Structure of Atom - NCERT Solutions

Chapter Overview

The Structure of Atom is a fundamental chapter in chemistry that introduces the basic concepts of atomic structure, subatomic particles, and quantum mechanical model of the atom. This chapter forms the basis for understanding chemical bonding and periodic properties.

Key Concepts:

• Subatomic Particles: Electrons, protons, and neutrons
• Atomic Models: Thomson, Rutherford, and Bohr models
• Quantum Numbers: n, l, m₁, and mₛ
• Electronic Configuration: Distribution of electrons in orbitals
• Photoelectric Effect: Demonstration of particle nature of light

NCERT Exercise Solutions

Exercise 2.1

Question: (i) Calculate the number of electrons which will together weigh one gram. (ii) Calculate the mass and charge of one mole of electrons.

Solution: Given:

• Mass of one electron = 9.109 × 10⁻³¹ g
• Charge of one electron = 1.602 × 10⁻¹⁹ C

(i) Number of electrons in one gram: Number of electrons = Total mass / Mass of one electron = 1 g / (9.109 × 10⁻³¹ g) = 1.098 × 10³⁰ electrons

(ii) Mass and charge of one mole of electrons:

• Mass of one mole of electrons = 6.022 × 10²³ × 9.109 × 10⁻³¹ g = 5.486 × 10⁻⁴ g = 0.0005486 g

• Charge of one mole of electrons = 6.022 × 10²³ × 1.602 × 10⁻¹⁹ C = 9.647 × 10⁴ C = 96470 C

Answer:

• (i) 1.098 × 10³⁰ electrons weigh one gram
• (ii) One mole of electrons has mass 0.0005486 g and charge 96470 C

Exercise 2.2

Question: (i) Calculate the total number of electrons present in one mole of methane. (ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of ¹⁴C. (Assume that mass of a neutron = 1.675 × 10⁻²⁷ kg).

Solution: (i) Electrons in one mole of methane (CH₄):

• Carbon (C) has atomic number 6 → 6 electrons
• Hydrogen (H) has atomic number 1 → 1 electron each
• CH₄ has: 6 + (4 × 1) = 10 electrons per molecule
• One mole of CH₄ = 6.022 × 10²³ molecules
• Total electrons = 10 × 6.022 × 10²³ = 6.022 × 10²⁴ electrons

(ii) Neutrons in 7 mg of ¹⁴C:

• ¹⁴C has mass number 14 and atomic number 6
• Number of neutrons = 14 - 6 = 8 neutrons per atom

(a) Total number of neutrons: First, find number of atoms in 7 mg: Moles of ¹⁴C = 7 mg / (14 g/mol) = 0.007 g / 14 g/mol = 5 × 10⁻⁴ mol Number of atoms = 5 × 10⁻⁴ × 6.022 × 10²³ = 3.011 × 10²⁰ atoms Total neutrons = 8 × 3.011 × 10²⁰ = 2.409 × 10²¹ neutrons

(b) Total mass of neutrons: Mass = Number × Mass of one neutron = 2.409 × 10²¹ × 1.675 × 10⁻²⁷ kg = 4.034 × 10⁻⁶ kg = 4.034 mg

Answer:

• (i) 6.022 × 10²⁴ electrons in one mole of methane
• (ii) (a) 2.409 × 10²¹ neutrons, (b) 4.034 mg mass of neutrons

Exercise 2.3

Question: Find the wavelength of the radiation emitted when an electron in a hydrogen atom makes a transition from n=4 to n=2.

Solution: For hydrogen atom, the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$

Where:

• $R_H$ = Rydberg constant = 109677 cm⁻¹
• $n_1$ = 2 (final state)
• $n_2$ = 4 (initial state)

$\frac{1}{\lambda} = 109677 \left(\frac{1}{2^2} - \frac{1}{4^2}\right)$ $\frac{1}{\lambda} = 109677 \left(\frac{1}{4} - \frac{1}{16}\right)$ $\frac{1}{\lambda} = 109677 \left(\frac{4-1}{16}\right)$ $\frac{1}{\lambda} = 109677 \times \frac{3}{16}$ $\frac{1}{\lambda} = 20564$ cm⁻¹

Therefore, $\lambda = \frac{1}{20564}$ cm = 4.86 × 10⁻⁵ cm = 486 nm

Answer: The wavelength of emitted radiation is 486 nm (blue-green light).

NCERT Highlights for JEE/NEET

🎯 Important Concepts for JEE:

1. de Broglie Wavelength: $\lambda = \frac{h}{mv} = \frac{h}{p}$
2. Heisenberg Uncertainty Principle: $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$
3. Energy of Hydrogen Atom: $E_n = -13.6 \frac{Z^2}{n^2}$ eV
4. Rydberg Formula: $\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$

🧬 Key Points for NEET:

1. Quantum Numbers:

   • Principal quantum number (n): 1, 2, 3, …
   • Azimuthal quantum number (l): 0 to n-1
   • Magnetic quantum number (m₁): -l to +l
   • Spin quantum number (mₛ): +½ or -½

2. Electronic Configuration Rules:

   • Aufbau principle: n + l rule
   • Pauli exclusion principle: Maximum 2 electrons per orbital
   • Hund’s rule: Maximum multiplicity

JEE/NEET Practice Questions

Question 1 (JEE Level)

Statement: The kinetic energy of an electron in the second Bohr orbit of hydrogen atom is:

Solution: For hydrogen atom in Bohr model:

• Radius of nth orbit: $r_n = 0.529 n^2$ Å
• Velocity of electron in nth orbit: $v_n = \frac{2.18 \times 10^6}{n}$ m/s

For n = 2:

• $v_2 = \frac{2.18 \times 10^6}{2} = 1.09 \times 10^6$ m/s

Kinetic energy = $\frac{1}{2}mv^2$ $KE = \frac{1}{2} \times 9.109 \times 10^{-31} \times (1.09 \times 10^6)^2$ $KE = 5.41 \times 10^{-19}$ J $KE = \frac{5.41 \times 10^{-19}}{1.602 \times 10^{-19}}$ eV = 3.38 eV

Answer: 3.38 eV

Question 2 (NEET Level)

Statement: Which of the following sets of quantum numbers is not possible? (A) n = 1, l = 0, m₁ = 0, mₛ = +½ (B) n = 2, l = 1, m₁ = -1, mₛ = -½ (C) n = 3, l = 2, m₁ = 3, mₛ = +½ (D) n = 4, l = 0, m₁ = 0, mₛ = -½

Solution: Check each option against quantum number rules:

(A) n = 1, l = 0, m₁ = 0, mₛ = + ✓

• l = 0 to n-1 (0 to 0) ✓
• m₁ = -l to +l (0 to 0) ✓
• mₛ = ±½ ✓

(B) n = 2, l = 1, m₁ = -1, mₛ = -½ ✓

• l = 0 to n-1 (0 to 1) ✓
• m₁ = -l to +l (-1 to +1) ✓
• mₛ = ±½ ✓

(C) n = 3, l = 2, m₁ = 3, mₛ = +½ ✗

• l = 0 to n-1 (0 to 2) ✓
• m₁ = -l to +l (-2 to +2) ✗ (3 is not in range)

(D) n = 4, l = 0, m₁ = 0, mₛ = -½ ✓

• l = 0 to n-1 (0 to 3) ✓
• m₁ = -l to +l (0 to 0) ✓
• mₛ = ±½ ✓

Answer: Option (C) is not possible because m₁ = 3 is outside the range -2 to +2 for l = 2.

Important Derivations

de Broglie Wavelength

According to de Broglie’s hypothesis:

• Every moving particle exhibits wave-like character
• Wavelength associated with particle: $\lambda = \frac{h}{p}$
• Where p = momentum = mv

For electron in hydrogen atom (Bohr model):

• The electron wave must be stationary in the orbit
• Circumference must contain an integral number of wavelengths
• $2\pi r = n\lambda$

Substituting de Broglie wavelength: $2\pi r = n \times \frac{h}{mv}$ $mvr = \frac{nh}{2\pi}$

This is Bohr’s quantization condition of angular momentum!

Electronic Configuration Rules

1. Aufbau Principle (n + l rule)

• Electrons occupy orbitals in order of increasing (n + l) value
• For same (n + l) value, orbital with lower n is filled first
• Order: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p

2. Pauli Exclusion Principle

• No two electrons in an atom can have the same set of four quantum numbers
• Maximum of 2 electrons per orbital with opposite spins

3. Hund’s Rule of Maximum Multiplicity

• Electrons occupy degenerate orbitals singly before pairing
• All electrons in singly occupied orbitals have parallel spins

Important Formulas

Success Tips

1. Memorize Quantum Number Rules: Essential for electronic configurations
2. Practice Spectral Series: Lyman, Balmer, Paschen, etc.
3. Understand Photoelectric Effect: Threshold frequency and work function
4. Master de Broglie Concept: Connect wave and particle nature

Master Atomic Structure

Understanding atomic structure is crucial for chemical bonding and periodic properties. Focus on quantum mechanical concepts!

Remember: The structure of atom forms the foundation of modern chemistry. Master these concepts to excel in physical chemistry!

For additional practice, check out our Chemistry Practice Problems section.