The chapter Class 11 Chemistry Chapter 2 Structure of Atom Get detailed question-answers and clear explanations from Class 11 Chemistry Chapter 2 – Structure of Atom. lays the foundation of atomic theory in modern chemistry.

Class 11 Chemistry Chapter 2 Structure of Atom

It covers the evolution of atomic models, the discovery of subatomic particles like electrons, protons, and neutrons, and explains their properties including mass and charge. This chapter introduces key concepts such as isotopes, isobars, orbitals, quantum numbers, and electronic configurations. It also explores the Bohr model, hydrogen spectra, and the concept of photon energy and ionization. Through numerical and conceptual questions, students learn to calculate wavelengths, frequencies, energies of electrons, and understand how quantum mechanics governs the arrangement of electrons in atoms. These problem-based solutions strengthen the understanding of atomic structure and are essential for further learning in physical and inorganic chemistry.

Class 11 Chemistry Chapter 2 Structure of Atom Textbook Answers

Q1. Calculate the number of electrons which will together weigh one gram.

Mass of 1 electron = 9.1 × 10⁻³¹ kg
1 g = 1 × 10⁻³ kg
Number of electrons = 1 × 10⁻³ ÷ 9.1 × 10⁻³¹ = 1.10 × 10²⁷ electrons

Q2. Calculate the mass and charge of one mole of electrons.

Mass of one electron = 9.1 × 10⁻³¹ kg
Mass of 1 mole electrons = 9.1 × 10⁻³¹ × 6.022 × 10²³ = 5.48 × 10⁻⁷ kg = 5.48 × 10⁻⁴ g
Charge of one electron = 1.6 × 10⁻¹⁹ C
Charge of 1 mole = 1.6 × 10⁻¹⁹ × 6.022 × 10²³ = 9.65 × 10⁴ C

Q3. The diameter of a carbon atom is 0.15 nm. Calculate how many such atoms can be placed side by side in a straight line 20 cm long.

Diameter = 0.15 nm = 0.15 × 10⁻⁹ m
Length = 20 cm = 0.2 m
Number of atoms = 0.2 ÷ 0.15 × 10⁻⁹ = 1.33 × 10⁹ atoms

Q4. 1 mole of atom weighs 2 g. What is its mass in amu?

1 mol = 6.022 × 10²³ atoms
Mass of 1 atom = 2 ÷ 6.022 × 10²³ = 3.32 × 10⁻²⁴ g
1 amu = 1.66 × 10⁻²⁴ g
Mass in amu = 3.32 × 10⁻²⁴ ÷ 1.66 × 10⁻²⁴ = 2 amu

Q5. An element has 2 electrons in K-shell and 1 in L-shell. Write its electronic configuration and find its atomic number.

Electronic configuration = 2, 1
Atomic number = 3 (Li)

Q6. The number of electrons, protons and neutrons in a species are 18, 16 and 16 respectively. Assign the proper symbol to the species.

Atomic number = 16 → element is S (Sulfur)
Mass number = 16 + 16 = 32
Charge = 18 – 16 = 2⁻
Symbol = ³²S²⁻

Q7. Write electronic configurations of any one pair of isotopes and isobars.

Isotopes: ¹H and ²H → Configuration: 1s¹
Isobars: ⁴⁰Ca and ⁴⁰Ar
Ca: 2, 8, 8, 2
Ar: 2, 8, 8

Q8. The wave number of radiation is 10⁷ m⁻¹. Calculate its wavelength.

Wave number = 1 ÷ wavelength
Wavelength = 1 ÷ 10⁷ = 10⁻⁷ m

Q9. What is the number of photons of light with wavelength 4000 pm that provide 1 J energy?

Wavelength = 4000 pm = 4 × 10⁻⁹ m
Energy of one photon = (6.626 × 10⁻³⁴ × 3 × 10⁸) ÷ 4 × 10⁻⁹ = 4.97 × 10⁻¹⁷ J
Number of photons = 1 ÷ 4.97 × 10⁻¹⁷ = 2.01 × 10¹⁶ photons

Q10. Light of frequency 1.5 × 10¹⁵ Hz is emitted. Calculate energy of the photon in eV.

Energy = h × frequency = 6.626 × 10⁻³⁴ × 1.5 × 10¹⁵ = 9.94 × 10⁻¹⁹ J
1 eV = 1.6 × 10⁻¹⁹ J
Energy in eV = 9.94 ÷ 1.6 = 6.21 eV

Q11. Which of the following are isotopes and which are isobars?

(i) ¹⁴C and ¹⁴N
(ii) ⁴⁰Ca and ⁴⁰Ar
(iii) ¹H, ²H, ³H

Answer:
(i) Isobars – same mass number (14), different atomic numbers (6 and 7)
(ii) Isobars – same mass number (40), different atomic numbers (20 and 18)
(iii) Isotopes – same atomic number (1), different mass numbers (1, 2, 3)

Q12. Write electronic configuration of:

(i) Cr (Z = 24)
(ii) Cu (Z = 29)

Answer:
(i) Cr: 1s2 2s2 2p6 3s2 3p6 3d5 4s1 (due to extra stability of half-filled d orbital)
(ii) Cu: 1s2 2s2 2p6 3s2 3p6 3d10 4s1 (due to extra stability of fully-filled d orbital)

Q13. Write electronic configuration of element with atomic number 35.

Answer:
Z = 35 (Bromine):
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5

Q14. Which of the following orbitals are not possible?

(i) 1p
(ii) 2s
(iii) 2p
(iv) 3f
(v) 3d

Answer:
Not possible: (i) 1p and (iv) 3f (because l must be less than n)
Possible: 2s, 2p, 3d

Q15. Find number of unpaired electrons in:

(i) P (Z = 15)
(ii) Cl (Z = 17)
(iii) Fe (Z = 26)
(iv) Kr (Z = 36)

Answer:
(i) P: 3 unpaired electrons (3p3)
(ii) Cl: 1 unpaired electron (3p5)
(iii) Fe: 4 unpaired electrons (3d6)
(iv) Kr: 0 unpaired (noble gas configuration)

Q16. Which has higher energy: 4s or 3d?

Answer:
4s is filled before 3d → lower energy initially
But after filling, 3d becomes more stable → lower in energy than 4s in many atoms.

Q17. What is the wavelength of a photon having energy 3.0 × 10⁻¹⁹ J?

Answer:
E = h × c ÷ λ
λ = h × c ÷ E = (6.626 × 10⁻³⁴ × 3 × 10⁸) ÷ 3.0 × 10⁻¹⁹
= 6.63 × 10⁻⁷ m = 663 nm

Q18. Calculate energy of 1 mole of photons of wavelength 400 nm.

Answer:
λ = 400 × 10⁻⁹ m
E (one photon) = h × c ÷ λ = 4.97 × 10⁻¹⁹ J
E (1 mole) = 4.97 × 10⁻¹⁹ × 6.022 × 10²³ = 2.99 × 10⁵ J/mol

Q19. Calculate wave number of light having wavelength 5000 Å.

Answer:
1 Å = 10⁻¹⁰ m → λ = 5000 × 10⁻¹⁰ m = 5 × 10⁻⁷ m
Wave number = 1 ÷ λ = 1 ÷ 5 × 10⁻⁷ = 2 × 10⁶ m⁻¹

Q20. What transition in hydrogen gives rise to Lyman series?

Answer:
Lyman series corresponds to transitions where final orbit is n = 1.
Examples: n = 2 to 1, n = 3 to 1, n = 4 to 1, etc.

Q21. How much energy is required to ionize a hydrogen atom if the electron is in n = 5 orbit?

Energy at nth level = -13.6 / n² eV
E = -13.6 / 25 = -0.544 eV
Ionization energy = 0 – (-0.544) = 0.544 eV

Q22. What is the frequency of radiation emitted when electron transitions from n = 3 to n = 2 in hydrogen atom?

Delta E = -13.6 (1/2² – 1/3²) = -13.6 (1/4 – 1/9) = -13.6 (5/36) = -1.89 eV
Energy in joules = 1.89 × 1.6 × 10⁻¹⁹ = 3.02 × 10⁻¹⁹ J
Frequency = E / h = 3.02 × 10⁻¹⁹ / 6.626 × 10⁻³⁴ = 4.56 × 10¹⁴ Hz

Q23. Write four quantum numbers for 19th electron of Cr (Z = 24).

Electronic configuration: 1s2 2s2 2p6 3s2 3p6 3d5 4s1
19th electron is in 3d
n = 3, l = 2, m = 0 (one of -2 to +2), s = +1/2
Answer: n = 3, l = 2, m = 0, s = +1/2

Q24. What is the maximum number of electrons in a subshell with l = 3?

l = 3 is f-subshell
Orbitals = 2l + 1 = 7
Electrons = 2 × 7 = 14 electrons

Q25. How many electrons can have quantum numbers n = 4, l = 2?

l = 2 is d-subshell → has 5 orbitals
Each orbital holds 2 electrons
Total = 5 × 2 = 10 electrons

Q26. Which quantum number determines the shape of orbital?

Answer: Azimuthal quantum number (l)

Q27. How many electrons can have quantum numbers n = 3 and m = +1?

Possible l values for n = 3: 0 (s), 1 (p), 2 (d)
m = +1 possible in l = 1 and l = 2 → 1 orbital each
Each orbital = 2 electrons
Total = 4 electrons

Q28. What are possible values of m for l = 2?

m = -l to +l = -2, -1, 0, +1, +2
Answer: -2, -1, 0, +1, +2

Q29. State Pauli Exclusion Principle.

Answer:
No two electrons in an atom can have the same set of four quantum numbers.

Q30. What is the difference between orbit and orbital?

Orbit: Circular fixed path of electron (Bohr’s model)
Orbital: Region in space with high probability of finding electron (Quantum model)

Q31. What are the values of n and l for 3p orbital?

n = 3 (principal quantum number)
l = 1 (for p orbital)

Q32. What is the maximum number of electrons in the principal energy level (shell) with n = 3?

Maximum electrons = 2n² = 2 × 3² = 18 electrons

Q33. How many orbitals are present in n = 2 shell?

l = 0 (s), l = 1 (p)
Orbitals in s = 1, orbitals in p = 3
Total = 1 + 3 = 4 orbitals

Q34. What is the maximum number of electrons that can be accommodated in all orbitals with n = 2?

n = 2 → total orbitals = 4 (from Q33)
Each orbital holds 2 electrons → 4 × 2 = 8 electrons

Q35. How many orbitals are possible for n = 3?

l = 0 (s): 1 orbital
l = 1 (p): 3 orbitals
l = 2 (d): 5 orbitals
Total = 1 + 3 + 5 = 9 orbitals

Q36. What is the number of radial and angular nodes for 3p orbital?

Radial nodes = n – l – 1 = 3 – 1 – 1 = 1
Angular nodes = l = 1

Q37. How many radial and angular nodes are present in 4d orbital?

Radial nodes = n – l – 1 = 4 – 2 – 1 = 1
Angular nodes = l = 2

Q38. Which has higher energy: 4p or 3d?

According to (n + l) rule:
4p → n = 4, l = 1 → n + l = 5
3d → n = 3, l = 2 → n + l = 5
Same value: compare n
Since 4 > 3 → 4p has higher energy

Q39. How many electrons in an atom can have quantum numbers n = 4 and l = 1?

l = 1 → p-orbital → 3 orbitals
Each orbital = 2 electrons
Total = 3 × 2 = 6 electrons

Q40. How many subshells are possible for n = 3?

Possible l values = 0 (s), 1 (p), 2 (d)
So, 3 subshells for n = 3

Q41. What is the energy of an electron in the third orbit of hydrogen atom?

E = -13.6 / n² = -13.6 / 9 = -1.51 eV

Q42. How many possible combinations of quantum numbers are there for electrons in 2p orbital?

n = 2, l = 1
m = -1, 0, +1
Each m can have s = +1/2 or -1/2
Total = 3 m values × 2 s values = 6 combinations

Q43. Write electronic configuration of Zn (Z = 30). How many unpaired electrons does it have?

Zn: 1s2 2s2 2p6 3s2 3p6 3d10 4s2
All orbitals fully filled → 0 unpaired electrons

Q44. How many orbitals are there in d-subshell?

l = 2 → m = -2, -1, 0, +1, +2 → 5 orbitals

Q45. Why 4s is filled before 3d?

Because 4s has lower (n + l) value (4 + 0 = 4)
3d has (3 + 2 = 5)
Lower (n + l) means lower energy → 4s filled before 3d

Q46. Define node and give types of nodes.

Node: A region where the probability of finding an electron is zero.
Types:

  • Radial nodes: depend on distance from nucleus
  • Angular nodes: depend on angle and shape of orbital

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