Structure of Atom — Class 11 Chemistry NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Structure of Atom" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Structure of Atom" — 8 important questions with detailed answers for CBSE board exam…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 11 Chemistry — Structure of Atom. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Structure of Atom — All 8 Questions Solved
Q1. An atom of element X has 12 protons and 14 neutrons. Write its atomic notation and identify the element.
Given:
Number of protons = 12
Number of neutrons = 14
Step 1: Determine atomic number.
Atomic number (Z) = Number of protons = 12
Step 2: Determine mass number.
Mass number (A) = Number of protons + Number of neutrons
Mass number (A) = 12 + 14 = 26
Step 3: Identify the element.
An atom with 12 protons is Magnesium (Mg).
Step 4: Write the atomic notation.
Atomic notation: ²⁶Mg₁₂ or Mg-26
Answer: The element is Magnesium (Mg), and its atomic notation is ²⁶Mg₁₂.
Q2. What are isotopes? Give an example and explain why they have similar chemical properties but different physical properties.
Definition of Isotopes:
Isotopes are atoms of the same element that have the same atomic number (same number of protons) but different mass numbers (different number of neutrons).
Example:
Carbon has three naturally occurring isotopes:
1. ¹²C (6 protons, 6 neutrons) — most abundant
2. ¹³C (6 protons, 7 neutrons) — less abundant
3. ¹⁴C (6 protons, 8 neutrons) — radioactive, used in dating
Why similar chemical properties?
→ Chemical properties depend on the electron configuration and number of valence electrons.
→ Since all isotopes have the same number of protons and electrons, they have identical electron configurations.
→ Therefore, they undergo the same chemical reactions.
Why different physical properties?
→ Physical properties (like atomic mass, density, melting point) depend on the mass of the nucleus.
→ Different numbers of neutrons → different atomic masses → different physical properties.
Example: ¹²C and ¹⁴C both form CO₂ and burn, but ¹⁴C has different density and half-life.
Q3. Describe the Bohr model of the atom and state its limitations.
Bohr Model of the Atom:
The Bohr model (1913) proposed that:
1. Electrons revolve around the nucleus in fixed circular orbits (called shells or energy levels).
2. Each orbit has a specific fixed energy, and electrons in an orbit do not radiate energy.
3. Electrons can jump between orbits by absorbing or emitting energy (photons) in fixed amounts.
4. The closer the orbit to the nucleus, the lower the energy of the electron.
Orbits are labeled as: K (n=1), L (n=2), M (n=3), etc.
Successes of Bohr Model:
→ Explained the hydrogen atom spectrum accurately
→ Introduced the concept of energy levels
→ Explained the stability of atoms
Limitations of Bohr Model:
1. Cannot explain the spectra of multi-electron atoms (atoms with more than one electron)
2. Does not account for the wave-particle nature of electrons
3. Cannot explain why electrons don't spiral into the nucleus if they are accelerating
4. Cannot explain the fine structure of spectral lines
5. Does not consider the orbital angular momentum correctly
6. Treats orbits as fixed paths, contradicting the uncertainty principle
Conclusion: The Bohr model was superseded by the quantum mechanical model (Schrödinger's model).
Q4. What is the maximum number of electrons that can be accommodated in the M shell? Also calculate the number of electrons in a neutral atom of Aluminum (Al, Z = 13).
Part 1: Maximum electrons in the M shell.
Using the formula: Maximum electrons = 2n²
where n = shell number
For M shell: n = 3
Maximum electrons = 2 × (3)² = 2 × 9 = 18 electrons
Answer: The M shell can accommodate a maximum of 18 electrons.
Part 2: Electron configuration of Aluminum (Z = 13).
Since Al has 13 protons, a neutral Al atom has 13 electrons.
Electron configuration (using shell notation):
K shell (n=1): 2 electrons (max = 2)
L shell (n=2): 8 electrons (max = 8)
M shell (n=3): 3 electrons (remaining = 13 - 2 - 8 = 3)
Electron configuration of Al: 2, 8, 3
Or in orbital notation: 1s² 2s² 2p⁶ 3s² 3p¹
Answer: Aluminum has 13 electrons distributed as 2, 8, 3 in the K, L, and M shells respectively.
Q5. Explain the dual nature of electrons and how it relates to the quantum mechanical model of the atom.
Dual Nature of Electrons:
Electrons exhibit both wave and particle properties, depending on how they are observed.
1. Particle Nature (Corpuscular):
→ Electrons have mass and charge
→ They can collide with other particles
→ They carry momentum (p = mv)
→ Demonstrated by the cathode ray experiment
2. Wave Nature:
→ Electrons can be diffracted and show interference patterns
→ They have a wavelength given by de Broglie's equation:
λ = h / p = h / (mv)
where h = Planck's constant (6.626 × 10⁻³⁴ J·s)
→ Demonstrated by electron diffraction experiments
Relation to Quantum Mechanical Model:
The quantum mechanical model incorporates both properties:
1. Wave-like behavior → Electrons are described by wave functions (ψ)
2. The probability of finding an electron in a region is given by |ψ|² (probability density)
3. Orbitals represent regions where electrons are likely to be found (not definite paths)
4. The uncertainty principle (Δx × Δp ≥ h/4π) arises from the wave nature
Key Difference from Bohr Model:
→ Bohr model: Electrons move in fixed circular orbits (particle-only view)
→ Quantum model: Electrons exist as probability distributions (wave-particle view)
This dual nature explains why the quantum model is more accurate in describing atomic structure.
Q6. Calculate the de Broglie wavelength of a neutron moving at 3 × 10⁵ m/s. (Mass of neutron = 1.675 × 10⁻²⁷ kg, h = 6.626 × 10⁻³⁴ J·s)
De Broglie wavelength formula:
λ = h / (m × v) = h / p
where:
h = Planck's constant = 6.626 × 10⁻³⁴ J·s
m = mass of neutron = 1.675 × 10⁻²⁷ kg
v = velocity = 3 × 10⁵ m/s
Step 1: Calculate momentum (p).
p = m × v
p = 1.675 × 10⁻²⁷ kg × 3 × 10⁵ m/s
p = 5.025 × 10⁻²² kg·m/s
Step 2: Calculate de Broglie wavelength.
λ = h / p
λ = (6.626 × 10⁻³⁴ J·s) / (5.025 × 10⁻²² kg·m/s)
λ = 1.32 × 10⁻¹² m
λ = 1.32 pm (picometers)
Answer: The de Broglie wavelength of the neutron is 1.32 × 10⁻¹² m or 1.32 pm.
Q7. What is the relationship between frequency (ν) and wavelength (λ) of electromagnetic radiation? If a photon has frequency 5 × 10¹⁴ Hz, calculate its wavelength. (c = 3 × 10⁸ m/s)
Relationship between frequency and wavelength:
The speed of light equation:
c = λ × ν
where:
c = speed of light = 3 × 10⁸ m/s
λ = wavelength (m)
ν = frequency (Hz or s⁻¹)
Therefore:
λ = c / ν or ν = c / λ
The frequency and wavelength are inversely proportional — as one increases, the other decreases.
Problem: Calculate wavelength for ν = 5 × 10¹⁴ Hz
Step 1: Apply the formula.
λ = c / ν
λ = (3 × 10⁸ m/s) / (5 × 10¹⁴ Hz)
λ = 0.6 × 10⁻⁶ m
λ = 6 × 10⁻⁷ m
λ = 600 nm (nanometers)
Step 2: Identify the region.
This wavelength falls in the visible light region (red light).
Answer: The wavelength is 6 × 10⁻⁷ m or 600 nm, which corresponds to red light.
Q8. Define quantum number and describe the four quantum numbers used to describe the position and energy of an electron in an atom.
Quantum Number Definition:
Quantum numbers are a set of numbers that describe the position, energy, and angular momentum of an electron in an atom.
The Four Quantum Numbers:
1. Principal Quantum Number (n):
→ Describes the shell or energy level
→ Values: n = 1, 2, 3, 4, ... (positive integers)
→ Represents size and energy of the orbital
→ n = 1 (K shell), n = 2 (L shell), etc.
2. Azimuthal/Angular Momentum Quantum Number (l):
→ Describes the subshell or orbital shape
→ Values: l = 0, 1, 2, ..., (n-1)
→ l = 0 → s orbital (spherical)
→ l = 1 → p orbital (dumbbell-shaped)
→ l = 2 → d orbital (cloverleaf-shaped)
→ l = 3 → f orbital (complex shape)
3. Magnetic Quantum Number (ml):
→ Describes the orientation of the orbital in space
→ Values: ml = -l, -(l-1), ..., 0, ..., (l-1), +l
→ For p orbital (l=1): ml = -1, 0, +1 (three orientations)
→ For d orbital (l=2): ml = -2, -1, 0, +1, +2 (five orientations)
4. Spin Quantum Number (ms):
→ Describes the intrinsic angular momentum (spin) of the electron
→ Values: ms = +1/2 or -1/2
→ +1/2 represents spin up (↑)
→ -1/2 represents spin down (↓)
→ Maximum two electrons per orbital (with opposite spins)
Example: Electron in 2p orbital
n = 2, l = 1, ml can be -1, 0, or +1, ms = +1/2 or -1/2
Structure of Atom — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which experiment led to the discovery of the atomic nucleus?
- A. Thomson's cathode ray experiment
- B. Millikan's oil drop experiment
- C. Rutherford's alpha-scattering experiment
- D. Planck's black-body radiation experiment
Answer: C. Rutherford's alpha-scattering experiment — Rutherford's alpha-scattering experiment, observing the deflection of alpha particles by a thin gold foil, led to the conclusion that atoms have a small, dense, positively charged nucleus.
MCQ 2. According to Bohr's model, the angular momentum of an electron in a stationary orbit is quantized and given by which expression?
- A. nh/2π
- B. 2π/nh
- C. nλ
- D. mv
Answer: A. nh/2π — Bohr's second postulate states that the angular momentum of an electron in a stable orbit is an integral multiple of h/2π, where 'n' is the principal quantum number.
MCQ 3. What is the maximum number of electrons that can be accommodated in a subshell for which l = 3?
- A. 6
- B. 10
- C. 14
- D. 18
Answer: C. 14 — For l = 3 (f-subshell), the possible values for m_l are -3, -2, -1, 0, +1, +2, +3, meaning there are 7 orbitals. Each orbital can hold a maximum of 2 electrons, so 7 * 2 = 14 electrons.
MCQ 4. If the kinetic energy of an electron is increased four times, how does its de Broglie wavelength change?
- A. It becomes half
- B. It becomes twice
- C. It becomes four times
- D. It remains unchanged
Answer: A. It becomes half — The de Broglie wavelength is given by λ = h/√(2mKE). If KE becomes 4 times, then λ' = h/√(2m * 4KE) = h/(2√(2mKE)) = λ/2. So, the wavelength becomes half.
MCQ 5. No two electrons in an atom can have all four quantum numbers identical. This statement is known as:
- A. Hund's Rule
- B. Aufbau Principle
- C. Pauli Exclusion Principle
- D. Heisenberg Uncertainty Principle
Answer: C. Pauli Exclusion Principle — The Pauli Exclusion Principle states that no two electrons in the same atom can have identical values for all four quantum numbers (n, l, m_l, m_s).
MCQ 6. What is the correct electronic configuration of Chromium (Cr), atomic number 24?
- A. [Ar] 3d⁴ 4s²
- B. [Ar] 3d⁵ 4s¹
- C. [Ar] 3d³ 4s³
- D. [Ar] 3d⁶ 4s⁰
Answer: B. [Ar] 3d⁵ 4s¹ — Chromium exhibits an anomalous electronic configuration due to the extra stability associated with half-filled (3d⁵) and fully-filled (3d¹⁰) subshells. Therefore, it is [Ar] 3d⁵ 4s¹.
MCQ 7. In the photoelectric effect, if the frequency of the incident radiation is less than the threshold frequency (ν₀), what happens?
- A. Electrons are emitted with higher kinetic energy
- B. Electrons are emitted instantaneously
- C. No electrons are emitted
- D. More electrons are emitted
Answer: C. No electrons are emitted — The photoelectric effect only occurs when the energy of the incident photon (hν) is greater than or equal to the work function (hν₀) of the metal. If ν < ν₀, no electrons are emitted, regardless of the intensity.
MCQ 8. Which series in the hydrogen spectrum lies in the visible region?
- A. Lyman series
- B. Balmer series
- C. Paschen series
- D. Brackett series
Answer: B. Balmer series — The Balmer series corresponds to electron transitions from higher energy levels to the second energy level (n=2). The spectral lines of the Balmer series fall within the visible region of the electromagnetic spectrum.
MCQ 9. What is the shape of an 's' orbital?
- A. Dumbbell
- B. Spherical
- C. Double dumbbell
- D. Complex
Answer: B. Spherical — An 's' orbital is spherically symmetrical, meaning the probability of finding the electron is the same in all directions from the nucleus.
MCQ 10. For a principal quantum number n=3, what are the possible values for the azimuthal (angular momentum) quantum number l?
- A. 0, 1, 2
- B. 1, 2, 3
- C. 0, 1, 2, 3
- D. 0, 1
Answer: A. 0, 1, 2 — For a given principal quantum number 'n', the azimuthal quantum number 'l' can take values from 0 to (n-1). So, for n=3, l can be 0, 1, 2.
MCQ 11. How many radial nodes are present in a 3p orbital?
- A. 0
- B. 1
- C. 2
- D. 3
Answer: B. 1 — The number of radial nodes is given by the formula (n - l - 1). For a 3p orbital, n=3 and l=1. So, radial nodes = 3 - 1 - 1 = 1.
MCQ 12. Which of the following has the highest frequency?
- A. Radio waves
- B. Microwaves
- C. X-rays
- D. Visible light
Answer: C. X-rays — The electromagnetic spectrum arranges radiation based on frequency and wavelength. X-rays have a much higher frequency (and shorter wavelength) than radio waves, microwaves, and visible light.
MCQ 13. According to the Aufbau principle, electrons fill atomic orbitals in order of:
- A. Decreasing energy
- B. Increasing energy
- C. Increasing spin
- D. Decreasing stability
Answer: B. Increasing energy — The Aufbau principle states that electrons first occupy the orbitals with the lowest energy and then proceed to occupy orbitals with higher energy.
MCQ 14. If the uncertainty in the position of an electron is zero, what can be concluded about its momentum?
- A. Its momentum is also zero
- B. Its momentum is precisely known
- C. Its momentum is completely uncertain
- D. Its momentum is quantized
Answer: C. Its momentum is completely uncertain — According to Heisenberg's Uncertainty Principle, it is impossible to simultaneously determine with perfect accuracy both the position and momentum of a subatomic particle. If Δx = 0, then Δp must be infinitely large (completely uncertain).
MCQ 15. Atoms of the same element that have different mass numbers are called:
- A. Isobars
- B. Isotones
- C. Isotopes
- D. Isomers
Answer: C. Isotopes — Isotopes are atoms of the same element (same atomic number, Z) but with different numbers of neutrons, leading to different mass numbers (A).
MCQ 16. One of the major limitations of Bohr's model was its inability to explain:
- A. The stability of atoms
- B. The quantization of energy
- C. The spectra of multi-electron atoms
- D. The existence of discrete energy levels
Answer: C. The spectra of multi-electron atoms — While Bohr's model successfully explained the spectrum of hydrogen and hydrogen-like ions, it failed to explain the spectra of multi-electron atoms and the splitting of spectral lines in magnetic or electric fields.
MCQ 17. What is the transition in the hydrogen spectrum that emits a photon of the highest energy?
- A. n=2 to n=1
- B. n=3 to n=2
- C. n=infinity to n=1
- D. n=4 to n=1
Answer: C. n=infinity to n=1 — The energy of an emitted photon is greatest when the electron transitions from a higher energy level to a lower one with the largest energy difference. This occurs when an electron falls from n=infinity to n=1, corresponding to the ionization energy and the series limit of the Lyman series.
MCQ 18. Who discovered the electron?
- A. Ernest Rutherford
- B. J.J. Thomson
- C. James Chadwick
- D. Niels Bohr
Answer: B. J.J. Thomson — J.J. Thomson discovered the electron in 1897 through his experiments with cathode ray tubes.
MCQ 19. How many orbitals are associated with n=4 and l=1?
- A. 1
- B. 3
- C. 5
- D. 7
Answer: B. 3 — For l=1 (p-subshell), the magnetic quantum number m_l can take values of -1, 0, +1. Each distinct m_l value corresponds to a different orbital. Therefore, there are 3 orbitals (4px, 4py, 4pz).
MCQ 20. Which of the following species is isoelectronic with O²⁻?
- A. N³⁻
- B. F⁻
- C. Na⁺
- D. All of the above
Answer: D. All of the above — O²⁻ has 8+2=10 electrons. N³⁻ (7+3=10), F⁻ (9+1=10), and Na⁺ (11-1=10) all have 10 electrons, making them isoelectronic with O²⁻.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.