NCERT Solutions for Class 11 Chemistry Chapter 2: Structure of Atom (NCERT 2026–27)

These Class 11 Chemistry Chapter 2 solutions cover Structure of Atom with every NCERT end-of-chapter exercise (2.1–2.67) reproduced verbatim and solved step by step. All numericals are worked with full units and cross-checked against the NCERT answer key, so you can see exactly how marks are earned in your CBSE exam. Topics include the discovery of sub-atomic particles, atomic models, Planck’s quantum theory, the photoelectric effect, Bohr’s model of hydrogen, de Broglie’s relation, Heisenberg’s uncertainty principle, quantum numbers and electronic configurations.

Class: 11 Subject: Chemistry Chapter: 2 Chapter Name: Structure of Atom Exercises: 2.1 – 2.67 Session: 2026–27
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Class 11 Chemistry Chapter 2 Solutions – Overview

Chapter 2, Structure of Atom, traces how the indivisible “atom” of Dalton was shown to be built from three fundamental particles — the electron (Thomson, Millikan), the proton and the neutron (Chadwick). It then follows the line of atomic models: Thomson’s plum-pudding model, Rutherford’s nuclear model and its drawbacks, and Bohr’s model of the hydrogen atom built on Planck’s quantum theory and the photoelectric effect. Finally the chapter introduces the quantum-mechanical model: de Broglie’s wave–particle duality, Heisenberg’s uncertainty principle, the four quantum numbers, the shapes of orbitals, and the rules (Aufbau principle, Pauli’s exclusion principle, Hund’s rule) used to write electronic configurations. The exercises are a rich mix of conceptual reasoning and numerical problems on energy, frequency, wavelength, spectral lines and quantum numbers.

Key Concepts & Definitions

Atomic number (Z): number of protons in the nucleus = number of electrons in a neutral atom.

Mass number (A): total number of nucleons, A = number of protons (Z) + number of neutrons.

Isotopes: atoms of the same element (same Z) with different mass numbers (different number of neutrons), e.g. 612C, 613C, 614C.

Isobars: atoms with the same mass number (A) but different atomic number (Z), e.g. 614C and 714N.

Isoelectronic species: atoms/ions having the same number of electrons.

Quantum: the smallest packet of energy that can be emitted or absorbed; E = hν.

Photoelectric effect: ejection of electrons from a metal surface when radiation of frequency ν > threshold frequency ν0 strikes it.

Quantum numbers: principal (n), azimuthal/orbital-angular-momentum (l), magnetic (ml) and spin (ms) numbers that completely describe an electron.

Aufbau principle: orbitals fill in order of increasing energy (lowest n + l first; if equal, lower n first).

Pauli’s exclusion principle: no two electrons in an atom can have the same set of four quantum numbers.

Hund’s rule: electrons occupy degenerate orbitals singly with parallel spins before pairing.

Important Formulas

Speed of light: c = νλ, where c = 3.0 × 108 m s−1

Wavenumber: ν̅ = 1/λ (units m−1 or cm−1)

Energy of a quantum: E = hν = hc/λ, where h = 6.626 × 10−34 J s

Photoelectric equation: hν = hν0 + ½mev2 (kinetic energy = hν − W0)

Bohr energy of H-like atom: En = −2.18 × 10−18 (Z2/n2) J atom−1

Bohr radius: rn = 52.9 × (n2/Z) pm = 0.529 (n2/Z) Å

Rydberg (spectral lines): ν̅ = 1.097 × 107 (1/n12 − 1/n22) m−1

de Broglie wavelength: λ = h/mv = h/p

Heisenberg uncertainty: Δx · Δp ≥ h/4π

Max. spectral lines (n → ground): number of lines = n(n − 1)/2

NCERT Exercise Solutions (2.1 – 2.67)

Questions are reproduced exactly as in the NCERT textbook. Constants used: h = 6.626 × 10−34 J s, c = 3.0 × 108 m s−1, e = 1.602 × 10−19 C, me = 9.11 × 10−31 kg, NA = 6.022 × 1023 mol−1.

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

ANSWER (i) Mass of one electron = 9.10939 × 10−31 kg = 9.10939 × 10−28 g. Number of electrons in 1 g = 1 / (9.10939 × 10−28) = 1.099 × 1027 electrons. (ii) Mass of 1 mole of electrons = 9.10939 × 10−31 kg × 6.022 × 1023 = 5.48 × 10−7 kg. Charge of 1 mole of electrons = 1.602 × 10−19 C × 6.022 × 1023 = 9.65 × 104 C (Faraday’s constant).

2.2 (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).(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at STP. Will the answer change if the temperature and pressure are changed?

ANSWER (i) One CH4 molecule has 6 + 4(1) = 10 electrons. In 1 mole = 10 × 6.022 × 1023 = 6.022 × 1024 electrons. (ii) 14 g of 14C contains NA atoms; neutrons per atom = 14 − 6 = 8. Moles in 7 mg = 7×10−3/14 = 5 × 10−4 mol. (a) Number of neutrons = 8 × 5 × 10−4 × 6.022 × 1023 = 2.4088 × 1021 neutrons. (b) Mass = 2.4088 × 1021 × 1.675 × 10−27 kg = 4.0347 × 10−6 kg. (iii) Molar mass of NH3 = 17 g; protons per molecule = 7 + 3(1) = 10. Moles in 34 mg = 34×10−3/17 = 2 × 10−3 mol. (a) Number of protons = 10 × 2 × 10−3 × 6.022 × 1023 = 1.2044 × 1022 protons. (b) Mass = 1.2044 × 1022 × 1.6726 × 10−27 kg = 2.015 × 10−5 kg. No, the answer does not change with temperature or pressure, since the number of protons depends only on the amount (mass) of substance.

2.3 How many neutrons and protons are there in the following nuclei?613C, 816O, 1224Mg, 2656Fe, 3888Sr

ANSWER Protons = Z; Neutrons = A − Z.
NucleusProtons (Z)Neutrons (A−Z)
613C67
816O88
1224Mg1212
2656Fe2630
3888Sr3850

2.4 Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A).(i) Z = 17, A = 35.  (ii) Z = 92, A = 233.  (iii) Z = 4, A = 9.

ANSWER (i) Z = 17 → chlorine: 1735Cl. (ii) Z = 92 → uranium: 92233U. (iii) Z = 4 → beryllium: 49Be.

2.5 Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wavenumber (ν̅) of the yellow light.

ANSWER λ = 580 nm = 580 × 10−9 m = 5.8 × 10−7 m. ν = c/λ = (3.0 × 108) / (5.8 × 10−7) = 5.17 × 1014 s−1. ν̅ = 1/λ = 1 / (5.8 × 10−7 m) = 1.72 × 106 m−1.

2.6 Find energy of each of the photons which(i) correspond to light of frequency 3 × 1015 Hz.  (ii) have wavelength of 0.50 Å.

ANSWER (i) E = hν = (6.626 × 10−34 J s)(3 × 1015 s−1) = 1.988 × 10−18 J. (ii) λ = 0.50 Å = 0.50 × 10−10 m = 5.0 × 10−11 m. E = hc/λ = (6.626 × 10−34)(3.0 × 108) / (5.0 × 10−11) = 3.98 × 10−15 J.

2.7 Calculate the wavelength, frequency and wavenumber of a light wave whose period is 2.0 × 10−10 s.

ANSWER Frequency ν = 1/period = 1 / (2.0 × 10−10 s) = 5.0 × 109 s−1. λ = c/ν = (3.0 × 108) / (5.0 × 109) = 6.0 × 10−2 m. ν̅ = 1/λ = 1 / (6.0 × 10−2 m) = 16.66 m−1.

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

ANSWER λ = 4000 pm = 4000 × 10−12 m = 4.0 × 10−9 m. Energy of one photon E = hc/λ = (6.626 × 10−34)(3.0 × 108) / (4.0 × 10−9) = 4.969 × 10−17 J. Number of photons = 1 J / (4.969 × 10−17 J) = 2.012 × 1016 photons.

2.9 A photon of wavelength 4 × 10−7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV = 1.6020 × 10−19 J).

ANSWER (i) E = hc/λ = (6.626 × 10−34)(3.0 × 108) / (4 × 10−7) = 4.97 × 10−19 J. In eV: E = (4.97 × 10−19) / (1.602 × 10−19) = 3.10 eV. (ii) K.E. = E − W0 = 3.10 − 2.13 = 0.97 eV (= 1.554 × 10−19 J). (iii) ½mev2 = 1.554 × 10−19 J. So v = √(2 × 1.554 × 10−19 / 9.11 × 10−31) = √(3.412 × 1011) = 5.84 × 105 m s−1.

2.10 Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol−1.

ANSWER λ = 242 nm = 2.42 × 10−7 m. Energy per atom E = hc/λ = (6.626 × 10−34)(3.0 × 108) / (2.42 × 10−7) = 8.21 × 10−19 J. Ionisation energy per mole = 8.21 × 10−19 × 6.022 × 1023 = 4.945 × 105 J mol−1 = 494 kJ mol−1.

2.11 A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 µm. Calculate the rate of emission of quanta per second.

ANSWER λ = 0.57 µm = 5.7 × 10−7 m. Power = 25 J s−1. Energy of one photon = hc/λ = (6.626 × 10−34)(3.0 × 108) / (5.7 × 10−7) = 3.487 × 10−19 J. Rate = 25 / (3.487 × 10−19) = 7.18 × 1019 s−1.

2.12 Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν0) and work function (W0) of the metal.

ANSWER At zero velocity, the incident frequency equals the threshold frequency. λ0 = 6800 Å = 6.8 × 10−7 m. ν0 = c/λ0 = (3.0 × 108) / (6.8 × 10−7) = 4.41 × 1014 s−1. W0 = hν0 = (6.626 × 10−34)(4.41 × 1014) = 2.92 × 10−19 J (≈ 1.82 eV).

2.13 What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

ANSWER ν̅ = 1.097 × 107 (1/22 − 1/42) = 1.097 × 107 (0.25 − 0.0625) = 1.097 × 107 × 0.1875 = 2.057 × 106 m−1. λ = 1/ν̅ = 1 / (2.057 × 106) = 4.86 × 10−7 m = 486 nm (Balmer series, visible blue-green).

2.14 How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n = 1 orbit).

ANSWER En = −2.18 × 10−18 / n2 J. Ionisation means raising the electron to n = ∞ (E = 0). From n = 5: ΔE = 0 − (−2.18 × 10−18 / 25) = 8.72 × 10−20 J. From n = 1: ΔE = 2.18 × 10−18 J. So ionisation from n = 5 needs only 1/25 (= 4%) of the energy needed from the ground state — the n = 5 electron is far more loosely bound.

2.15 What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?

ANSWER Number of spectral lines = n(n − 1)/2 = 6 × 5 / 2 = 15 emission lines.

2.16 (i) The energy associated with the first orbit in the hydrogen atom is −2.18 × 10−18 J atom−1. What is the energy associated with the fifth orbit?(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

ANSWER (i) E5 = −2.18 × 10−18 / 52 = −2.18 × 10−18 / 25 = −8.72 × 10−20 J. (ii) rn = 52.9 × n2 pm = 52.9 × 25 = 1322.5 pm = 1.3225 nm.

2.17 Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

ANSWER Longest wavelength = smallest energy gap = transition n = 3 → n = 2. ν̅ = 1.097 × 107 (1/22 − 1/32) = 1.097 × 107 (0.25 − 0.1111) = 1.097 × 107 × 0.1389 = 1.523 × 106 m−1.

2.18 What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is −2.18 × 10−11 ergs.

ANSWER E1 = −2.18 × 10−11 erg = −2.18 × 10−18 J (1 erg = 10−7 J). ΔE = E5 − E1 = (−2.18 × 10−18/25) − (−2.18 × 10−18) = 2.18 × 10−18 (1 − 1/25) = 2.18 × 10−18 × 0.96 = 2.0928 × 10−18 J ≈ 2.08 × 10−11 ergs. When the electron returns (n = 5 → 1), the same energy is emitted. λ = hc/ΔE = (6.626 × 10−34)(3.0 × 108) / (2.0928 × 10−18) = 9.498 × 10−8 m = 950 Å.

2.19 The electron energy in hydrogen atom is given by En = (−2.18 × 10−18)/n2 J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

ANSWER ΔE = E∞ − E2 = 0 − (−2.18 × 10−18/4) = 5.45 × 10−19 J. Longest wavelength corresponds to this minimum energy: λ = hc/ΔE = (6.626 × 10−34)(3.0 × 108) / (5.45 × 10−19) = 3.647 × 10−7 m = 3.647 × 10−5 cm (≈ 3647 Å).

2.20 Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 m s−1.

ANSWER λ = h/mev = (6.626 × 10−34) / [(9.11 × 10−31)(2.05 × 107)] = (6.626 × 10−34) / (1.868 × 10−23) = 3.55 × 10−11 m.

2.21 The mass of an electron is 9.1 × 10−31 kg. If its K.E. is 3.0 × 10−25 J, calculate its wavelength.

ANSWER K.E. = ½mv2 ⇒ v = √(2·K.E./m) = √(2 × 3.0 × 10−25 / 9.1 × 10−31) = √(6.593 × 105) = 811.9 m s−1. λ = h/mv = (6.626 × 10−34) / [(9.1 × 10−31)(811.9)] = (6.626 × 10−34) / (7.388 × 10−28) = 8.967 × 10−7 m = 8967 Å.

2.22 Which of the following are isoelectronic species i.e., those having the same number of electrons?Na+, K+, Mg2+, Ca2+, S2−, Ar.

ANSWER Count electrons: Na+ = 10, K+ = 18, Mg2+ = 10, Ca2+ = 18, S2− = 18, Ar = 18. Two isoelectronic groups: (i) Na+ and Mg2+ (10 electrons each); (ii) K+, Ca2+, S2− and Ar (18 electrons each).

2.23 (i) Write the electronic configurations of the following ions: (a) H− (b) Na+ (c) O2− (d) F−(ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s1 (b) 2p3 and (c) 3p5?(iii) Which atoms are indicated by the following configurations? (a) [He] 2s1 (b) [Ne] 3s2 3p3 (c) [Ar] 4s2 3d1.

ANSWER (i) (a) H− (2 e−): 1s2. (b) Na+ (10 e−): 1s22s22p6. (c) O2− (10 e−): 1s22s22p6. (d) F− (10 e−): 1s22s22p6. (ii) (a) 3s1 → [Ne]3s1 = Na, Z = 11. (b) 2p3 → 1s22s22p3 = N, Z = 7. (c) 3p5 → [Ne]3s23p5 = Cl, Z = 17. (iii) (a) [He]2s1 = Li (Z = 3). (b) [Ne]3s23p3 = P (Z = 15). (c) [Ar]4s23d1 = Sc (Z = 21).

2.24 What is the lowest value of n that allows g orbitals to exist?

ANSWER g orbitals have l = 4. Since l can be 0 to (n − 1), we need n − 1 ≥ 4, i.e. n ≥ 5. The lowest value is n = 5.

2.25 An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.

ANSWER For 3d: n = 3, l = 2, and ml = −2, −1, 0, +1, +2 (any one of these values for a given electron).

2.26 An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.

ANSWER (i) For a neutral atom, protons = electrons = 29 (this is copper, Cu). (ii) Electronic configuration: 1s2 2s2 2p6 3s2 3p6 3d10 4s1, i.e. [Ar] 3d10 4s1 (the exceptional, more stable fully-filled-3d configuration).

2.27 Give the number of electrons in the species H2+, H2 and O2+.

ANSWER H2+: (1 + 1) − 1 = 1 electron. H2: 1 + 1 = 2 electrons. O2+: (8 + 8) − 1 = 15 electrons.

2.28 (i) An atomic orbital has n = 3. What are the possible values of l and ml?(ii) List the quantum numbers (ml and l) of electrons for 3d orbital.(iii) Which of the following orbitals are possible? 1p, 2s, 2p and 3f

ANSWER (i) For n = 3, l = 0, 1, 2. ml values: for l = 0 → 0; for l = 1 → −1, 0, +1; for l = 2 → −2, −1, 0, +1, +2. (ii) For 3d, l = 2 and ml = −2, −1, 0, +1, +2. (iii) Possible: 2s and 2p. Not possible: 1p (for n = 1, only l = 0 i.e. s is allowed) and 3f (for n = 3, l can be at most 2, so f needs n ≥ 4).

2.29 Using s, p, d notations, describe the orbital with the following quantum numbers. (a) n = 1, l = 0; (b) n = 3, l = 1; (c) n = 4, l = 2; (d) n = 4, l = 3.

ANSWER (a) 1s; (b) 3p; (c) 4d; (d) 4f.

2.30 Explain, giving reasons, which of the following sets of quantum numbers are not possible.(a) n = 0, l = 0, ml = 0, ms = +½  (b) n = 1, l = 0, ml = 0, ms = −½  (c) n = 1, l = 1, ml = 0, ms = +½(d) n = 2, l = 1, ml = 0, ms = −½  (e) n = 3, l = 3, ml = −3, ms = +½  (f) n = 3, l = 1, ml = 0, ms = +½

ANSWER (a) Not possible — n cannot be 0 (the principal quantum number must be ≥ 1). (b) Possible (1s). (c) Not possible — for n = 1, l can only be 0, so l = 1 is not allowed. (d) Possible (2p). (e) Not possible — for n = 3, l can be 0, 1, 2 only, so l = 3 is not allowed. (f) Possible (3p). Hence (a), (c) and (e) are not possible.

2.31 How many electrons in an atom may have the following quantum numbers? (a) n = 4, ms = −½  (b) n = 3, l = 0

ANSWER (a) Total electrons in n = 4 shell = 2n2 = 32; half have ms = −½ → 16 electrons. (b) n = 3, l = 0 is the 3s orbital, which holds 2 electrons.

2.32 Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

ANSWER Bohr’s quantisation of angular momentum: mvr = nh/2π ⇒ 2πr = nh/mv. de Broglie wavelength: λ = h/mv. Substituting: 2πr = nλ. Thus the circumference (2πr) of the nth Bohr orbit equals n times the de Broglie wavelength, i.e. only orbits that accommodate a whole number of electron waves (standing waves) are allowed. Hence proved.

2.33 What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum?

ANSWER For He+ (Z = 2): ν̅ = R·Z2(1/22 − 1/42) = R × 4 × (0.25 − 0.0625) = R × 4 × 0.1875 = 0.75 R. For H (Z = 1) we need 1/n12 − 1/n22 = 0.75 = 1/12 − 1/22 (= 1 − 0.25). So the matching transition in hydrogen is n = 2 → n = 1 (a Lyman-series line).

2.34 Calculate the energy required for the process He+ (g) → He2+ (g) + e−. The ionization energy for the H atom in the ground state is 2.18 × 10−18 J atom−1.

ANSWER Energy of a one-electron species in the ground state ∝ Z2. For He+, Z = 2. E = (2.18 × 10−18 J) × Z2/n2 = 2.18 × 10−18 × 4/1 = 8.72 × 10−18 J per atom.

2.35 If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.

ANSWER Diameter = 0.15 nm = 0.15 × 10−9 m = 1.5 × 10−10 m. Length = 20 cm = 0.2 m = 2 × 10−1 m. Number of atoms = (2 × 10−1) / (1.5 × 10−10) = 1.33 × 109 atoms.

2.36 2 × 108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.

ANSWER Length per atom (diameter) = 2.4 cm / (2 × 108) = 1.2 × 10−8 cm. Radius = diameter/2 = 0.6 × 10−8 cm = 6 × 10−9 cm = 6 × 10−11 m = 0.06 nm (= 60 pm).

2.37 The diameter of zinc atom is 2.6 Å. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.

ANSWER (a) Radius = 2.6/2 = 1.3 Å = 1.3 × 10−10 m = 1.3 × 102 pm (= 130 pm). (b) Diameter = 2.6 Å = 2.6 × 10−8 cm. Number of atoms = 1.6 cm / (2.6 × 10−8 cm) = 6.15 × 107 atoms.

2.38 A certain particle carries 2.5 × 10−16 C of static electric charge. Calculate the number of electrons present in it.

ANSWER Number of electrons = total charge / charge on one electron = (2.5 × 10−16) / (1.602 × 10−19) = 1560 electrons.

2.39 In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is −1.282 × 10−18 C, calculate the number of electrons present on it.

ANSWER Number of electrons = (1.282 × 10−18) / (1.602 × 10−19) = 8 electrons.

2.40 In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?

ANSWER Lighter atoms have nuclei with a smaller positive charge and smaller size. Therefore the repulsion experienced by the α-particles is weaker. As a result, more α-particles would pass straight through the foil undeflected, and fewer would be deflected (and far fewer bounced back), compared with a heavy-atom (gold) foil.

2.41 Symbols 3579Br and 79Br can be written, whereas symbols 7935Br and 35Br are not acceptable. Answer briefly.

ANSWER For a given element the atomic number (Z) is fixed, so writing only the mass number (79Br) is enough and acceptable. Writing both as 3579Br is also correct. However, writing only the atomic number (35Br) is meaningless because Z is already implied by the symbol Br and does not identify the particular isotope — the mass number is the information that varies. Hence 35Br (and a misplaced superscript) is not acceptable.

2.42 An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.

ANSWER Let protons = p, then neutrons = p + 0.317p = 1.317p. Also p + 1.317p = 81 (mass number). 2.317p = 81 ⇒ p = 34.96 ≈ 35. So Z = 35 (bromine), neutrons = 81 − 35 = 46. Symbol: 3581Br.

2.43 An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.

ANSWER Let electrons = e. The ion has one extra electron (1− charge), so protons = e − 1. Neutrons = e + 0.111e = 1.111e. Mass number = protons + neutrons = (e − 1) + 1.111e = 37 ⇒ 2.111e = 38 ⇒ e = 18. So electrons = 18, protons (Z) = 17, neutrons = 1.111 × 18 = 20. Z = 17 is chlorine. Symbol: 1737Cl−.

2.44 An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.

ANSWER Let electrons = e. Ion has 3+ charge, so protons = e + 3. Neutrons = e + 0.304e = 1.304e. Mass number = protons + neutrons = (e + 3) + 1.304e = 56 ⇒ 2.304e = 53 ⇒ e = 23. So electrons = 23, protons (Z) = 26, neutrons = 1.304 × 23 = 30. Z = 26 is iron. Symbol: 2656Fe3+.

2.45 Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.

ANSWER Frequency increases as wavelength decreases. Increasing order of frequency: FM radio < microwave < amber (visible) light < X-rays < cosmic rays.

2.46 Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 × 1024, calculate the power of this laser.

ANSWER Energy of one photon = hc/λ = (6.626 × 10−34)(3.0 × 108) / (337.1 × 10−9) = 5.896 × 10−19 J. Total energy (power, since photons given per emission) = 5.896 × 10−19 × 5.6 × 1024 = 3.3 × 106 J.

2.47 Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance traveled by this radiation in 30 s, (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.

ANSWER (a) ν = c/λ = (3.0 × 108) / (616 × 10−9) = 4.87 × 1014 s−1. (b) Distance = c × t = (3.0 × 108)(30) = 9.0 × 109 m. (c) E = hν = (6.626 × 10−34)(4.87 × 1014) = 3.227 × 10−19 J (= 32.27 × 10−20 J). (d) Number of quanta = 2 J / (3.227 × 10−19 J) = 6.2 × 1018 quanta.

2.48 In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 × 10−18 J from the radiations of 600 nm, calculate the number of photons received by the detector.

ANSWER Energy of one photon = hc/λ = (6.626 × 10−34)(3.0 × 108) / (600 × 10−9) = 3.313 × 10−19 J. Number of photons = (3.15 × 10−18) / (3.313 × 10−19) = 10 photons (approx.).

2.49 Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 × 1015, calculate the energy of the source.

ANSWER Frequency ν = 1/duration = 1 / (2 × 10−9 s) = 5 × 108 s−1. Energy of one photon = hν = (6.626 × 10−34)(5 × 108) = 3.313 × 10−25 J. Energy of source = 3.313 × 10−25 × 2.5 × 1015 = 8.28 × 10−10 J.

2.50 The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calcualte the frequency of each transition and energy difference between two excited states.

ANSWER ν1 = c/λ1 = (3.0 × 108) / (589 × 10−9) = 5.093 × 1014 s−1. ν2 = c/λ2 = (3.0 × 108) / (589.6 × 10−9) = 5.088 × 1014 s−1. ΔE = h(ν1 − ν2) = (6.626 × 10−34)(5.093 × 1014 − 5.088 × 1014) = (6.626 × 10−34)(5.2 × 1011) = 3.45 × 10−22 J.

2.51 The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

ANSWER W0 = 1.9 eV = 1.9 × 1.602 × 10−19 = 3.044 × 10−19 J. (a) λ0 = hc/W0 = (6.626 × 10−34)(3.0 × 108) / (3.044 × 10−19) = 6.53 × 10−7 m = 652.46 nm. (b) ν0 = c/λ0 = (3.0 × 108) / (6.5246 × 10−7) = 4.598 × 1014 s−1. (c) Energy at 500 nm = hc/λ = (6.626 × 10−34)(3.0 × 108) / (500 × 10−9) = 3.976 × 10−19 J. K.E. = 3.976 × 10−19 − 3.044 × 10−19 = 9.29 × 10−20 J. v = √(2·K.E./m) = √(2 × 9.29 × 10−20 / 9.11 × 10−31) = √(2.039 × 1011) = 4.516 × 105 m s−1.

2.52 Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.

ANSWER
λ (nm)500450400
v × 10−5 (cm s−1)2.554.355.35
Using ½mv2 = hc(1/λ − 1/λ0), v2 is linear in (1/λ). Taking the data for 500 nm and 400 nm (v in m s−1): v1 = 2.55 × 103, v3 = 5.35 × 103. (a) Threshold wavelength: solving the simultaneous equations gives λ0 ≈ 531 nm. (b) From the slope of v2 versus (1/λ), ½m × (slope) = hc gives h ≈ 6.6 × 10−34 J s, in good agreement with the accepted value.

2.53 The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.

ANSWER Energy of incident photon = hc/λ = (6.626 × 10−34)(3.0 × 108) / (256.7 × 10−9) = 7.74 × 10−19 J = 4.83 eV. Maximum K.E. = eV0 = 0.35 eV. Work function W0 = E − K.E. = 4.83 − 0.35 = 4.48 eV.

2.54 If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 × 107 m s−1, calculate the energy with which it is bound to the nucleus.

ANSWER Energy of photon = hc/λ = (6.626 × 10−34)(3.0 × 108) / (150 × 10−12) = 1.325 × 10−15 J = 8.27 × 103 eV. K.E. of electron = ½mv2 = ½(9.11 × 10−31)(1.5 × 107)2 = 1.025 × 10−16 J = 0.64 × 103 eV. Binding energy = Ephoton − K.E. = (8.27 − 0.64) × 103 = 7.63 × 103 eV ≈ 7.6 × 103 eV.

2.55 Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as ν = 3.29 × 1015 (Hz) [1/32 − 1/n2]. Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

ANSWER ν = c/λ = (3.0 × 108) / (1285 × 10−9) = 2.335 × 1014 s−1. 2.335 × 1014 = 3.29 × 1015 (1/9 − 1/n2) ⇒ (1/9 − 1/n2) = 0.07097. 1/n2 = 0.1111 − 0.07097 = 0.04014 ⇒ n2 = 24.9 ⇒ n = 5. The Paschen series lies in the infrared region.

2.56 Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

ANSWER Radius rn = 52.9 n2 pm. For r = 1.3225 nm = 1322.5 pm: n2 = 1322.5/52.9 = 25 ⇒ n = 5. For r = 211.6 pm: n2 = 211.6/52.9 = 4 ⇒ n = 2. Transition: n = 5 → n = 2 (Balmer series). ν̅ = 1.097 × 107 (1/4 − 1/25) = 1.097 × 107 × 0.21 = 2.304 × 106 m−1. λ = 1/ν̅ = 4.34 × 10−7 m = 434 nm — Balmer series, visible region.

2.57 Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 × 106 m s−1, calculate de Broglie wavelength associated with this electron.

ANSWER λ = h/mev = (6.626 × 10−34) / [(9.11 × 10−31)(1.6 × 106)] = (6.626 × 10−34) / (1.458 × 10−24) = 4.55 × 10−10 m = 455 pm.

2.58 Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.

ANSWER Mass of neutron m = 1.675 × 10−27 kg, λ = 800 pm = 8.0 × 10−10 m. v = h/mλ = (6.626 × 10−34) / [(1.675 × 10−27)(8.0 × 10−10)] = (6.626 × 10−34) / (1.34 × 10−36) = 494.5 m s−1.

2.59 If the velocity of the electron in Bohr’s first orbit is 2.19 × 106 m s−1, calculate the de Broglie wavelength associated with it.

ANSWER λ = h/mev = (6.626 × 10−34) / [(9.11 × 10−31)(2.19 × 106)] = (6.626 × 10−34) / (1.995 × 10−24) = 3.32 × 10−10 m = 332 pm.

2.60 The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 105 m s−1. If the hockey ball of mass 0.1 kg is moving with this velocity, calcualte the wavelength associated with this velocity.

ANSWER λ = h/mv = (6.626 × 10−34) / [(0.1)(4.37 × 105)] = (6.626 × 10−34) / (4.37 × 104) = 1.516 × 10−38 m. (So small it is undetectable — macroscopic objects show no wave nature.)

2.61 If the position of the electron is measured within an accuracy of ± 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/4πm × 0.05 nm, is there any problem in defining this value.

ANSWER Δx = 0.002 nm = 2 × 10−12 m. By Heisenberg: Δp ≥ h/(4πΔx) = (6.626 × 10−34) / (4 × 3.14 × 2 × 10−12) = 2.638 × 10−23 kg m s−1. Actual momentum = h/(4πm × 0.05 nm) order ≈ (6.626 × 10−34)/(4π × 9.11 × 10−31 × 5 × 10−11) × m… ⇒ the stated momentum (≈ 1.057 × 10−24 kg m s−1) is smaller than the uncertainty. So the value cannot be defined — the actual magnitude is smaller than the uncertainty in it.

2.62 The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:1. n = 4, l = 2, ml = −2, ms = −1/2  2. n = 3, l = 2, ml = 1, ms = +1/2  3. n = 4, l = 1, ml = 0, ms = +1/24. n = 3, l = 2, ml = −2, ms = −1/2  5. n = 3, l = 1, ml = −1, ms = +1/2  6. n = 4, l = 1, ml = 0, ms = +1/2

ANSWER Energy ranked by (n + l) value (lower = lower energy; ties broken by lower n):
ElectronOrbitaln + l
14d6
23d5
34p5
43d5
53p4
64p5
Order: 5 < 2 = 4 < 3 = 6 < 1. (Electrons 2 and 4 are both 3d → equal energy; 3 and 6 are both 4p → equal energy.)

2.63 The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective nuclear charge?

ANSWER The farther an electron is from the nucleus (higher n), the more it is shielded by inner electrons, so it experiences less effective nuclear charge. Among 2p, 3p and 4p, the 4p electrons (the outermost) experience the lowest effective nuclear charge.

2.64 Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p.

ANSWER (i) 2s — it is closer to the nucleus (lower n) and less shielded. (ii) 4d — d penetrates more than f, so it is less shielded and feels a larger Zeff. (iii) 3p — p orbitals penetrate closer to the nucleus than d orbitals, so 3p feels a larger Zeff.

2.65 The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?

ANSWER Silicon (Z = 14) has one more proton than aluminium (Z = 13) but the added electron goes into the same 3p sub-shell, giving little extra shielding. So the 3p electrons of silicon experience the greater effective nuclear charge.

2.66 Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.

ANSWER (a) P [Ne]3s23p3 → 3 unpaired. (b) Si [Ne]3s23p2 → 2 unpaired. (c) Cr [Ar]3d54s1 → 6 unpaired. (d) Fe [Ar]3d64s2 → 4 unpaired. (e) Kr [Ar]3d104s24p6 (fully filled) → zero unpaired.

2.67 (a) How many subshells are associated with n = 4? (b) How many electrons will be present in the subshells having ms value of −1/2 for n = 4?

ANSWER (a) For n = 4, l = 0, 1, 2, 3 → 4 subshells (4s, 4p, 4d, 4f). (b) Total electrons in n = 4 = 2n2 = 32. Half have ms = −1/2 → 16 electrons.

Extra Practice Questions

Short Answer Type Questions

Q1. State the four quantum numbers and what each describes.

ANSWERPrincipal (n) — size and energy of the shell; azimuthal (l) — shape of the subshell; magnetic (ml) — orientation of the orbital in space; spin (ms) — spin direction of the electron (+½ or −½).

Q2. Why could Rutherford’s model not explain the stability of the atom?

ANSWERAn electron revolving in an orbit is constantly accelerating, so by Maxwell’s theory it must continuously radiate energy, spiral inward and collapse into the nucleus in about 10−8 s. Since atoms are stable, the model fails.

Q3. Define threshold frequency and work function.

ANSWERThreshold frequency (ν0) is the minimum frequency of incident light needed to eject an electron from a metal. Work function (W0 = hν0) is the minimum energy needed to remove an electron from the metal surface.

Q4. Calculate the energy of one mole of photons of radiation of frequency 5 × 1014 Hz.

ANSWERE (per photon) = hν = (6.626 × 10−34)(5 × 1014) = 3.313 × 10−19 J. Per mole = 3.313 × 10−19 × 6.022 × 1023 = 199.5 kJ mol−1.

Q5. Why is the electronic configuration of chromium [Ar]3d54s1 and not [Ar]3d44s2?

ANSWERHalf-filled (3d5) and fully-filled subshells have extra stability due to symmetrical electron distribution and maximum exchange energy. So one 4s electron shifts to 3d, giving the more stable [Ar]3d54s1.

Long Answer Type Questions

Q1. State the postulates of Bohr’s model of the hydrogen atom and one limitation.

ANSWER(1) The electron revolves around the nucleus only in certain fixed circular orbits (stationary states) of definite energy. (2) Energy is absorbed or emitted only when the electron jumps between orbits, with ΔE = hν. (3) The angular momentum of the electron is quantised: mvr = nh/2π, where n = 1, 2, 3…. These explain the line spectrum of hydrogen and the stability of the atom. Limitation: it cannot explain the spectra of multi-electron atoms, the splitting of spectral lines (Zeeman/Stark effects), and it violates the Heisenberg uncertainty principle by fixing both the position and momentum of the electron.

Q2. Explain the photoelectric effect and how Einstein’s explanation supports the particle nature of light.

ANSWERWhen light of suitable frequency strikes a metal, electrons are ejected instantly. Three observations — no time lag, a characteristic threshold frequency, and kinetic energy rising with frequency (not intensity) — cannot be explained by the wave theory. Einstein proposed that light consists of photons of energy E = hν. A photon transfers all its energy to one electron: part (W0) frees the electron and the rest becomes kinetic energy, hν = W0 + ½mev2. This explains every observation and confirms light’s particle (quantised) nature.

Q3. Compare Thomson’s and Rutherford’s atomic models and explain how the α-scattering experiment overturned the plum-pudding model.

ANSWERThomson pictured the atom as a uniform sphere of positive charge with electrons embedded in it (plum-pudding), so mass and charge were spread evenly. Rutherford bombarded a thin gold foil with α-particles: most passed straight through, a few were deflected, and about 1 in 20,000 bounced back nearly 180°. A uniform charge distribution could never repel particles that strongly. Rutherford concluded that the positive charge and almost all the mass are concentrated in a tiny central nucleus (radius ≈ 10−15 m) while electrons revolve in the largely empty space around it. This nuclear model replaced Thomson’s, though it could not yet explain atomic stability or spectra.

MCQs & Assertion–Reason

1. The charge to mass ratio (e/me) of the electron was determined by:

(a) Millikan    (b) J.J. Thomson    (c) Rutherford    (d) Chadwick

2. The number of neutrons in 2656Fe is:

(a) 26    (b) 56    (c) 30    (d) 82

3. Which set of species is isoelectronic?

(a) Na+, Mg2+    (b) Na+, K+    (c) Ca2+, Mg2+    (d) Ar, Na+

4. The maximum number of emission lines for an electron falling from n = 6 to the ground state is:

(a) 6    (b) 10    (c) 15    (d) 21

5. The lowest value of n that allows g orbitals is:

(a) 3    (b) 4    (c) 5    (d) 6

6. The de Broglie wavelength is given by:

(a) λ = hν    (b) λ = mv/h    (c) λ = h/mv    (d) λ = hc/E

7. The number of unpaired electrons in chromium (Z = 24) is:

(a) 4    (b) 5    (c) 6    (d) 1

8. Which quantum number determines the shape of an orbital?

(a) n    (b) l    (c) ml    (d) ms

9. The Balmer series of hydrogen lies in which region of the spectrum?

(a) ultraviolet    (b) visible    (c) infrared    (d) microwave

10. The energy of an electron in the n = 1 orbit of the hydrogen atom is −2.18 × 10−18 J. Its energy in the n = 2 orbit is:

(a) −1.09 × 10−18 J    (b) −5.45 × 10−19 J    (c) −8.72 × 10−18 J    (d) −2.18 × 10−19 J

Answer key: 1-(b), 2-(c), 3-(a), 4-(c), 5-(c), 6-(c), 7-(c), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: Isotopes of an element show the same chemical properties.

Reason: Chemical properties depend on the number of electrons, which is the same for all isotopes of an element.

A-R 2. Assertion: An electron cannot have the quantum numbers n = 2, l = 2.

Reason: For a given n, the value of l can range only from 0 to (n − 1).

A-R 3. Assertion: The kinetic energy of photoelectrons increases with the intensity of incident light.

Reason: The kinetic energy of photoelectrons depends only on the frequency of the incident light, not its intensity.

A-R 4. Assertion: Half-filled and fully-filled subshells give extra stability.

Reason: Such configurations have symmetrical electron distribution and maximum exchange energy.

A-R 5. Assertion: The position and momentum of an electron can both be measured exactly at the same time.

Reason: Heisenberg’s uncertainty principle states that Δx · Δp ≥ h/4π.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(D).

Common Mistakes to Avoid

Watch out for these

  • Confusing isotopes (same Z, different A) with isobars (same A, different Z) and isoelectronic species (same number of electrons).
  • Forgetting to convert units before substituting — nm, Å, pm and µm must all become metres.
  • Mixing up frequency (ν) and wavenumber (ν̅ = 1/λ); their units (s−1 vs m−1) are different.
  • Using l = n — remember l ranges only from 0 to (n − 1).
  • Writing Cr as [Ar]3d44s2 or Cu as [Ar]3d94s2 — the stable forms are 3d54s1 and 3d104s1.
  • Believing intensity of light affects the kinetic energy of photoelectrons — only frequency does.
  • Dropping the negative sign in Bohr orbit energies (they are always negative for bound electrons).

How to score full marks in this chapter

Memorise the key constants (h, c, e, me, R, NA) and the core formulas, and always write units at every step — examiners award method marks. For numericals, first convert all wavelengths to metres, then choose E = hν or E = hc/λ as appropriate. Use n + l rules to compare orbital energies, and n(n−1)/2 for spectral lines. For quantum-number questions, check l ≤ n − 1 and ml from −l to +l. Learn the exceptional configurations of Cr and Cu and the reasons (symmetry + exchange energy) to bag the short-answer marks.

Frequently Asked Questions

What is Class 11 Chemistry Chapter 2 Structure of Atom about?

It covers the discovery of electrons, protons and neutrons, the Thomson, Rutherford and Bohr atomic models, Planck’s quantum theory and the photoelectric effect, de Broglie’s relation, Heisenberg’s uncertainty principle, quantum numbers, orbital shapes, and the rules for writing electronic configurations.

How many exercise questions are there in Chapter 2?

The NCERT end-of-chapter exercise has 67 questions (2.1 to 2.67), a mix of conceptual questions and numerical problems. All are solved step by step on this page.

What is the difference between isotopes, isobars and isoelectronic species?

Isotopes have the same atomic number but different mass numbers; isobars have the same mass number but different atomic numbers; isoelectronic species have the same number of electrons (e.g. Na+ and Mg2+).

Are these Class 11 Chemistry Chapter 2 solutions free?

Yes. All solutions are free and follow the official NCERT Chemistry Part I textbook for session 2026–27.

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