CBSE Class 11 Chemistry HOTs Structure of Atom Set 03

Higher Order Thinking Questions

Question 1. A hydrogen atom in the ground state absorbs just enough energy to reach n = 2. (a) Calculate this energy in eV. (b) What fraction of the ionisation energy (13.6 eV) is it?
Answer:
(a) ΔE = 13.6 × (1/1² − 1/2²) = 13.6 × 3/4 = 10.2 eV.
(b) 10.2 ÷ 13.6 = 3/4, that is 75%.
So three-quarters of the energy needed to remove the electron completely is already used in reaching the first excited state.
In simple words: Going from n = 1 to n = 2 takes most of the energy needed to escape completely.

Teacher's Note:
1. Energy levels crowd together as n increases.
2. Ionisation is n = 1 to n = ∞, which is 13.6 eV.

Question 2. Case study. Hydrogen atoms in a discharge tube are excited to n = 4, and then return to lower levels. (a) How many different spectral lines can be emitted? (b) How many of them lie in the visible region? (c) Find the wavelength of the longest-wavelength line (R = 1.097 × 10⁷ m⁻¹).
Answer:
(a) Number of lines = n(n − 1)/2 = 4 × 3/2 = 6: 4 to 3, 4 to 2, 4 to 1, 3 to 2, 3 to 1 and 2 to 1.
(b) Visible lines end at n = 2: 4 to 2 and 3 to 2. So 2 lines (486 nm and 656 nm).
(c) The smallest energy gap is 4 to 3: 1/λ = R(1/9 − 1/16) = R × 7/144, so λ = 1.88 × 10⁻⁶ m (1875 nm), in the infrared (Paschen series).
In simple words: The smallest energy jump gives the longest wavelength.

Teacher's Note:
1. Count each possible jump once.
2. Longest wavelength means lowest energy, which is the jump between the two highest levels.

Question 3. Assertion (A): The radius of the first Bohr orbit of Li²⁺ is smaller than that of hydrogen.
Reason (R): The radius of the nth Bohr orbit is proportional to n²/Z.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Answer: (a).
r = 52.9 n²/Z pm. For n = 1, hydrogen (Z = 1) has 52.9 pm and Li²⁺ (Z = 3) has 17.6 pm. The stronger pull of the larger nuclear charge draws the electron closer. R explains A.
In simple words: A larger nuclear charge pulls the electron closer.

Teacher's Note:
1. Z is in the denominator of the radius formula.
2. Energy goes as Z², radius as 1/Z.

Question 4. Assertion (A): The wave nature of a moving cricket ball is not observed.
Reason (R): The de Broglie wavelength of a cricket ball is extremely small because its mass is large.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Answer: (a).
λ = h/mv. For a 150 g ball at 30 m s⁻¹, λ is about 1.5 × 10⁻³⁴ m, many orders smaller than any measurable length. R explains A.
In simple words: Heavy objects have wavelengths too tiny to detect.

Teacher's Note:
1. Electrons have large enough wavelengths because their mass is tiny.
2. That is why electron diffraction is observed.

Question 5. Assertion (A): The angular momentum of an electron in a 2s orbital is zero.
Reason (R): For an s orbital the azimuthal quantum number l is zero, and the orbital angular momentum is √[l(l + 1)] h/2π.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Answer: (a).
For 2s, l = 0, so √[0 × 1] × h/2π = 0. R gives the formula that proves A.
In simple words: No angular momentum for any s electron.

Teacher's Note:
1. For 2p, l = 1, so the value is √2 h/2π.
2. For 3d, l = 2, so the value is √6 h/2π.

Question 6. Assertion (A): Nitrogen has three unpaired electrons in its 2p subshell.
Reason (R): An orbital can hold at most two electrons, which must have opposite spins.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Answer: (b).
A is true by Hund's rule: the three 2p electrons occupy 2p_x, 2p_y and 2p_z singly with parallel spins. R is also true (Pauli exclusion principle), but it does not explain why the electrons stay unpaired; Hund's rule does.
In simple words: Hund's rule, not Pauli's, keeps the three p electrons apart.

Teacher's Note:
1. Pauli: maximum two per orbital. Hund: fill singly first.
2. Both are true here, yet R is not the reason.

Question 7. Assertion (A): The electronic configuration of chromium is [Ar] 3d⁵ 4s¹.
Reason (R): A half-filled subshell has no extra stability compared with a partly filled subshell.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Answer: (c).
A is true. R is false: half-filled and completely filled subshells are especially stable because of symmetry and exchange energy, which is exactly why Cr is 3d⁵ 4s¹.
In simple words: Cr is an exception because half-filled d is extra stable.

Teacher's Note:
1. Check whether the reason supports or contradicts the assertion.
2. Cu, 3d¹⁰ 4s¹, is the second exception.

Question 8. Which of the following transitions of the hydrogen atom lie in the visible region? (more than one may be correct)
(a) n = 3 to n = 2
(b) n = 4 to n = 2
(c) n = 2 to n = 1
(d) n = 4 to n = 3
Answer:
Answer: (a) and (b).
Visible lines (Balmer series) end at n = 2. A fall to n = 1 is the Lyman series (ultraviolet) and a fall to n = 3 is the Paschen series (infrared).
In simple words: Look for the jumps that land on n = 2.

Teacher's Note:
1. Lyman: UV; Balmer: visible; Paschen and above: infrared.
2. The Balmer series has four visible lines (656, 486, 434, 410 nm).

Question 9. A proton and an α-particle are accelerated from rest through the same potential difference. Find the ratio of their de Broglie wavelengths (m_α = 6.646 × 10⁻²⁷ kg, m_p = 1.673 × 10⁻²⁷ kg).
Answer:
K.E. = qV and p = √(2m K.E.) = √(2mqV), so λ = h/√(2mqV).
λ_p/λ_α = √(m_α q_α/(m_p q_p)) = √(4 × 2/(1 × 1)) = √8.
The ratio is 2.82 : 1 (2√2 : 1).
In simple words: The α-particle is four times as heavy and has twice the charge, so its wavelength is shorter.

Teacher's Note:
1. Charge as well as mass matters when the same voltage is used.
2. For the same speed, the ratio would be 4 : 1 instead.

Question 10. A student lists these sets of quantum numbers (n, l, m_l, m_s) for electrons in an atom: (i) 3, 2, +2, +1/2; (ii) 2, 2, 0, −1/2; (iii) 4, 3, −3, −1/2; (iv) 2, 1, +1, 0. Which are acceptable, and what is wrong with the others?
Answer:
(i) Acceptable (a 3d electron).
(ii) Not acceptable: l must be less than n, so for n = 2, l cannot be 2.
(iii) Acceptable (a 4f electron with m_l = −3).
(iv) Not acceptable: m_s can only be +1/2 or −1/2, never 0.
In simple words: Check l < n, |m_l| ≤ l and m_s = ±1/2.

Teacher's Note:
1. Apply the rules one at a time.
2. Two valid sets means two electrons in different subshells.

Question 11. Compare the 1s and 2s orbitals in terms of shape, size, nodes and energy.
Answer:
Shape: both are spherical and non-directional, with l = 0.
Size: 2s is larger than 1s because n is larger.
Nodes: 1s has no node; 2s has one radial (spherical) node, since n − l − 1 = 1.
Energy: 2s is higher in energy than 1s, because its electron is on average farther from the nucleus.
Both have zero orbital angular momentum.
In simple words: Same shape, but 2s is bigger, has a spherical node and has higher energy.

Teacher's Note:
1. Radial nodes = n − l − 1.
2. Each extra shell adds one radial node to the s orbital.