This MCQ module is based on: NCERT Exercises and Solutions: Structure of Atom
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NCERT Exercises and Solutions: Structure of Atom
🎓 Class 11
Chemistry
CBSE
Theory
Ch 2 – Structure of Atom
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NCERT Exercises and Solutions: Structure of Atom
Chapter 2 Summary — Structure of Atom
| Topic | Key point / formula |
|---|---|
| Sub-atomic particles | Electron (−1.602×10⁻¹⁹ C, 9.109×10⁻³¹ kg); Proton (+1.602×10⁻¹⁹ C, 1.672×10⁻²⁷ kg); Neutron (0 C, 1.675×10⁻²⁷ kg). |
| Thomson's e/m | 1.758820 × 10¹¹ C kg⁻¹; Millikan's e = 1.602 × 10⁻¹⁹ C. |
| Atomic models | Dalton → Thomson (plum pudding) → Rutherford (nuclear, unstable per classical EM) → Bohr. |
| Atomic numbers | Z = protons; A = p + n; notation AZX. Isotopes (same Z), isobars (same A), isotones (same n). |
| EM waves | c = νλ; \(\bar{\nu}=1/\lambda\); c = 3 × 10⁸ m s⁻¹. |
| Planck | E = hν = hc/λ, h = 6.626 × 10⁻³⁴ J·s. |
| Photoelectric | hν = W₀ + ½mv²; KEmax = h(ν − ν₀). |
| Rydberg | \(\bar{\nu}=R_H\big[\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\big]\), RH = 109 677 cm⁻¹. |
| Bohr (H-like, charge Z) | rn = 52.9 n²/Z pm; En = −13.6 Z²/n² eV; vn = 2.188 × 10⁶ Z/n m s⁻¹. |
| de Broglie | λ = h/(mv). |
| Heisenberg | Δx · Δp ≥ h/4π. |
| Quantum numbers | n (shell); ℓ (subshell, 0…n−1); mℓ (−ℓ…+ℓ); ms (±½). |
| Rules | Aufbau (lowest n+ℓ first); Pauli (max 2 e per orbital, opposite spins); Hund (max multiplicity). |
| Exceptions | Cr = [Ar] 3d⁵ 4s¹; Cu = [Ar] 3d¹⁰ 4s¹ (half/full-filled stability). |
Keywords
ElectronNegatively charged subatomic particle
ProtonPositively charged nuclear particle
NeutronNeutral nuclear particle
Cathode rayElectron beam in discharge tube
Canal rayPositive ray from perforated cathode
NucleusTiny central massive positive core
Atomic number (Z)Number of protons
Mass number (A)Protons + neutrons
IsotopesSame Z, different A
IsobarsSame A, different Z
IsotonesSame number of neutrons
IsoelectronicSame number of electrons
QuantumDiscrete packet of energy
PhotonQuantum of light
Planck's constanth = 6.626 × 10⁻³⁴ J·s
Work functionMinimum energy to eject electron
Threshold frequencyν₀ = W₀/h
SpectrumSeparation of radiation by wavelength
Line spectrumDiscrete atomic emission lines
Rydberg constant109 677 cm⁻¹
Bohr orbitQuantised stationary shell
Ionisation energyEnergy to remove electron (H: 13.6 eV)
de Broglie wavelengthλ = h/mv
Uncertainty principleΔx · Δp ≥ h/4π
OrbitalRegion of 90% electron probability
Quantum numbern, ℓ, mℓ, ms
AufbauFill lowest-energy orbitals first
Pauli principleNo two electrons same 4 QN
Hund's ruleMaximise unpaired spins
NodeSurface where ψ = 0
NCERT Exercises — Worked Solutions
Constants used: h = 6.626×10⁻³⁴ J·s; c = 3×10⁸ m s⁻¹; me = 9.11×10⁻³¹ kg; e = 1.602×10⁻¹⁹ C; NA = 6.022×10²³ mol⁻¹; RH = 109 677 cm⁻¹; 1 eV = 1.602×10⁻¹⁹ J.
2.1 Calculate the number of protons, neutrons and electrons in (a) 8035Br, (b) 80Br (Z = 35), (c) 23Na (Z = 11).
(a) Z = 35 ⇒ p = 35, e = 35; n = A − Z = 80 − 35 = 45.
(b) Same as (a) since A and Z are identical: p = 35, e = 35, n = 45.
(c) p = e = 11; n = 23 − 11 = 12.
2.2 The number of electrons in an element X is 15 and the number of neutrons is 16. Write the symbol with mass and atomic numbers.
Z = 15 (phosphorus); A = p + n = 15 + 16 = 31. Symbol: 3115P.
2.3 How many protons and electrons are present in (a) O²⁻, (b) Al³⁺?
(a) O (Z = 8): p = 8, e = 8 − (−2) = 10.
(b) Al (Z = 13): p = 13, e = 13 − 3 = 10. Both are isoelectronic with Ne.
2.4 Arrange the following species in order of increasing number of electrons: Na, Na⁺, Mg²⁺, O²⁻.
Electrons: Na⁺ = 10, Mg²⁺ = 10, O²⁻ = 10, Na = 11. So Na⁺ = Mg²⁺ = O²⁻ < Na.
2.5 Write the electronic configurations of (a) 16O²⁻, (b) 24Mg, (c) 56Fe, (d) Br⁻ (Z = 35).
(a) O²⁻ (10 e): 1s² 2s² 2p⁶.
(b) Mg (12 e): 1s² 2s² 2p⁶ 3s².
(c) Fe (26 e): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² = [Ar] 3d⁶ 4s².
(d) Br⁻ (36 e): [Ar] 3d¹⁰ 4s² 4p⁶ = [Kr].
2.6 Calculate the frequency and wave number of radiation of wavelength 5800 Å.
λ = 5800 × 10⁻¹⁰ m = 5.8 × 10⁻⁷ m.
ν = c/λ = 3×10⁸ / 5.8×10⁻⁷ = 5.17 × 10¹⁴ Hz.
\(\bar{\nu}\) = 1/λ = 1/(5.8×10⁻⁵ cm) = 1.72 × 10⁴ cm⁻¹.
2.7 Calculate the energy of one photon of light of wavelength 4000 Å and the energy per mole.
λ = 4 × 10⁻⁷ m.
E = hc/λ = (6.626×10⁻³⁴)(3×10⁸)/(4×10⁻⁷) = 4.97 × 10⁻¹⁹ J.
Per mole: E·NA = 4.97×10⁻¹⁹ × 6.022×10²³ ≈ 2.99 × 10⁵ J mol⁻¹ = 299 kJ mol⁻¹.
2.8 A 25-watt bulb emits monochromatic yellow light of wavelength 0.57 μm. Calculate the number of photons emitted per second.
λ = 0.57 × 10⁻⁶ m. Energy per photon: E = hc/λ = (6.626×10⁻³⁴)(3×10⁸)/(5.7×10⁻⁷) = 3.49 × 10⁻¹⁹ J.
n = P/E = 25 / 3.49×10⁻¹⁹ ≈ 7.17 × 10¹⁹ photons s⁻¹.
2.9 Electromagnetic radiation of wavelength 242 nm is just enough to ionise a sodium atom. Calculate the ionisation energy of Na in kJ mol⁻¹.
E = hc/λ = (6.626×10⁻³⁴)(3×10⁸)/(242×10⁻⁹) = 8.21 × 10⁻¹⁹ J/atom.
Per mole: 8.21×10⁻¹⁹ × 6.022×10²³ = 4.945 × 10⁵ J = 494.5 kJ mol⁻¹.
2.10 A 60-watt bulb emits 95 % of its energy as heat and the remaining 5 % as light of wavelength 625 nm. How many photons are emitted per second by the light output?
Light power = 0.05 × 60 = 3 W. E per photon = hc/λ = (6.626×10⁻³⁴)(3×10⁸)/(625×10⁻⁹) = 3.18 × 10⁻¹⁹ J.
n = 3 / 3.18×10⁻¹⁹ = 9.43 × 10¹⁸ photons s⁻¹.
2.11 The threshold frequency for a metal is 7.0 × 10¹⁴ Hz. Calculate the kinetic energy of an electron emitted when radiation of frequency 1.0 × 10¹⁵ Hz hits the metal.
KE = h(ν − ν₀) = (6.626×10⁻³⁴)(1.0×10¹⁵ − 7.0×10¹⁴)
= (6.626×10⁻³⁴)(3.0×10¹⁴) = 1.99 × 10⁻¹⁹ J ≈ 1.24 eV.
2.12 The work function of caesium is 1.9 eV. Calculate (a) the threshold wavelength and (b) the kinetic energy of the photoelectron ejected by light of wavelength 500 nm.
(a) W₀ = 1.9 × 1.602×10⁻¹⁹ = 3.044 × 10⁻¹⁹ J. λ₀ = hc/W₀ = (1.986×10⁻²⁵)/(3.044×10⁻¹⁹) = 6.53 × 10⁻⁷ m ≈ 653 nm.
(b) Photon energy at 500 nm: E = 3.97×10⁻¹⁹ J = 2.48 eV. KE = 2.48 − 1.9 = 0.58 eV ≈ 9.28 × 10⁻²⁰ J.
2.13 Calculate the wavelength of the spectral line emitted when an electron in hydrogen falls from n = 4 to n = 2.
\(\bar{\nu} = 109677(1/4 - 1/16) = 109677(3/16) = 20\,564\) cm⁻¹.
λ = 1/\(\bar{\nu}\) = 4.86 × 10⁻⁵ cm = 486 nm (Hβ, blue-green).
2.14 How much energy (in joules and in kJ mol⁻¹) is required to ionise a hydrogen atom from its ground state?
IE = 2.18 × 10⁻¹⁸ J/atom (since E₁ = −2.18 × 10⁻¹⁸ J).
Per mole = 2.18×10⁻¹⁸ × 6.022×10²³ = 1.312 × 10⁶ J mol⁻¹ = 1312 kJ mol⁻¹.
2.15 What is the maximum number of emission lines when the excited electron of a hydrogen atom in n = 6 drops to the ground state?
Number of lines = n(n − 1)/2 = 6(5)/2 = 15 lines.
2.16 (i) Energy associated with the fifth orbit of H. (ii) Radius of the fifth orbit of H.
(i) E₅ = −2.18×10⁻¹⁸ /25 = −8.72 × 10⁻²⁰ J/atom (−0.544 eV).
(ii) r₅ = 52.9 × 25 = 1322.5 pm = 1.32 nm.
2.17 Calculate the wave number for the longest-wavelength transition in the Balmer series of hydrogen.
Longest λ ↔ smallest energy gap ↔ n₂ = 3, n₁ = 2.
\(\bar{\nu} = 109677(1/4 - 1/9) = 109677(5/36) = 15\,233\) cm⁻¹ ⇒ λ = 656 nm (Hα).
2.18 What is the energy in joules, wavelength (nm) and spectral region of the photon emitted when an electron in Li²⁺ falls from n = 2 to n = 1?
For Li²⁺ (Z = 3): En = −2.18×10⁻¹⁸ × 9/n² J.
ΔE = E₁ − E₂ = −2.18×10⁻¹⁸ × 9 × (1 − 1/4) = −2.18×10⁻¹⁸ × 9 × 0.75 = −1.472 × 10⁻¹⁷ J.
Photon energy = 1.472 × 10⁻¹⁷ J. λ = hc/E = (1.986×10⁻²⁵)/(1.472×10⁻¹⁷) = 1.35 × 10⁻⁸ m ≈ 13.5 nm (far UV).
2.19 The electron energy in a hydrogen atom is given by En = (−2.18 × 10⁻¹⁸)/n² J. Calculate the energy required to remove an electron completely from n = 2. What is the longest wavelength of light in cm that can be used to bring about this transition?
ΔE = 0 − E₂ = 2.18×10⁻¹⁸/4 = 5.45 × 10⁻¹⁹ J.
λ = hc/ΔE = 1.986×10⁻²⁵ / 5.45×10⁻¹⁹ = 3.64 × 10⁻⁷ m = 3.64 × 10⁻⁵ cm (364 nm, UV).
2.20 Calculate the wavelength of an electron moving with a velocity of 2.05 × 10⁷ m s⁻¹.
λ = h/(mv) = (6.626×10⁻³⁴)/((9.11×10⁻³¹)(2.05×10⁷))
= 6.626×10⁻³⁴ / 1.868×10⁻²³ = 3.55 × 10⁻¹¹ m = 35.5 pm.
2.21 The mass of an electron is 9.1 × 10⁻³¹ kg. If its kinetic energy is 3.0 × 10⁻²⁵ J, calculate its wavelength.
v = √(2KE/m) = √(2·3×10⁻²⁵/9.1×10⁻³¹) = √(6.593×10⁵) = 811.8 m s⁻¹.
λ = h/mv = 6.626×10⁻³⁴/(9.1×10⁻³¹ × 811.8) = 8.97 × 10⁻⁷ m ≈ 897 nm.
2.22 A golf ball of mass 40 g moves at 45 m s⁻¹. If the uncertainty in velocity is 2 %, what is the minimum uncertainty in the position of the ball?
Δv = 0.02 × 45 = 0.9 m s⁻¹. Δp = mΔv = 0.04 × 0.9 = 0.036 kg m s⁻¹.
Δx ≥ h/(4π·Δp) = 6.626×10⁻³⁴ / (4π × 0.036) = 1.46 × 10⁻³³ m — utterly negligible for a macroscopic object.
2.23 An electron is moving with a velocity of 600 m s⁻¹, accurate up to 0.005 %. Calculate the minimum uncertainty in its position.
Δv = 0.00005 × 600 = 0.03 m s⁻¹. Δp = 9.11×10⁻³¹ × 0.03 = 2.73×10⁻³² kg m s⁻¹.
Δx ≥ 6.626×10⁻³⁴/(4π × 2.73×10⁻³²) = 1.93 × 10⁻³ m ≈ 1.93 mm — enormous compared with an atomic dimension.
2.24 Which of the following sets of quantum numbers are not possible? (a) n = 1, ℓ = 0, mℓ = 0, ms = +½ (b) n = 1, ℓ = 1, mℓ = 0, ms = +½ (c) n = 2, ℓ = 1, mℓ = 0, ms = −½ (d) n = 3, ℓ = 3, mℓ = −3, ms = +½.
(b) is invalid — ℓ can only go up to n − 1 = 0 when n = 1.
(d) is invalid — ℓ can only go up to n − 1 = 2 when n = 3.
(a) and (c) are valid.
2.25 How many electrons in an atom may have the following quantum numbers? (a) n = 4, ms = −½ (b) n = 3, ℓ = 0.
(a) For n = 4 there are 2n² = 32 electrons; half have ms = −½, so 16 electrons.
(b) n = 3, ℓ = 0 refers to the 3s orbital, which can hold 2 electrons.
2.26 An atom of an element contains 29 electrons and 35 neutrons. Deduce (a) the number of protons and (b) the electronic configuration of the element.
(a) In a neutral atom p = e = 29 (copper).
(b) Cu: [Ar] 3d¹⁰ 4s¹ (full-filled d-shell is more stable than [Ar] 3d⁹ 4s²).
2.27 Give the number of electrons in the species: H2⁺, H2 and O2⁺.
H2⁺: 2 − 1 = 1 e. H2: 2 e. O2⁺: 16 − 1 = 15 e.
2.28 An atomic orbital has n = 3. What are the possible values of ℓ and mℓ? List all the orbitals by symbol.
ℓ = 0, 1, 2. Orbitals: 3s (mℓ = 0); 3p (mℓ = −1, 0, +1); 3d (mℓ = −2, −1, 0, +1, +2). Total 1 + 3 + 5 = 9 orbitals, 18 electrons max.
2.29 Using s, p, d, f 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.
(a) 1s (b) 3p (c) 4d (d) 4f.
2.30 Explain the Pauli exclusion principle, Aufbau principle and Hund's rule in one line each.
Pauli: No two electrons in an atom can have all four quantum numbers identical.
Aufbau: Electrons occupy orbitals in order of increasing energy (lowest (n + ℓ) first).
Hund: Within a subshell, electrons occupy orbitals singly with parallel spin before pairing.
2.31 Which orbital is filled next after 4s according to the Aufbau principle?
3d (n + ℓ = 5 same as 4p but lower n means lower energy — actually 4s has n + ℓ = 4 and fills first, followed by 3d with n + ℓ = 5, then 4p).
2.32 Write the electron configuration of Cr (Z = 24) and explain any anomaly.
Expected: [Ar] 3d⁴ 4s². Actual: [Ar] 3d⁵ 4s¹. The half-filled 3d⁵ together with half-filled 4s¹ lowers the total energy through greater exchange stability, more than compensating for the small 4s→3d promotion energy.
2.33 An element with mass number 81 contains 31.7 % more neutrons than protons. Identify the element and write its symbol.
Let protons = p. Neutrons = 1.317p. Total A = p + 1.317p = 2.317p = 81 ⇒ p = 35. Element is Br. A = 81 ⇒ 8135Br.
2.34 An ion of mass number 56 carries 3 units of positive charge and has 30.4 % more neutrons than electrons. Identify the ion.
Let electrons = e; protons = e + 3; neutrons = 1.304 e. Mass: (e + 3) + 1.304 e = 56 ⇒ 2.304 e = 53 ⇒ e = 23; protons = 26 (Fe), neutrons = 30. Ion = 5626Fe³⁺.
2.35 Neon gas is generally used in sign boards. If it emits strongly at 616 nm, calculate (a) the frequency, (b) the distance travelled by this radiation in 30 s, (c) energy of one photon, (d) number of photons present in 2 J of energy.
(a) ν = c/λ = 3×10⁸ /(616×10⁻⁹) = 4.87 × 10¹⁴ Hz.
(b) Distance = c × t = 3 × 10⁸ × 30 = 9 × 10⁹ m.
(c) E = hν = 6.626×10⁻³⁴ × 4.87×10¹⁴ = 3.23 × 10⁻¹⁹ J.
(d) n = 2 / 3.23×10⁻¹⁹ = 6.19 × 10¹⁸ photons.
2.36 In astronomical observations, signals from distant stars are generally weak. If the photon detector receives a total of 3.15 × 10⁻¹⁸ J from radiation of wavelength 600 nm, calculate the number of photons received.
E per photon = hc/λ = (6.626×10⁻³⁴)(3×10⁸)/(6×10⁻⁷) = 3.313×10⁻¹⁹ J.
n = 3.15×10⁻¹⁸ / 3.313×10⁻¹⁹ ≈ 9.5 ≈ 10 photons.
2.37 A dye absorbs radiation of wavelength 5500 Å and then fluoresces at 6000 Å. What percentage of the absorbed energy is re-emitted as fluorescence?
Since E ∝ 1/λ, ratio Eemit/Eabs = λabs/λemit = 5500/6000 = 0.9167.
Percentage re-emitted = 91.67 %; the remaining 8.33 % appears as heat (Stokes shift).
2.38 The longest wavelength observed in the Lyman series is 1216 Å. Calculate RH.
Longest Lyman: n₁ = 1, n₂ = 2. \(\bar{\nu} = R_H(1 - 1/4) = (3/4)R_H\).
λ = 1216 Å = 1216 × 10⁻⁸ cm ⇒ \(\bar{\nu}\) = 1/1216×10⁻⁸ = 8.22 × 10⁴ cm⁻¹.
RH = (4/3) × 8.22 × 10⁴ = 1.096 × 10⁵ cm⁻¹ — matches 109 677 cm⁻¹.
2.39 Calculate the velocity of an electron in the first Bohr orbit of H. What is its de Broglie wavelength?
v₁ = 2.188 × 10⁶ m s⁻¹.
λ = h/(mv) = 6.626×10⁻³⁴/(9.11×10⁻³¹ × 2.188×10⁶) = 3.32 × 10⁻¹⁰ m = 332 pm = 2π r₁. The wave fits exactly once around the first orbit — confirming the standing-wave picture.
2.40 The position of both an electron and a helium atom is known within 1.0 nm. What is the minimum uncertainty in velocity of each? (mHe = 6.64 × 10⁻²⁷ kg)
Δv ≥ h/(4π·m·Δx).
Electron: Δv = 6.626×10⁻³⁴/(4π × 9.11×10⁻³¹ × 10⁻⁹) = 5.79 × 10⁴ m s⁻¹.
Helium: Δv = 6.626×10⁻³⁴/(4π × 6.64×10⁻²⁷ × 10⁻⁹) = 7.94 m s⁻¹. Macroscopic helium atoms are essentially un-disturbed by the uncertainty relation.
2.41 Write the electronic configuration and find the number of unpaired electrons in (a) Fe³⁺ (b) Ni²⁺ (c) Mn²⁺ (d) Cu⁺.
(a) Fe³⁺ (23 e): [Ar] 3d⁵ — 5 unpaired.
(b) Ni²⁺ (26 e): [Ar] 3d⁸ — 2 unpaired.
(c) Mn²⁺ (23 e): [Ar] 3d⁵ — 5 unpaired.
(d) Cu⁺ (28 e): [Ar] 3d¹⁰ — 0 unpaired (diamagnetic).
2.42 Calculate the energy required for the process He⁺(g) → He²⁺(g) + e⁻. Given RH = 2.18 × 10⁻¹⁸ J/atom.
For He⁺ (Z = 2, n = 1): E₁ = −2.18 × 10⁻¹⁸ × 4 = −8.72 × 10⁻¹⁸ J.
Energy to ionise = 8.72 × 10⁻¹⁸ J/atom = 5253 kJ mol⁻¹.
2.43 If the first excitation energy of a hydrogen-like ion is 27.2 eV, identify the ion.
ΔE(1→2) = 13.6 Z²(1 − 1/4) = 10.2 Z² eV. Given 10.2 Z² = 27.2 ⇒ Z² ≈ 2.67 — the question intends Z² = 4, ΔE = 40.8 eV for He⁺. For Z = 2, first excitation = 40.8 eV. Checking for Li²⁺: 10.2 × 9 = 91.8 eV. So if 27.2 eV corresponds to the photon emitted when electron in n = 2 falls to n = 1 of He⁺: actually |E₁| of He⁺ is 54.4 eV and E(n=1→∞) transitions give different values. The cleanest matching species: He⁺ has first excitation (n=1 → n=2) = 40.8 eV; the number 27.2 eV matches 2 × |EH,1| = 2 × 13.6 i.e. the second ionisation step of H-like hydrogen. So consult numerical setup carefully. If given exactly 27.2 eV as a first excitation energy, no simple hydrogen-like ion fits — the nearest consistent identification is Z = √(27.2/10.2) ≈ √2.67 — not integer, hence the data must refer to a different transition (often used as He²⁺ ionisation of He⁺ → He²⁺ when E₁(He⁺) = −54.4 eV and 2 × 13.6 = 27.2 eV is the H ground state energy, showing that He⁺ is 4× more tightly bound than H).
2.44 A 1 g mass moves at 0.1 cm s⁻¹. Calculate the de Broglie wavelength.
m = 10⁻³ kg, v = 10⁻³ m s⁻¹.
λ = 6.626×10⁻³⁴/(10⁻³ × 10⁻³) = 6.626 × 10⁻²⁸ m — far below any possible measurement; wave nature negligible.
2.45 The energy of a photon is 3.03 × 10⁻¹⁹ J. Find its wavelength (nm).
λ = hc/E = 1.986×10⁻²⁵ / 3.03×10⁻¹⁹ = 6.55 × 10⁻⁷ m = 655 nm (red). This is the Hα line energy.
2.46 Calculate the total number of angular nodes and radial nodes present in a 3p orbital.
Angular nodes = ℓ = 1.
Radial nodes = n − ℓ − 1 = 3 − 1 − 1 = 1.
Total nodes = n − 1 = 2.
2.47 Write the complete electron configuration of (a) As (Z = 33), (b) I (Z = 53), (c) Pd (Z = 46, an exception).
(a) As: [Ar] 3d¹⁰ 4s² 4p³ — 3 unpaired (half-filled 4p).
(b) I: [Kr] 4d¹⁰ 5s² 5p⁵ — 1 unpaired.
(c) Pd: [Kr] 4d¹⁰ (5s⁰) — fully-filled 4d; a well-known anomaly.
2.48 Calculate the frequency and wavelength of the photon emitted when the electron in a He⁺ ion drops from n = 3 to n = 2.
\(\bar{\nu} = Z^2 R_H(1/n_1^2 - 1/n_2^2) = 4 × 109677 × (1/4 - 1/9) = 4 × 109677 × 5/36 = 60\,932\) cm⁻¹.
λ = 1/\(\bar{\nu}\) = 1.641 × 10⁻⁵ cm = 164.1 nm (UV).
ν = c/λ = 1.83 × 10¹⁵ Hz.
2.49 What transition in the hydrogen spectrum will have the same wavelength as the Balmer transition n = 4 → n = 2 of He⁺?
For He⁺: \(\bar{\nu} = 4R_H(1/4 - 1/16) = 4R_H(3/16) = (3/4)R_H\).
For H we need \(\bar{\nu} = R_H(1/n_1^2 - 1/n_2^2) = (3/4)R_H\) ⇒ n₁ = 1, n₂ = 2 (Lyman first line).
Result: the He⁺ Balmer 4 → 2 transition has the same wavelength as the H Lyman 2 → 1 transition.
2.50 Calculate the ionisation enthalpy (in kJ mol⁻¹) of atomic hydrogen in its first excited state.
E₂ = −13.6/4 = −3.40 eV. IE = +3.40 eV/atom.
Per mole = 3.40 × 1.602×10⁻¹⁹ × 6.022×10²³ = 3.28 × 10⁵ J = 328 kJ mol⁻¹ — one quarter of the ground-state IE (1312 kJ mol⁻¹).
Study tip: Chapter 2 numericals cluster into five types — (i) counting particles (p, n, e), (ii) photon energies, (iii) photoelectric effect, (iv) Rydberg / Bohr transitions, (v) de Broglie & uncertainty. If you can solve one example from each cluster cleanly, any board-exam problem is a minor variation.
Frequently Asked Questions — NCERT Exercises and Solutions: Structure of Atom
What are the most important formulas for Class 11 Chemistry Chapter 2 exercises?
Key formulas for NCERT Class 11 Chemistry Chapter 2 include: E = hν = hc/λ (photon energy), 1/λ = R(1/n₁² − 1/n₂²) (Rydberg), E_n = −13.6/n² eV (hydrogen energy levels), r_n = 0.529 × n² Å (Bohr radius), λ = h/mv (de Broglie), and Δx · Δp ≥ h/4π (Heisenberg). For electron configurations use the Aufbau, Pauli and Hund rules. Memorising these and understanding their derivations is essential to solve every Chapter 2 numerical and conceptual question in the CBSE board exam.
How do you calculate energy of an electron transition in hydrogen?
To calculate the energy emitted or absorbed during an electron transition in hydrogen, use ΔE = −13.6 (1/n_f² − 1/n_i²) eV or ΔE = R_H · h · c (1/n₁² − 1/n₂²). For example, the transition from n = 3 to n = 2 (Balmer series) gives ΔE = −13.6 (1/4 − 1/9) = −1.89 eV, emitting a photon of wavelength λ = hc/|ΔE| = 656 nm (red). NCERT Class 11 Chemistry exercises feature many such transition problems for the Lyman, Balmer and Paschen series. Always check the sign of ΔE — negative means emission, positive means absorption.
How are quantum numbers assigned in NCERT exercises?
Quantum numbers in NCERT Class 11 Chemistry exercises follow strict rules: principal n = 1, 2, 3,…; azimuthal l = 0 to n−1 (0=s, 1=p, 2=d, 3=f); magnetic m_l = −l to +l including 0; spin m_s = +½ or −½. For example, a 3d electron has n = 3, l = 2, m_l ∈ {−2,−1,0,1,2}, m_s = ±½. Exercise questions test which sets of quantum numbers are valid, the maximum number of electrons in a subshell (2(2l+1)) and total electrons in a shell (2n²). Always verify each quantum number against these rules.
How do you write electron configurations for ions and exceptions?
To write electron configuration of an ion: start from the neutral atom configuration, then add electrons (for anions) or remove electrons from the outermost shell first (for cations). For example, Fe (Z = 26) is [Ar] 3d⁶ 4s², so Fe²⁺ is [Ar] 3d⁶ (4s lost first), and Fe³⁺ is [Ar] 3d⁵. NCERT Class 11 Chemistry exercises also include exceptions like Cr ([Ar] 3d⁵ 4s¹) and Cu ([Ar] 3d¹⁰ 4s¹) which arise from extra stability of half-filled and fully-filled d-subshells. Memorise these exceptions for the board exam.
What is the standard approach for de Broglie wavelength problems?
To solve de Broglie wavelength problems in NCERT Class 11 Chemistry: (1) identify the particle's mass m and velocity v (or kinetic energy KE), (2) if KE is given, use v = √(2KE/m), (3) apply λ = h/(mv) with h = 6.626 × 10⁻³⁴ J·s. Always convert mass to kg and velocity to m/s. For an electron (m = 9.1 × 10⁻³¹ kg) at 10⁶ m/s, λ ≈ 7.27 × 10⁻¹⁰ m or 0.727 nm. Compare with classical objects — a cricket ball has negligible λ. This is why wave nature matters only for subatomic particles.
What types of questions appear in Chapter 2 CBSE board exams?
CBSE Class 11 Chemistry Chapter 2 board questions include: 1-mark MCQs on subatomic particles and atomic models, 2-mark short questions on quantum numbers and electron configurations, 3-mark problems on Bohr model energy calculations, photoelectric effect and Rydberg formula, and 5-mark numerical problems combining de Broglie wavelength, Heisenberg uncertainty and hydrogen spectrum. Diagrams of orbitals (s, p, d shapes) are frequently asked. The MyAiSchool exercise set covers every CBSE pattern question with model solutions, marking schemes and conceptual tips for high scores.
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Chemistry Class 11 Part I – NCERT (2025-26)
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