    K 1 \[ 2(1)^2 = 2 \]
    L 2 \[ 2(2)^2 = 8 \]
    M 3 \[ 2(3)^2 = 18 \]
    N 4 \[ 2(4)^2 = 32 \]
  5. Shells are Filled from Inner to Outer Region

  6. Electrons always fill the innermost shell first because it has the lowest energy. Only after a shell is filled does the next shell begin to receive electrons.

    Therefore, the filling order is:

    K → L → M → N → ...

  7. Limitation for the Outermost Shell

  8. Another important rule is that the outermost shell of an atom cannot contain more than 8 electrons.

    Therefore, although the M-shell can theoretically accommodate 18 electrons, for the first eighteen elements it contains at most 8 electrons.

  9. Distribution for the First Eighteen Elements

  10. For the first eighteen elements:

    • The K-shell can have a maximum of 2 electrons.
    • The L-shell can have a maximum of 8 electrons.
    • The remaining electrons are placed in the M-shell up to a maximum of 8 electrons.
    • Electrons always fill lower-energy shells first.
  11. Rules for Distribution of Electrons
    1. The maximum number of electrons present in a shell is given by: \[ 2n^2 \]
    2. The outermost shell cannot contain more than 8 electrons.
    3. Electrons are filled progressively from inner shells to outer shells.
    4. A shell is occupied only after the inner shell has been filled according to its capacity.
    5. For the first eighteen elements, electron distribution generally follows: K (2), L (8), M (up to 8).
💡 Answer Final Answer ›

The distribution of electrons in various shells follows certain rules. The maximum number of electrons that can be accommodated in a shell is given by \[ 2n^2 \] where \(n\) is the shell number. The outermost shell cannot contain more than 8 electrons. Electrons fill shells in order of increasing energy, starting from the innermost shell. For the first eighteen elements, electrons are distributed mainly in the K, L, and M shells, with capacities of 2, 8, and up to 8 electrons, respectively.

🎯 Exam Significance Exam Significance ›
  • This is one of the most important theory questions from the chapter.
  • The electron distribution rules are directly used in numerical and conceptual questions.
  • Forms the foundation for understanding valency and chemical bonding.
  • Frequently asked in CBSE board examinations and school tests.
  • Very important for NTSE, Olympiads, Foundation courses, and other competitive examinations.
  • Knowledge of these rules is essential for writing electronic configurations correctly.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  8 points

  1. Electrons occupy shells around the nucleus.

  2. Shells are represented as K, L, M, N, etc.

  3. The maximum electron capacity of a shell is given by \(\mathbf{2n^2}\).

  4. K-shell can hold 2 electrons.

  5. L-shell can hold 8 electrons.

  6. For the first eighteen elements, the M-shell contains a maximum of 8 electrons.

  7. The outermost shell cannot have more than 8 electrons.

  8. Electrons always fill lower-energy shells first.
← Q5
6 / 19  ·  32%
Q7 →
Q7
NUMERIC3 marks
Define valency by taking examples of silicon and oxygen
📘 Concept & Theory Concept Builder ›

Atoms tend to attain a stable electronic configuration similar to that of noble gases. Noble gases are chemically stable because their outermost shell is completely filled.

During chemical reactions, atoms gain, lose, or share electrons to complete their outermost shell. The tendency of an atom to combine with other atoms depends on the number of electrons required to achieve this stable configuration.

This combining capacity of an atom is called its valency.

For elements having less than four electrons in the outermost shell, valency is usually equal to the number of electrons lost or shared. For elements having more than four electrons in the outermost shell, valency is equal to:

\[ 8 - \text{Number of valence electrons} \]

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Define valency.

  2. Determine the electronic configuration of silicon.

  3. Find the number of electrons needed for silicon to attain stability.

  4. Determine the electronic configuration of oxygen.

  5. Find the number of electrons needed for oxygen to attain stability.

  6. State the valency of both elements.

📊 Graph / Figure Graph / Figure ›
VALENCY OF SILICON & OXYGEN A comparison of atomic orbits, electron configurations, and outer-shell valency Silicon (Si) 14 Si Oxygen (O) 8 O LEGEND Core Electron (Filled) Valence Electron Electron Vacancy Silicon (Si) Configuration: 2, 8, 4 Valency: 4 (Tetravalent) Shares 4 valence e⁻ in M shell Oxygen (O) Configuration: 2, 6 Valency: 2 (Divalent) Needs 2 e⁻ to fill L shell
Understanding Valency of Silicon and Oxygen
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. Definition of Valency

  2. Valency is the combining capacity of an atom. It is the number of electrons an atom loses, gains, or shares to attain a stable electronic configuration.

    In simple words, valency indicates how many chemical bonds an atom can form.

  3. Valency of Silicon

  4. Silicon has atomic number 14.

    Therefore, its electronic configuration is:

    \[ 2,\;8,\;4 \]

    Thus, silicon has 4 electrons in its outermost shell.

    To attain a stable octet, silicon requires:

    \[ 8 - 4 = 4 \]

    electrons.

    Therefore, silicon can share or effectively gain four electrons during bonding.

  5. Valency of Silicon = 4
  6. Valency of Oxygen

  7. Oxygen has atomic number 8.

    Therefore, its electronic configuration is:

    \[ 2,\;6 \]

    Thus, oxygen has 6 electrons in its outermost shell.

    To complete its octet, oxygen requires:

    \[ 8 - 6 = 2 \]

    electrons.

    Therefore, oxygen gains or shares two electrons while forming chemical bonds.

    Valency of Oxygen = 2

  8. Summary Table
  9. Element Atomic Number Electronic Configuration Valence Electrons Electrons Required for Octet Valency
    Silicon (Si) 14 2, 8, 4 4 4 4
    Oxygen (O) 8 2, 6 6 2 2
💡 Answer Final Answer ›

Valency is the combining capacity of an atom and is equal to the number of electrons lost, gained, or shared by the atom to attain a stable electronic configuration. Silicon has electronic configuration \[ 2,\;8,\;4 \] and requires four more electrons to complete its octet. Therefore, its valency is 4. Oxygen has electronic configuration \[ 2,\;6 \] and requires two more electrons to complete its octet. Therefore, its valency is 2.

🎯 Exam Significance Exam Significance ›
  • Valency is one of the most fundamental concepts in Chemistry.
  • Frequently asked in CBSE board examinations as a definition-based question.
  • Forms the basis for writing chemical formulae correctly.
  • Important for understanding chemical bonding and compound formation.
  • Frequently appears in NTSE, Olympiads, Foundation courses, and scholarship examinations.
  • Many MCQs test the relationship between electronic configuration and valency.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  7 points

  1. Valency is the combining capacity of an atom.

  2. Atoms attain stability by completing their outermost shell.

  3. Valency depends on the number of valence electrons.

  4. Silicon has electronic configuration 2, 8, 4 and valency 4.

  5. Oxygen has electronic configuration 2, 6 and valency 2.

  6. Valency determines how atoms combine to form compounds.

  7. Elements with complete outer shells generally have valency 0.
← Q6
7 / 19  ·  37%
Q8 →
Q8
NUMERIC3 marks
Explain with examples:
(i) Atomic number,
(ii) Mass number,
(iii) Isotopes and
(iv) Isobars.
Give any two uses of isotopes.
📘 Concept & Theory Concept Builder ›

The identity and properties of an atom depend on the number of subatomic particles present in it. Scientists use terms such as atomic number, mass number, isotopes, and isobars to describe and classify atoms.

These concepts help us understand the structure of atoms, the periodic table, nuclear chemistry, radioactive substances, and many practical applications in medicine, agriculture, industry, and archaeology.

Before studying isotopes and isobars, it is important to remember:

\[ \text{Mass Number (A)} = \text{Number of Protons (Z)} + \text{Number of Neutrons (N)} \]

\[ A = Z + N \]

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Define atomic number and explain it with examples.

  2. Define mass number and calculate it using protons and neutrons.

  3. Explain isotopes with suitable examples.

  4. Explain isobars with suitable examples.

  5. List important applications of isotopes.

📊 Graph / Figure Graph / Figure ›
CHEMISTRY REFERENCE Atomic Structures & Nuclear Relations Understanding Atomic Number, Mass Number, Isotopes, and Isobars ATOMIC NUMBER (Z) The number of protons in the nucleus of an atom. Z = PROTONS 6 Protons 6 Neutrons NUCLEAR NOTATION C 12 6 Mass Number (A) Atomic Number (Z) Carbon-12 Example MASS NUMBER (A) The total number of protons and neutrons in a nucleus. A = Z + N 12 Nucleons (Protons + Neutrons) NUCLEON SUMMATION A = Z + N Protons (Z) : 6 Neutrons (N) : 6 Mass Number (A) = 12 ISOTOPES Same element (same Z) with different mass numbers (different N). SAME Z, DIFFERENT A Carbon-12 Stable Abundance Carbon-14 Radioactive Trace CARBON-12 PROPERTIES CARBON-14 6 Protons Same Z (6) 6 Protons 6 Neutrons Different 8 Neutrons A = 12 Different A = 14 ISOBARS Different elements (different Z) with the same mass number (same A). SAME A, DIFFERENT Z Argon-40 Noble Gas (Z = 18) Calcium-40 Reactive Metal (Z = 20) ARGON-40 PROPERTIES CALCIUM-40 18 Protons Different 20 Protons 22 Neutrons Different 20 Neutrons A = 40 Same A (40) A = 40 Educational Chemistry Blueprint • High-Fidelity Vector Asset • Representational Scale Only
Atomic Number, Mass Number, Isotopes and Isobars
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. (i) Atomic Number

  2. The atomic number of an element is the number of protons present in the nucleus of an atom.

    It is represented by the symbol:

    \[ Z \]

    The atomic number determines the identity of an element because no two elements can have the same number of protons.

    Examples:

    • Hydrogen contains 1 proton, so its atomic number is 1.
    • Carbon contains 6 protons, so its atomic number is 6.
    • Oxygen contains 8 protons, so its atomic number is 8.
  3. (ii) Mass Number

  4. The mass number of an atom is the total number of protons and neutrons present in its nucleus.

    It is represented by:

    \[ A \]

    Mathematical relation:

    \[ \text{Mass Number} = \text{Number of Protons} + \text{Number of Neutrons} \]

    \[ A = Z + N \]

    Example 1: Carbon-12

    Number of protons = 6
    Number of neutrons = 6

    Therefore,

    \[ A = 6 + 6 = 12 \]

    Hence, the mass number of carbon-12 is 12.

    Example 2: Sodium-23

    Number of protons = 11
    Number of neutrons = 12

    \[ A = 11 + 12 = 23 \]

    Hence, the mass number of sodium is 23.

  5. (iii) Isotopes

  6. Isotopes are atoms of the same element having the same atomic number but different mass numbers.

    Since they belong to the same element, they have:

    • Same number of protons.
    • Different number of neutrons.

    Examples:

    Isotope Atomic Number (Z) Mass Number (A) Neutrons
    Carbon-12 6 12 6
    Carbon-14 6 14 8

    Both atoms have atomic number 6 but different mass numbers (12 and 14). Therefore, they are isotopes.

    Other examples include:

    \[ ^1H,\; ^2H,\; ^3H \]

    (Isotopes of hydrogen)

  7. (iv) Isobars

  8. Isobars are atoms of different elements having the same mass number but different atomic numbers.

    Therefore:

    • Mass number is the same.
    • Atomic number is different.
    • They belong to different elements.

    Example:

    Element Atomic Number (Z) Mass Number (A)
    Argon-40 18 40
    Calcium-40 20 40

    Since both have mass number 40 but different atomic numbers, they are isobars.

  9. Uses of Isotopes

  10. Isotopes have numerous applications in science, medicine, industry, and agriculture.

    Any Two Uses:

    1. Cobalt-60 (\(^{60}Co\)) is used in the treatment of cancer through radiation therapy.
    2. Carbon-14 (\(^{14}C\)) is used in radiocarbon dating to determine the age of fossils, archaeological remains, and ancient artifacts.

    Additional examples often asked in examinations:

    • \(^{131}I\) is used in the diagnosis and treatment of thyroid disorders.
    • \(^{235}U\) is used as fuel in nuclear reactors.
💡 Answer Final Answer ›

Atomic Number: The number of protons present in the nucleus of an atom. Example: Carbon has atomic number 6.

Mass Number: The sum of protons and neutrons present in the nucleus. Example: Carbon-12 has mass number 12.

Isotopes: Atoms of the same element having the same atomic number but different mass numbers. Example: Carbon-12 and Carbon-14.

Isobars: Atoms of different elements having the same mass number but different atomic numbers. Example: Argon-40 and Calcium-40.

Two Uses of Isotopes:

  1. Cobalt-60 is used in cancer treatment.
  2. Carbon-14 is used for determining the age of fossils and archaeological remains.
🎯 Exam Significance Exam Significance ›
  • This is one of the most important long-answer questions from the chapter.
  • Frequently asked in CBSE board examinations.
  • Definitions of atomic number, mass number, isotopes, and isobars are common MCQ topics.
  • Applications of isotopes are frequently asked in school and competitive examinations.
  • Forms the foundation for higher studies in atomic and nuclear chemistry.
  • Important for NTSE, Olympiads, Foundation courses, and scholarship examinations.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  7 points

  1. Atomic number equals the number of protons in the nucleus.

  2. Mass number equals the sum of protons and neutrons.

  3. \(A = Z + N\) is the fundamental relation connecting atomic number and mass number.

  4. Isotopes have the same atomic number but different mass numbers.

  5. Isobars have the same mass number but different atomic numbers.

  6. Isotopes have important applications in medicine, archaeology, agriculture, and industry.

  7. Cobalt-60 and Carbon-14 are among the most commonly cited examples in examinations.
← Q7
8 / 19  ·  42%
Q9 →
Q9
NUMERIC3 marks
\(\mathrm{Na^+}\) has completely filled K and L shells. Explain.
📘 Concept & Theory Concept Builder ›

Atoms tend to attain a stable electronic configuration similar to that of noble gases. Noble gases possess completely filled outermost shells and are therefore chemically stable.

Sodium (\(\mathrm{Na}\)) is an alkali metal with atomic number 11. It can attain stability by losing one electron from its outermost shell and forming a positively charged ion called the sodium ion (\(\mathrm{Na^+}\)).

To understand why \(\mathrm{Na^+}\) has completely filled K and L shells, we must examine the electronic configuration of both the sodium atom and the sodium ion.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Determine the atomic number of sodium.

  2. Write the electronic configuration of a neutral sodium atom.

  3. Understand the formation of the sodium ion.

  4. Write the electronic configuration of \(\mathrm{Na^+}\).

  5. Verify whether the K and L shells are completely filled.

  6. Explain the resulting stability of the ion.

📊 Graph / Figure Graph / Figure ›
FORMATION OF SODIUM ION (Na⁺) Loss of a valence electron results in a stable, positively charged sodium cation Sodium Atom Neutral (Na) K L M 11 p⁺ Na e⁻ LOST ELECTRON IONIZATION - 1 e⁻ (Loss) Sodium Ion Cation (Na⁺) K L M 11 p⁺ Na⁺ Sodium Atom (Na) Configuration: 2, 8, 1 Valency: 1 (Monovalent) Unstable valence shell (1 e⁻) Na → Na⁺ + e⁻ CHEMICAL EQUATION OXIDATION Sodium Ion (Na⁺) Configuration: 2, 8 Valency: 0 (Stable Cation) Stable octet shell (L filled)
Formation of \(\mathrm{Na^+}\) Ion
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  7 steps
  1. Atomic Number of Sodium

  2. Sodium (\(\mathrm{Na}\)) has atomic number:

    \[ Z = 11 \]

    Therefore, a neutral sodium atom contains:

    \[ 11 \text{ electrons} \]

  3. Electronic Configuration of Sodium Atom

  4. The 11 electrons are distributed according to shell capacities:

    \[ K = 2,\quad L = 8,\quad M = 1 \]

    Hence, the electronic configuration of sodium is:

    \[ 2,\;8,\;1 \]

    The single electron in the M-shell is the valence electron.

  5. Formation of the Sodium Ion

  6. Sodium loses its outermost electron to attain a stable noble-gas configuration.

    The process can be represented as:

    \[ \mathrm{Na \rightarrow Na^+ + e^-} \]

    Thus, one electron is removed from the M-shell.

  7. Electronic Configuration of \(\mathrm{Na^+}\)

  8. After losing one electron:

    \[ 11 - 1 = 10 \]

    electrons remain in the ion.

    These 10 electrons are distributed as:

    \[ 2,\;8 \]

    Therefore:

    • K-shell contains 2 electrons.
    • L-shell contains 8 electrons.
    • M-shell becomes empty.
  9. Verification of Filled Shells

  10. The maximum capacity of the K-shell is:

    \[ 2 \]

    Since it contains 2 electrons, the K-shell is completely filled.

    The maximum capacity of the L-shell is:

    \[ 8 \]

    Since it contains 8 electrons, the L-shell is also completely filled.

    Thus, \(\mathrm{Na^+}\) possesses completely filled K and L shells.

  11. Stability of the Sodium Ion

  12. The electronic configuration

    \[ 2,\;8 \]

    is identical to that of neon (\(\mathrm{Ne}\)), a noble gas.

    Since noble gases have stable electronic configurations, the sodium ion also becomes highly stable after losing one electron.

  13. Summary Table
  14. Species Number of Electrons Electronic Configuration
    \(\mathrm{Na}\) 11 2, 8, 1
    \(\mathrm{Na^+}\) 10 2, 8
    \(\mathrm{Ne}\) 10 2, 8
💡 Answer Final Answer ›

Sodium has atomic number 11 and electronic configuration \[ 2,\;8,\;1 \] . During the formation of \(\mathrm{Na^+}\), sodium loses one electron from its outermost shell: \[ \mathrm{Na \rightarrow Na^+ + e^-} \] . The sodium ion then contains 10 electrons with electronic configuration \[ 2,\;8 \] . Since the K-shell contains its maximum of 2 electrons and the L-shell contains its maximum of 8 electrons, both shells are completely filled. Therefore, \(\mathrm{Na^+}\) has completely filled K and L shells and attains a stable noble-gas configuration similar to neon.

🎯 Exam Significance Exam Significance ›
  • Frequently asked in CBSE examinations as a conceptual question on ion formation.
  • Tests understanding of electronic configuration and stability.
  • Important for understanding valency and chemical bonding.
  • Forms the basis for explaining the formation of ionic compounds such as sodium chloride.
  • Commonly appears in NTSE, Olympiads, Foundation courses, and scholarship examinations.
  • Often asked as an MCQ involving noble-gas configuration.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  8 points

  1. Sodium has atomic number 11 and electronic configuration 2, 8, 1.

  2. Sodium loses one electron to form \(\mathrm{Na^+}\).

  3. \(\mathrm{Na^+}\) contains 10 electrons.

  4. The electronic configuration of \(\mathrm{Na^+}\) is 2, 8.

  5. K-shell becomes completely filled with 2 electrons.

  6. L-shell becomes completely filled with 8 electrons.

  7. \(\mathrm{Na^+}\) attains the stable electronic configuration of neon.

  8. Filled outer shells are responsible for the stability of ions and noble gases.
← Q8
9 / 19  ·  47%
Q10 →
Q10
NUMERIC2 marks
If bromine atom is available in the form of two isotopes \(^{79}_{35}\mathrm{Br}\) (49.7%) and \(^{81}_{35}\mathrm{Br}\) (50.3%), calculate the average atomic mass of bromine atom.
📘 Concept & Theory Concept Builder ›

Most elements occur in nature as a mixture of isotopes. Since isotopes of an element have different mass numbers, the atomic mass shown in the periodic table is not the mass of a single isotope. Instead, it is the weighted average of the masses of all naturally occurring isotopes.

The contribution of each isotope depends on its percentage abundance in nature. Therefore, isotopes present in greater abundance contribute more to the average atomic mass.

The formula used is:

\[ \text{Average Atomic Mass} = \frac{\sum(\text{Isotopic Mass} \times \text{Percentage Abundance})}{100} \]

Bromine naturally exists as a mixture of two isotopes:

  • \(^{79}_{35}\mathrm{Br}\) with abundance 49.7%
  • \(^{81}_{35}\mathrm{Br}\) with abundance 50.3%
🗺️ Solution Roadmap Step-by-step Plan ›

  1. Write the isotopic masses and their percentage abundances.

  2. Apply the weighted average formula.

  3. Multiply each isotopic mass by its percentage abundance.

  4. Add the contributions of both isotopes.

  5. Divide the result by 100.

  6. Write the final average atomic mass with proper units.

📊 Graph / Figure Graph / Figure ›
Average Atomic Mass of Bromine Weighted Average Based on Isotopic Abundance ⁷⁹Br BROMINE- 79 Mass: 79 u Abundance: 49.7% CONTRIBUTION 39.263 u (79 u × 49. 7%) 49. 7% ABUND. ⁸¹Br BROMINE- 81 Mass: 81 u Abundance: 50.3% CONTRIBUTION 40.743 u (81 u × 50. 3%) 50. 3% ABUND. WEIGHTED AVERAGE MASS CALCULATION Average Atomic Mass = (79 × 0.497) + (81 × 0. 503) = 39.263 u + 40.743 u = 80.006 u
Weighted Average Atomic Mass of Bromine
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  6 steps
  1. Given Data

  2. First isotope:

    \[ ^{79}_{35}\mathrm{Br} \]

    Percentage abundance:

    \[ 49.7\% \]

    Second isotope:

    \[ ^{81}_{35}\mathrm{Br} \]

    Percentage abundance:

    \[ 50.3\% \]

  3. Formula for Average Atomic Mass
  4. \[ \text{Average Atomic Mass} = \frac{ (79 \times 49.7)+(81 \times 50.3) }{100} \]
  5. Calculate Contribution of Each Isotope

  6. Contribution of \(^{79}_{35}\mathrm{Br}\):

    \[ 79 \times 49.7 = 3926.3 \]

  7. Contribution of \(^{81}_{35}\mathrm{Br}\):

    \[ 81 \times 50.3 = 4074.3 \]

  8. Add the Contributions
  9. \[3926.3 + 4074.3 = 8000.6\]
  10. Divide by 100
  11. \[ \begin{aligned} \text{Average Atomic Mass} &=\frac{8000.6}{100}\\ &=80.006 \end{aligned} \]
💡 Answer Final Answer ›
Therefore, the average atomic mass of bromine is: \(80.006\ \text{u}\)
🎯 Exam Significance Exam Significance ›
  • This is one of the most important numerical questions from the chapter.
  • Tests understanding of isotopes and relative abundance.
  • Frequently asked in CBSE examinations and school assessments.
  • Weighted-average calculations are commonly asked in Olympiads and scholarship examinations.
  • Helps students understand why atomic masses in the periodic table are often fractional.
  • Forms the foundation for higher studies in atomic and nuclear chemistry.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  6 points

  1. Most elements exist as mixtures of isotopes.

  2. Average atomic mass is a weighted average based on isotopic abundance.

  3. More abundant isotopes contribute more to the average atomic mass.

  4. Bromine occurs mainly as \(^{79}\mathrm{Br}\) and \(^{81}\mathrm{Br}\).

  5. The average atomic mass of bromine is \(80.006\;\text{u}\).

  6. Fractional atomic masses arise because isotopes occur in different proportions.
← Q9
10 / 19  ·  53%
Q11 →
Q11
NUMERIC3 marks
The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes \(^{16}_{8}\mathrm{X}\) and \(^{18}_{8}\mathrm{X}\) in the sample?
📘 Concept & Theory Concept Builder ›

The average atomic mass of an element is the weighted average of the masses of its naturally occurring isotopes.

If the percentage abundance of one isotope is known, the abundance of the other isotope can be determined because the total abundance of all isotopes must add up to:

\[ 100\% \]

In this question, the average atomic mass lies closer to 16 u than to 18 u. Therefore, we can predict that isotope \(^{16}_{8}\mathrm{X}\) is present in a greater proportion than \(^{18}_{8}\mathrm{X}\).

We use the weighted-average formula:

\[ \text{Average Atomic Mass} = \frac{\sum(\text{Isotopic Mass} \times \text{Percentage Abundance})}{100} \]

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Assume the percentage abundance of one isotope.

  2. Express the abundance of the second isotope in terms of the first.

  3. Apply the weighted-average formula.

  4. Solve the resulting linear equation.

  5. Find the percentage abundance of both isotopes.

  6. Verify the answer using the average atomic mass.

📊 Graph / Figure Graph / Figure ›
FINDING THE PERCENTAGE OF ISOTOPES AVERAGE ATOMIC MASS 16.2 u ¹⁶X ABUNDANCE 90% ¹⁸X ABUNDANCE 10% 16.2 u 16.0 u (¹⁶X) 17.0 u 18.0 u (¹⁸X) Average = (16 × 90 + 18 × 10) ÷ 100 = 16.2 u
Determining Isotopic Abundance
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  9 steps
  1. Let the Percentage of \(^{16}_{8}\mathrm{X}\) be \(x\%\)

  2. Then the percentage of \(^{18}_{8}\mathrm{X}\) will be:

    \[ (100-x)\% \]

  3. Write the Given Data
    • Average atomic mass = 16.2 u
    • Mass of first isotope = 16 u
    • Mass of second isotope = 18 u
  5. Apply the Weighted Average Formula

  6. \[ 16.2 = \frac{ (16\times x)+(18\times(100-x)) }{100} \]

  7. Simplify the Equation
  8. \[ \begin{aligned} 16.2 & = \frac{16x+1800-18x}{100}\\ 16.2 & = \frac{1800-2x}{100}\\ 16.2\times 100 & = 1800-2x\\ \Rightarrow 2\times(900-x)&=16.2\times 100\\ 900-x&=8.1\times 100\\ \Rightarrow x&=900-810\\ x&=90 \end{aligned} \]
  9. Therefore,\[ ^{16}_{8}\mathrm{X}=90\%\]
  10. and
  11. \[\begin{aligned}^{18}_{8}\mathrm{X}&=100-90\\ &=10\%\end{aligned}\]
  12. Verification
  13. Checking the result: \[ \begin{aligned} \frac{(16\times90)+(18\times10)}{100} &=\frac{1440+180}{100}\\ &=\frac{1620}{100}\\ &=16.2\;\text{u} \end{aligned} \]
  14. The calculated value matches the given average atomic mass. Hence, the answer is correct.
💡 Answer Final Answer ›

Let the percentage of \(^{16}_{8}\mathrm{X}\) be \(x\%\). Then the percentage of \(^{18}_{8}\mathrm{X}\) is \((100-x)\%\).

Using the weighted-average formula:

\[16.2 = \frac{16x+18(100-x)}{100}\]

Solving gives:

\[x=90\]

Therefore,

\[\boxed{^{16}_{8}\mathrm{X}=90\%}\]

\[\boxed{^{18}_{8}\mathrm{X}=10\%}\]

🎯 Exam Significance Exam Significance ›
  • This is a standard NCERT numerical based on isotopic abundance.
  • Frequently asked in CBSE examinations and school tests.
  • Tests understanding of weighted averages and isotopes.
  • Commonly appears in NTSE, Olympiads, Foundation courses, and scholarship examinations.
  • Develops skills required for solving atomic-mass calculations in higher classes.
  • Students should be comfortable setting up linear equations from percentage data.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  7 points

  1. Average atomic mass is a weighted average of isotopic masses.

  2. The total percentage abundance of all isotopes is always 100%.

  3. Letting one abundance be \(x\%\) simplifies the calculation.

  4. The isotope closer to the average mass is generally present in larger proportion.

  5. \(^{16}_{8}\mathrm{X}\) constitutes 90% of the sample.

  6. \(^{18}_{8}\mathrm{X}\) constitutes 10% of the sample.

  7. Weighted-average problems are among the most important isotope numericals in Class 9 Chemistry.
← Q10
11 / 19  ·  58%
Q12 →
Q12
NUMERIC3 marks
If Z = 3, what would be the valency of the element? Also, name the element.
📘 Concept & Theory Concept Builder ›

The atomic number (Z) of an element represents the number of protons present in its nucleus. In a neutral atom, the number of electrons is equal to the number of protons.

To determine the valency of an element, we first write its electronic configuration. The valency depends on the number of electrons present in the outermost shell (valence shell).

Atoms tend to attain a stable electronic configuration similar to that of noble gases by losing, gaining, or sharing electrons.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Identify the element using its atomic number.

  2. Determine the total number of electrons.

  3. Write the electronic configuration.

  4. Find the number of valence electrons.

  5. Determine how many electrons must be lost or gained for stability.

  6. Calculate the valency and write the name of the element.

📊 Graph / Figure Graph / Figure ›
Valency of Lithium (Z = 3) Lithium achieves stability by losing its single outermost valence electron Lithium Atom (Li) 3p⁺ 4n⁰ Configuration: 2, 1 1 Valence Electron (Unstable) e⁻ Loses 1 Electron Lithium Ion (Li⁺) 3p⁺ 4n⁰ Configuration: 2 Stable Duet State Lithium Loses 1 Valence Electron ➔ Valency = 1
Determining the Valency of Lithium
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  6 steps
  1. Identify the Element

  2. The atomic number is given as:

    \[ Z = 3 \]

    The element having atomic number 3 is:

    \[ \text{Lithium (Li)} \]

  3. Determine the Number of Electrons

  4. In a neutral atom:

    \[ \text{Number of Electrons} = \text{Atomic Number} \]

    Therefore,

    \[ \text{Number of Electrons} = 3 \]

  5. Write the Electronic Configuration

  6. Electrons are filled according to shell capacities:

    • K-shell can hold a maximum of 2 electrons.
    • The remaining electron goes into the L-shell.

    Therefore, the electronic configuration of lithium is:

    \[ 2,\;1 \]

  7. Find the Valence Electron

  8. The outermost shell (L-shell) contains:

    \[ 1 \]

    electron.

    Therefore, lithium has one valence electron.

  9. Determine the Valency

  10. Lithium can become stable by losing its single valence electron.

    After losing one electron:

    \[ \mathrm{Li \rightarrow Li^+ + e^-} \]

    The electronic configuration becomes:

    \[ 2 \]

    which is a stable duplet configuration similar to helium.

    Hence, lithium loses one electron and its valency is:

    \[ \boxed{1} \]

  11. Quick Summary Table
  12. Property Value
    Atomic Number (Z) 3
    Element Lithium (Li)
    Number of Electrons 3
    Electronic Configuration 2, 1
    Valence Electrons 1
    Valency 1
💡 Answer Final Answer ›

If \[ Z = 3 \] the element is Lithium (Li).

Its electronic configuration is:

\[ 2,\;1 \]

Since lithium has one electron in its outermost shell, it loses that electron to attain a stable configuration. Therefore, the valency of lithium is:

\[ \boxed{\text{Valency} = 1} \]

Element Name: Lithium (Li)

🎯 Exam Significance Exam Significance ›
  • This is a frequently asked short-answer question from electronic configuration and valency.
  • Tests understanding of atomic number and electron distribution.
  • Important for learning how valency is determined from electronic configuration.
  • Forms the basis for understanding ionic bond formation.
  • Common in CBSE examinations, NTSE, Olympiads, and scholarship tests.
  • Often appears as a one-mark or MCQ-based question.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  7 points

  1. Atomic number 3 corresponds to Lithium.

  2. Lithium contains 3 electrons.

  3. Its electronic configuration is 2, 1.

  4. The outermost shell contains one electron.

  5. Lithium attains stability by losing one electron.

  6. The valency of lithium is 1.

  7. Elements with one valence electron usually have valency 1.
← Q11
12 / 19  ·  63%
Q13 →
Q13
NUMERIC3 marks
The composition of the nuclei of two atomic species X and Y are given as under: \[ \begin{array}{|c|c|c|} \hline & \mathrm{X} & \mathrm{Y}\\ \hline \text{Protons} & 6 & 6\\ \hline \text{Neutrons} & 6 & 8\\ \hline \end{array} \] Give the mass numbers of X and Y. What is the relation between the two species?
📘 Concept & Theory Concept Builder ›

The nucleus of an atom contains protons and neutrons. Two important quantities associated with an atom are its atomic number and mass number.

The atomic number is equal to the number of protons present in the nucleus, whereas the mass number is equal to the total number of protons and neutrons.

The mass number is calculated using:

\[ \text{Mass Number (A)} = \text{Number of Protons (Z)} + \text{Number of Neutrons (N)} \]

\[ A = Z + N \]

When two atoms have the same number of protons but different numbers of neutrons, they are known as isotopes.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Identify the number of protons and neutrons in X.

  2. Calculate the mass number of X.

  3. Identify the number of protons and neutrons in Y.

  4. Calculate the mass number of Y.

  5. Compare the proton numbers of X and Y.

  6. Determine the relationship between the two species.

📊 Graph / Figure Graph / Figure ›
ISOTOPES OF CARBON SAME ATOMIC NUMBER • DIFFERENT NEUTRON COUNT VS SPECIES X CARBON-12 6 Protons 6 Neutrons Mass Number = 12 SPECIES Y CARBON-14 6 Protons 8 Neutrons Mass Number = 14 Same Number of Protons (6) = Same Element THEREFORE: ISOTOPES OF CARBON
Identifying Isotopes of Carbon
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  7 steps
  1. Calculate the Mass Number of X

  2. For species X:

    • Number of protons = 6
    • Number of neutrons = 6

    Using the formula:

    \[ A = Z + N \]

    Substituting the values:

    \[ A = 6 + 6 \]

    \[ A = 12 \]

    Therefore, the mass number of X is:

    \[ \boxed{12} \]

  3. Calculate the Mass Number of Y

  4. For species Y:

    • Number of protons = 6
    • Number of neutrons = 8

    Using the same formula:

    \[ A = Z + N \]

    Substituting the values:

    \[ A = 6 + 8 \]

    \[ A = 14 \]

    Therefore, the mass number of Y is:

    \[ \boxed{14} \]

  5. Determine the Relationship Between X and Y

  6. Compare the number of protons:

    \[ \mathrm{X}: 6 \text{ protons} \]

    \[ \mathrm{Y}: 6 \text{ protons} \]

    Since both species have the same number of protons, they belong to the same element.

    Atomic number 6 corresponds to:

    \[ \text{Carbon (C)} \]

    However, the numbers of neutrons are different:

    \[ \mathrm{X}: 6 \text{ neutrons} \]

    \[ \mathrm{Y}: 8 \text{ neutrons} \]

    Consequently, their mass numbers are different:

    \[ 12 \neq 14 \]

  7. Atoms having the same atomic number but different mass numbers are called isotopes.
  8. Therefore:
  9. \[\boxed{\text{X and Y are isotopes of carbon}}\]
  10. Summary Table
  11. Property X Y
    Protons 6 6
    Neutrons 6 8
    Mass Number 12 14
    Element Carbon Carbon
    Relationship Isotopes
💡 Answer Final Answer ›

For species X:

\[ \text{Mass Number} = 6+6 = 12 \]

For species Y:

\[ \text{Mass Number} = 6+8 = 14 \]

Thus,

\[ \boxed{\text{Mass Number of X} = 12} \]

\[ \boxed{\text{Mass Number of Y} = 14} \]

Since both species have the same number of protons (atomic number = 6) but different mass numbers, they are isotopes of carbon.

🎯 Exam Significance Exam Significance ›
  • This is a very important conceptual question on isotopes.
  • Frequently asked in CBSE examinations and school tests.
  • Tests understanding of atomic number and mass number.
  • Helps distinguish between isotopes and isobars.
  • Commonly appears as MCQs and short-answer questions.
  • Important for NTSE, Olympiads, Foundation courses, and scholarship examinations.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  8 points

  1. Mass number = Number of protons + Number of neutrons.

  2. Species X has mass number 12.

  3. Species Y has mass number 14.

  4. Both species have atomic number 6.

  5. Atomic number 6 corresponds to carbon.

  6. Different neutron numbers lead to different mass numbers.

  7. Atoms with the same atomic number but different mass numbers are called isotopes.

  8. X and Y are isotopes of carbon (\(^{12}\mathrm{C}\) and \(^{14}\mathrm{C}\)).
← Q12
13 / 19  ·  68%
Q14 →
Q14
NUMERIC3 marks
For the following statements, write T for True and F for False.
(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
(c) The mass of an electron is about \(\frac{1}{2000}\) times that of proton.
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
📘 Concept & Theory Concept Builder ›

True/False questions test conceptual understanding of atomic structure and the contributions of various scientists to atomic theory.

To answer such questions correctly, it is important to remember:

  • J.J. Thomson proposed the "plum pudding model" and did not propose the nucleus.
  • The nucleus was discovered later by Rutherford.
  • Neutrons were discovered by James Chadwick.
  • The electron is extremely light compared to the proton.
  • Radioactive isotopes of iodine are used in medicine, while tincture iodine is simply a solution of iodine used as an antiseptic.
🗺️ Solution Roadmap Step-by-step Plan ›

  1. Analyze each statement individually.

  2. Compare it with established atomic theory.

  3. Identify whether the statement is scientifically correct or incorrect.

  4. Provide justification for each answer.

📊 Graph / Figure Graph / Figure ›
ATOMIC STRUCTURE CONCEPTS J.J. Thomson Model - - - - - Positive Pudding Electrons (- ) ● Diffuse positive charge ● Embedded negative electrons ● NO nucleus present Neutron Concept p⁺ Proton e⁻ Electron n⁰ Neutron ● Neutron is independent ● NOT a proton-electron bind ● Discovered by Chadwick Electron Mass e⁻ p⁺ mₑ ≈ 1/1836 mₚ ● Electron is extremely light ● ~1/2000 of proton mass ● Most mass is in nucleus Iodine Uses ¹³¹I Isotope ● Used in thyroid therapy ● Tincture iodine = elemental I₂ ● Tincture ≠ ¹³¹I radioactive Key Concepts Tested in Atomic Structure Exam Syllabus
Important Concepts Behind the True/False Statements
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  4 steps
  1. (a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.

  2. J.J. Thomson did not propose the existence of a nucleus.

    According to Thomson's plum pudding model, the positive charge was spread uniformly throughout the atom and electrons were embedded in it.

    The concept of a central nucleus was introduced later by Rutherford after the gold foil experiment.

    Therefore, the statement is:

    \[ \boxed{\text{False (F)}} \]

  3. (b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.

  4. This explanation is not correct according to modern atomic theory.

    A neutron is an independent subatomic particle discovered by James Chadwick in 1932.

    Although an early hypothesis suggested that a neutron might consist of a proton and an electron, this idea was later proven incorrect.

    Therefore, the statement is:

    \[ \boxed{\text{False (F)}} \]

  5. (c) The mass of an electron is about \(\frac{1}{2000}\) times that of proton.

  6. The actual mass of an electron is approximately:

    \[ \frac{1}{1836} \]

    times the mass of a proton.

    For school-level calculations, this is often approximated as:

    \[ \frac{1}{2000} \]

    Hence, the statement is considered:

    \[ \boxed{\text{True (T)}} \]

  7. (d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.

  8. Tincture iodine is a solution of ordinary iodine dissolved in alcohol and water.

    It is used as an antiseptic for wounds and cuts.

    Radioactive isotopes of iodine such as \(\mathrm{^{131}I}\) are used in the diagnosis and treatment of thyroid disorders, but they are not used to prepare tincture iodine.

    Therefore, the statement is:

    \[ \boxed{\text{False (F)}} \]

💡 Answer Final Answer ›
Statement Answer
J.J. Thomson proposed that the nucleus of an atom contains only nucleons. F
A neutron is formed by an electron and a proton combining together. F
The mass of an electron is about \(\frac{1}{2000}\) times that of proton. T
An isotope of iodine is used for making tincture iodine. F
🎯 Exam Significance Exam Significance ›
  • Frequently asked as MCQs and True/False questions in CBSE examinations.
  • Tests conceptual understanding of atomic models and subatomic particles.
  • Important for distinguishing the contributions of Thomson, Rutherford, Bohr, and Chadwick.
  • Helps avoid common misconceptions regarding neutrons and isotopes.
  • Important for NTSE, Olympiads, Foundation courses, and scholarship examinations.
  • Conceptual questions of this type are commonly used in competitive entrance tests.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  6 points

  1. Thomson proposed the plum pudding model and not the nucleus.

  2. Rutherford discovered the nucleus through the gold foil experiment.

  3. Neutrons are independent particles discovered by James Chadwick.

  4. The mass of an electron is approximately \(\frac{1}{1836}\) of a proton's mass.

  5. \(\mathrm{^{131}I}\) is used in thyroid treatment and diagnosis.

  6. Tincture iodine is an antiseptic solution and does not contain radioactive iodine isotopes.
← Q13
14 / 19  ·  74%
Q15 →
Q15
NUMERIC3 marks
Rutherford’s alpha-particle scattering experiment was responsible for the discovery of
(a) Atomic Nucleus
(b) Electron
(c) Proton
(d) Neutron
📘 Concept & Theory Concept Builder ›

One of the most important experiments in the history of atomic theory was Rutherford's alpha-particle scattering experiment, also known as the gold foil experiment.

In this experiment, fast-moving alpha particles were directed towards a very thin sheet of gold foil. The observations made during this experiment helped scientists understand the internal structure of the atom.

Before Rutherford's work, J.J. Thomson's atomic model suggested that positive charge was spread uniformly throughout the atom. Rutherford's experiment proved this idea incorrect.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Recall the purpose of Rutherford's experiment.

  2. Analyze the observations made during the experiment.

  3. Identify the conclusion drawn by Rutherford.

  4. Select the correct option based on the conclusion.

📊 Graph / Figure Graph / Figure ›
RUTHERFORD'S ALPHA-PARTICLE SCATTERING EXPERIMENT DISCOVERY OF THE ATOMIC NUCLEUS · GEIGER & MARSDEN · 1911 ① Experimental Setup (Macro View) ZnS Screen (Detector) α Source Slit Gold Foil (~1000 atoms) α beam Rare rebound Rebound ~99% straight ② Atomic-Scale View (Scattering) 79+ Nucleus α Undeflected α Undeflected α α α Large deflection α Head-on rebound Gold Nucleus Alpha (α) particle OBSERVATION 1 99% of α-particles pass straight through the foil. → Atom is mostly empty space. OBSERVATION 2 Some α-particles deflect at small / medium angles. → Positive charge is concentrated. OBSERVATION 3 1 in ~20,000 bounce back at angles > 90°. → Nucleus is tiny, dense & heavy.
Rutherford's Alpha-Particle Scattering Experiment
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. Observation of Alpha Particles

  2. Rutherford observed that:

    • Most alpha particles passed straight through the gold foil.
    • A small number were deflected through small angles.
    • Very few were deflected through large angles or bounced back.
  3. Interpretation of the Observations

  4. These observations indicated that:

    • Most of the atom is empty space.
    • The positive charge is not spread throughout the atom.
    • Almost all the mass and positive charge are concentrated in a very small central region.
  5. Rutherford's Conclusion

  6. Rutherford concluded that every atom contains a tiny, dense, positively charged center called the nucleus.

    This discovery led to the nuclear model of the atom.

  7. Analyze the Options
  8. Option Analysis
    (a) Atomic Nucleus Correct. Rutherford discovered the nucleus.
    (b) Electron Incorrect. Electron was discovered by J.J. Thomson.
    (c) Proton Incorrect. Rutherford later identified the proton, but this experiment specifically established the nucleus.
    (d) Neutron Incorrect. Neutron was discovered by James Chadwick.
  9. Important Scientists and Discoveries
  10. Scientist Discovery
    J.J. Thomson Electron
    Ernest Rutherford Atomic Nucleus
    Ernest Rutherford Proton
    James Chadwick Neutron
💡 Answer Final Answer ›

Rutherford's alpha-particle scattering experiment showed that the positive charge and most of the mass of an atom are concentrated in a very small central region called the nucleus.

Therefore, the correct answer is:

✓ (a) Atomic Nucleus

🎯 Exam Significance Exam Significance ›
  • This is one of the most frequently asked MCQs from the chapter.
  • Tests knowledge of important scientific discoveries.
  • Students often confuse the discoveries of Rutherford, Thomson, and Chadwick.
  • Very common in CBSE examinations, NTSE, Olympiads, and scholarship tests.
  • The gold foil experiment is considered a landmark experiment in atomic physics.
  • Knowledge of the observations and conclusions of the experiment is essential for higher classes.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  7 points

  1. Rutherford performed the alpha-particle scattering experiment.

  2. Most alpha particles passed through the foil, showing that atoms are mostly empty space.

  3. A few alpha particles were strongly deflected.

  4. The experiment led to the discovery of the atomic nucleus.

  5. Electron was discovered by J.J. Thomson.

  6. Neutron was discovered by James Chadwick.

  7. The nucleus contains most of the mass and positive charge of an atom.
← Q14
15 / 19  ·  79%
Q16 →
Q16
NUMERIC3 marks
Isotopes of an element have
(a) the same physical properties
(b) different chemical properties
(c) different number of neutrons
(d) different atomic numbers.
📘 Concept & Theory Concept Builder ›

Isotopes are atoms of the same element having the same atomic number but different mass numbers.

Since atomic number represents the number of protons, isotopes of an element contain the same number of protons. However, their mass numbers differ because they contain different numbers of neutrons.

Since chemical properties depend mainly on the electronic configuration, isotopes generally have identical chemical properties. However, differences in mass may cause slight differences in certain physical properties.

Examples:

\[ ^{12}_{6}\mathrm{C}, \quad ^{13}_{6}\mathrm{C}, \quad ^{14}_{6}\mathrm{C} \]

All three are isotopes of carbon because they have the same atomic number (6) but different numbers of neutrons.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Recall the definition of isotopes.

  2. Compare the properties mentioned in the options.

  3. Identify the characteristic feature that distinguishes isotopes.

  4. Select the correct option.

📊 Graph / Figure Graph / Figure ›
Isotopes of Carbon SAME PROTON COUNT · DIFFERENT NEUTRON COUNT · DIFFERENT MASS CARBON · 12 12 6 C Carbon-12 n=1 n=2 6 Protons 6 Neutrons Mass Number = 12 6 protons + 6 neutrons = 12 Electron config: 2, 4 · Stable isotope CARBON · 14 14 6 C Carbon-14 n=1 n=2 6 Protons 8 Neutrons Mass Number = 14 6 protons + 8 neutrons = 14 Electron config: 2, 4 · Radioactive (β⁻ decay) Z = 6 SAME ELEMENT DIFFERENT MASS No. Same number of Protons (Z = 6) Different number of Neutrons
Understanding Isotopes
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  4 steps
  1. Definition of Isotopes

  2. Isotopes are atoms of the same element having:

    • Same atomic number
    • Same number of protons
    • Different mass numbers
    • Different numbers of neutrons
  3. Analyze Each Option
  4. Option Analysis
    (a) Same physical properties Incorrect. Physical properties may differ because isotopes have different masses.
    (b) Different chemical properties Incorrect. Chemical properties are nearly identical because isotopes have the same electronic configuration.
    (c) Different number of neutrons Correct. This is the defining characteristic of isotopes.
    (d) Different atomic numbers Incorrect. Isotopes always have the same atomic number.
  5. Select the Correct Answer
  6. Since isotopes differ only in the number of neutrons present in their nuclei, the correct option is:

    ✓ (c) Different number of neutrons

  7. Illustrative Example
  8. Isotope Protons Neutrons Mass Number
    \(^{12}_{6}\mathrm{C}\) 6 6 12
    \(^{14}_{6}\mathrm{C}\) 6 8 14

    Both atoms have the same atomic number (6), but the number of neutrons is different. Therefore, they are isotopes of carbon.

💡 Answer Final Answer ›

Isotopes of an element have the same atomic number and the same number of protons, but they differ in the number of neutrons present in their nuclei.

Therefore, the correct answer is:

✓ (c) Different number of neutrons

🎯 Exam Significance Exam Significance ›
  • This is one of the most important MCQs from the topic of isotopes.
  • Frequently asked in CBSE examinations and school tests.
  • Tests the fundamental definition of isotopes.
  • Students often confuse isotopes with isobars; this question helps distinguish them.
  • Very common in NTSE, Olympiads, Foundation courses, and scholarship examinations.
  • Knowledge of isotopes is essential for understanding atomic mass calculations.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  7 points

  1. Isotopes have the same atomic number.

  2. Isotopes contain the same number of protons.

  3. Isotopes have different numbers of neutrons.

  4. Different neutron numbers produce different mass numbers.

  5. Chemical properties of isotopes are nearly identical.

  6. Some physical properties may differ because of mass differences.

  7. The defining characteristic of isotopes is the difference in neutron count.
← Q15
16 / 19  ·  84%
Q17 →
Q17
NUMERIC3 marks
Number of valence electrons in \(\mathrm{Cl^-}\) ion are:
(a) 16
(b) 8
(c) 17
(d) 18
📘 Concept & Theory Concept Builder ›

Valence electrons are the electrons present in the outermost shell of an atom or ion. They play a crucial role in chemical bonding and determine the chemical properties of an element.

Chlorine is a non-metal belonging to Group 17 of the periodic table. It tends to gain one electron to achieve a stable noble-gas configuration.

When a chlorine atom gains one electron, it forms a chloride ion:

\[ \mathrm{Cl + e^- \rightarrow Cl^-} \]

The resulting chloride ion possesses a complete octet (8 electrons) in its outermost shell, making it highly stable.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Find the atomic number of chlorine.

  2. Write the electronic configuration of a chlorine atom.

  3. Determine the number of valence electrons in chlorine.

  4. Understand the formation of the chloride ion.

  5. Calculate the number of valence electrons in \(\mathrm{Cl^-}\).

  6. Select the correct option.

📊 Graph / Figure Graph / Figure ›
Formation of Chloride Ion (Cl⁻) Chlorine Atom Cl (Z = 17) M L K Cl Electronic Configuration 2 , 8 , 7 7 Valence Electrons Gains 1 Electron e⁻ Energy released (exothermic) Chloride Ion (Cl⁻) 18 electrons | charge = −1 M L K Cl⁻ Electronic Configuration 2 , 8 , 8 ✓ Octet Complete (8 e⁻) ● Blue = existing electrons ● Orange = gained electron | Correct answer: 8 valence electrons
Formation of Chloride Ion (\(\mathrm{Cl^-}\))
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  6 steps
  1. Atomic Number of Chlorine

  2. Chlorine has atomic number:

    \[ Z = 17 \]

    Therefore, a neutral chlorine atom contains:

    \[ 17 \text{ electrons} \]

  3. Electronic Configuration of Chlorine

  4. The distribution of 17 electrons is:

    \[ 2,\;8,\;7 \]

    Thus, chlorine has:

    \[ 7 \]

    valence electrons.

  5. Formation of Chloride Ion

  6. Chlorine gains one electron to attain a stable octet:

    \[ \mathrm{Cl + e^- \rightarrow Cl^-} \]

    After gaining one electron, the total number of electrons becomes:

    \[ 17 + 1 = 18 \]

  7. Electronic Configuration of \(\mathrm{Cl^-}\)

  8. The electronic configuration of the chloride ion becomes:

    \[ 2,\;8,\;8 \]

    Therefore, the outermost shell contains:

    \[ 8 \]

    electrons.

    Hence, the number of valence electrons in \(\mathrm{Cl^-}\) is:

    \[ \boxed{8} \]

  9. Check the Options
  10. Option Analysis
    (a) 16 Incorrect
    (b) 8 Correct
    (c) 17 Incorrect
    (d) 18 Incorrect
  11. Electronic Configuration Summary
  12. Species Electronic Configuration Valence Electrons
    \(\mathrm{Cl}\) 2, 8, 7 7
    \(\mathrm{Cl^-}\) 2, 8, 8 8
💡 Answer Final Answer ›

Chlorine has electronic configuration:

\[ 2,\;8,\;7 \]

After gaining one electron, it forms the chloride ion:

\[ \mathrm{Cl^-} \]

with electronic configuration:

\[ 2,\;8,\;8 \]

Therefore, the outermost shell contains 8 electrons.

✓ (b) 8

🎯 Exam Significance Exam Significance ›
  • This is a frequently asked MCQ from electronic configuration and valency.
  • Tests understanding of ion formation and octet stability.
  • Students often confuse total electrons with valence electrons.
  • Important for understanding ionic bonding and noble-gas configurations.
  • Common in CBSE examinations, NTSE, Olympiads, and scholarship tests.
  • Forms the foundation for learning chemical bonding in higher classes.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  8 points

  1. Chlorine has atomic number 17.

  2. Electronic configuration of chlorine is 2, 8, 7.

  3. Chlorine gains one electron to form \(\mathrm{Cl^-}\).

  4. \(\mathrm{Cl^-}\) has 18 total electrons.

  5. Electronic configuration of \(\mathrm{Cl^-}\) is 2, 8, 8.

  6. The outermost shell contains 8 electrons.

  7. Valence electrons in \(\mathrm{Cl^-}\) = 8.

  8. A complete octet makes the chloride ion stable.
← Q16
17 / 19  ·  89%
Q18 →
Q18
NUMERIC3 marks
Which one of the following is a correct electronic configuration of sodium?
(a) 2,8
(b) 8,2,1
(c) 2,1,8
(d) 2,8,1
📘 Concept & Theory Concept Builder ›

The electronic configuration of an atom shows how its electrons are distributed among different shells (K, L, M, N, etc.).

To write the electronic configuration, we use the atomic number of the element, which gives the total number of electrons in a neutral atom.

The maximum number of electrons that can be accommodated in a shell is given by:

\[ 2n^2 \]

where \(n\) is the shell number.

  • K-shell (\(n=1\)) can hold a maximum of 2 electrons.
  • L-shell (\(n=2\)) can hold a maximum of 8 electrons.
  • M-shell (\(n=3\)) can hold a maximum of 18 electrons.

Electrons always fill the inner shells first before occupying outer shells.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Find the atomic number of sodium.

  2. Determine the total number of electrons.

  3. Distribute electrons according to shell capacities.

  4. Compare the obtained configuration with the given options.

  5. Select the correct answer.

📊 Graph / Figure Graph / Figure ›
Electronic Configuration of Sodium (Na) M L K Na 11p⁺, 12n⁰ Valence Electron (3s¹) SHELL DATA K Shell = 2 L Shell = 8 M Shell = 1 Total Electrons: 11 Electronic Configuration = 2, 8, 1 ● Blue = Core Electrons (10) ● Yellow = Valence Electron (1) | Atomic Number (Z) = 11
Electronic Configuration of Sodium
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  6 steps
  1. Atomic Number of Sodium

  2. Sodium (\(\mathrm{Na}\)) has atomic number:

    \[ Z = 11 \]

    Therefore, a neutral sodium atom contains:

    \[ 11 \text{ electrons} \]

  3. Fill the K-Shell

  4. The K-shell can hold a maximum of:

    \[ 2 \text{ electrons} \]

    After filling the K-shell:

    \[ 11-2=9 \]

    electrons remain.

  5. Fill the L-Shell

  6. The L-shell can hold a maximum of:

    \[ 8 \text{ electrons} \]

    After filling the L-shell:

    \[ 9-8=1 \]

    electron remains.

  7. Fill the M-Shell

  8. The remaining electron occupies the M-shell.

    Therefore:

    \[ K = 2,\quad L = 8,\quad M = 1 \]

    Hence, the electronic configuration of sodium is:

    \[ \boxed{2,\;8,\;1} \]

  9. Check the Given Options
  10. Option Analysis
    (a) 2,8 Incorrect. Represents only 10 electrons.
    (b) 8,2,1 Incorrect. Shell filling order is wrong.
    (c) 2,1,8 Incorrect. Electrons cannot fill the M-shell before the L-shell is complete.
    (d) 2,8,1 Correct.
  11. Shell-wise Distribution of Sodium
  12. Shell Number of Electrons
    K 2
    L 8
    M 1
💡 Answer Final Answer ›

Sodium has atomic number 11 and therefore contains 11 electrons. Distributing these electrons into different shells gives:

\[ 2,\;8,\;1 \]

Therefore, the correct option is:

✓ (d) 2,8,1

🎯 Exam Significance Exam Significance ›
  • This is one of the most frequently asked MCQs from electronic configuration.
  • Tests understanding of shell filling and electron distribution.
  • Important for determining valency and chemical properties.
  • Forms the basis for understanding ion formation and chemical bonding.
  • Frequently asked in CBSE examinations, NTSE, Olympiads, and scholarship tests.
  • Knowledge of sodium's configuration is useful in many later chemistry chapters.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  8 points

  1. Sodium has atomic number 11.

  2. A neutral sodium atom contains 11 electrons.

  3. Electrons fill inner shells before outer shells.

  4. K-shell contains 2 electrons.

  5. L-shell contains 8 electrons.

  6. M-shell contains 1 electron.

  7. Electronic configuration of sodium is 2,8,1.

  8. Sodium has one valence electron and valency 1.
← Q17
18 / 19  ·  95%
Q19 →
Q19
NUMERIC3 marks
Complete the following table. Atomic Structure Data Table Reference values and fill-in-the-blank chemical characteristics Atomic Number (Z) Mass Number (A) Number of Neutrons Number of Protons Number of Electrons Name of the Atomic Species 9 — 10 — — — 16 32 — — — Sulphur — 24 — 12 — — — 2 — 1 — — — 1 0 1 0 — Fill in the missing values based on: Mass Number (A) = Protons (Z) + Neutrons (N)
📘 Concept & Theory Concept Builder ›

To complete this table, we use the relationships between atomic number, mass number, protons, neutrons, and electrons.

Important formulas:

\[ \text{Atomic Number (Z)} = \text{Number of Protons} \]

\[ \text{Mass Number (A)} = \text{Number of Protons} + \text{Number of Neutrons} \]

\[ \text{Number of Neutrons} = \text{Mass Number} - \text{Atomic Number} \]

For a neutral atom:

\[ \text{Number of Electrons} = \text{Number of Protons} \]

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Use atomic number to determine the number of protons.

  2. Calculate neutrons using \(A-Z\).

  3. For neutral atoms, set electrons equal to protons.

  4. Identify the element from its atomic number.

  5. Fill all missing entries systematically.

📊 Graph / Figure Graph / Figure ›
Completing Atomic Structure Tables Mass Number = Protons + Neutrons A = Z + N A = Mass Number Z = Atomic Number (Protons) N = Neutrons p⁺ Atomic Number = Number of Protons Identifies the element (Z) e⁻ = p⁺ Neutral Atom Electrons = Protons Net charge is zero (0) n⁰ Neutrons (N) = Mass (A) − Protons (Z) N = A − Z These Relations Help Complete Any Atomic Structure Table p⁺ = Protons (Positive Charge) | n⁰ = Neutrons (Neutral Charge) | e⁻ = Electrons (Negative Charge)
Relationship Between Atomic Number, Mass Number and Neutrons
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  6 steps
  1. Row 1: Fluorine

  2. Atomic number = 9

    Therefore,

    \[ \text{Protons}=9 \]

    \[ \text{Electrons}=9 \]

    Mass number = 19

    \[ \text{Neutrons}=19-9=10 \]

    Species = Fluorine

  3. Row 2: Sulphur

  4. Atomic number = 16

    \[ \text{Protons}=16 \]

    \[ \text{Electrons}=16 \]

    Mass number = 32

    \[ \text{Neutrons}=32-16=16 \]

    Species = Sulphur

  5. Row 3: Magnesium

  6. Protons = 12

    \[ \text{Atomic Number}=12 \]

    Mass number = 24

    \[ \text{Neutrons}=24-12=12 \]

    \[ \text{Electrons}=12 \]

    Species = Magnesium

  7. Row 4: Deuterium

  8. Protons = 1

    \[ \text{Atomic Number}=1 \]

    Mass number = 2

    \[ \text{Neutrons}=2-1=1 \]

    \[ \text{Electrons}=1 \]

    Species = Deuterium (\(^{2}_{1}\mathrm{H}\))

  9. Row 5: Protium

  10. Protons = 1

    \[ \text{Atomic Number}=1 \]

    Mass number = 1

    \[ \text{Neutrons}=1-1=0 \]

    Since Protium is a neutral atom:

    \[ \text{Electrons}=1 \]

    The value 0 under electrons in the given table is incorrect and should be corrected to 1.

  11. Completed Table
  12. Atomic Number Mass Number Neutrons Protons Electrons Species
    9 19 10 9 9 Fluorine
    16 32 16 16 16 Sulphur
    12 24 12 12 12 Magnesium
    1 2 1 1 1 Deuterium
    1 1 0 1 1 Protium
🎯 Exam Significance Exam Significance ›
  • Frequently asked as a table-completion question in CBSE examinations.
  • Tests understanding of atomic number, mass number, protons, neutrons, and electrons.
  • Important for distinguishing isotopes such as Protium and Deuterium.
  • Strengthens numerical skills involving atomic structure.
  • Useful for NTSE, Olympiads, Foundation courses, and scholarship examinations.
  • Often appears as short-answer or MCQ-based questions.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  8 points

  1. Atomic number equals the number of protons.

  2. For neutral atoms, electrons equal protons.

  3. Mass number equals protons plus neutrons.

  4. Neutrons can be calculated using \(A-Z\).

  5. Protium has no neutron.

  6. Deuterium contains one proton and one neutron.

  7. Fluorine has atomic number 9.

  8. Sulphur and Magnesium have atomic numbers 16 and 12 respectively.
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NCERT Class 9 Science Ch 4 Exercise Solutions | Atom
NCERT Class 9 Science Ch 4 Exercise Solutions | Atom — Complete Notes & Solutions · academia-aeternum.com
This page provides clear, comprehensive answers to textbook exercise questions from NCERT Class 9 Science Chapter 4: Structure of the Atom. It covers foundational concepts about atomic structure, atomic models, electronic configurations, valency, isotopes, isobars, and subatomic particles. Whether you’re preparing for school exams or aiming for deeper understanding of atomic theory, these explanations and tables help you master every topic step by step.
🎓 Class 9 📐 Science 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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    Structure of Atom — Learning Resources

    Frequently Asked Questions

    Bonds are formed by gaining, losing, or sharing outer electrons.

    It’s the specific arrangement of electrons in atomic shells or energy levels.

    It represents whole protons and neutrons, which are counted as whole particles.

    The electronic arrangement with electrons in the lowest possible energy levels.

    A period is a row; a group is a column, indicating similar electronic configuration patterns.

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