Previous year questions: Structure of Atom

Structure of Atom30 min readLesson 8 of 8

Q1 (JEE Main 2026, 4 Apr Shift 1, 4 marks) Which set of the four quantum numbers belongs to the 19th electron of chromium, atomic number 24, when the orbitals are filled by the Aufbau principle? Answer: Chromium has 24 electrons. Filling in the Aufbau order puts 18 electrons into argon's configuration: 1s2 2s2 2p6 3s2 3p6. The 19th electron enters the next orbital in the order, which is 4s, not 3d. For a 4s electron the principal quantum number is 4, the azimuthal quantum number for an s orbital is 0, the magnetic quantum number is 0, and the spin is written as +1/2. The correct set is therefore n = 4, l = 0, m_l = 0, m_s = +1/2, which is option D in the official key. The sets with n = 3 and l = 2 describe 3d electrons, which are filled only after 4s.

Q2 (JEE Main 2026, 4 Apr Shift 2, 4 marks) Under Bohr's theory, which two species have identical radii: (A) H in the first orbit, (B) a hydrogen-like ion in the first orbit, (C) a hydrogen-like ion in the second orbit, (D) Li2+ in the first orbit, (E) Be3+ in the second orbit? Answer: The Bohr radius of a hydrogen-like species is 52.9 n squared divided by Z pm. For H in the first orbit, Z is 1 and n is 1, so the radius is 52.9 pm. For Be3+ in the second orbit, Z is 4 and n is 2, so the radius is 52.9 x 4/4 = 52.9 pm. These two values are identical. The other options give different values: He+ in the first orbit would be 26.45 pm, and Li2+ in the first orbit would be 17.63 pm. The correct pair is A and E, which is option B in the official key.

r_{H, 1} = 52.9 \times \frac{1 ^{2}}{1} = 52.9 pm r_{Be^{3 +}, 2} = 52.9 \times \frac{2 ^{2}}{4} = 52.9 pm

Bohr radius check for H and Be3+

Q3 (JEE Main 2026, 5 Apr Shift 1, 4 marks) Identify the correct statements. (A) The Heisenberg uncertainty principle applies to electrons. (B) A 2p_x orbital is smaller than a 3p_x orbital. (C) The energy of the 2s orbital of hydrogen equals the energy of the 2s orbital of lithium. (D) The ground state configuration of chromium is [Ar] 3d5 4s1. Answer: Statement A is correct: the uncertainty principle applies to every particle, and its effect matters for electrons because their mass is tiny. Statement B is correct: for the same subshell, a larger principal quantum number means a larger, more diffuse orbital, so 3p_x is larger than 2p_x. Statement C is wrong: the energy depends on the nuclear charge as well as on n, and lithium's higher nuclear charge lowers its 2s energy compared with hydrogen. Statement D is correct: chromium has a half-filled 3d subshell and the configuration [Ar] 3d5 4s1, which is more stable than the simple filling order prediction. The correct statements are A, B and D.

Q4 (JEE Main 2026, 5 Apr Shift 2, 4 marks) Which of these statements are correct? (A) When two orbitals share the value of n + l, the orbital with the smaller n lies lower in energy. (B) Within one subshell, the orbital energies rise as the atomic number rises. (C) A 2p_x orbital is smaller than a 3p_x orbital. (D) Among 5f, 6s, 4d, 5p and 5d, no orbital has exactly two radial nodes. Answer: Statement A is true: the n + l rule breaks ties by choosing the lower n, which lies closer to the nucleus. Statement B is false: orbitals in one subshell are degenerate in a free atom, and their energy relative to other subshells is set by the electron configuration, not by the atomic number alone. Statement C is true, for the same reason as in the previous question. Statement D is false: the number of radial nodes is n minus l minus 1, and for 5d this gives 5 minus 2 minus 1 = 2, so 5d has exactly two radial nodes. The true statements are A and C.

Q5 (JEE Main 2026, 6 Apr Shift 1, 4 marks) If the shortest wavelength of the Lyman series of hydrogen is x, what is the longest wavelength of the Balmer series of He+? Answer: The shortest Lyman wavelength comes from the largest energy jump, from n = infinity to n = 1, so its photon energy is 13.6 eV for hydrogen and the wavelength is hc divided by 13.6 eV. For He+ the nuclear charge is 2, so every energy is multiplied by Z squared, which is 4, and the levels are E_n = -13.6 x 4/n squared eV. The longest Balmer line is the smallest jump in that series, from n = 3 to n = 2. Its energy is 54.4 x (1/4 minus 1/9), which is 54.4 x 5/36 = 7.56 eV. The corresponding wavelength is hc divided by 7.56 eV, and since x corresponds to hc divided by 13.6 eV, the new wavelength is x multiplied by 13.6/7.56, which is 1.80 x. In fraction form this is 9x/5, which is option A in the official key.

Δ E = 13.6 \times 4 (\frac{1}{4} - \frac{1}{9}) = 7.56 eV λ^{'} = \frac{h c}{7.56} = \frac{13.6}{7.56} x = \frac{9 x}{5}

Energy of the longest He+ Balmer line and the ratio of wavelengths

Q6 (JEE Main 2026, 6 Apr Shift 1, 4 marks) Match each orbital in List I with its radial nodes and nodal planes in List II. List I: (A) 2s, (B) 3s, (C) 3p, (D) 4d. List II: (I) 1 radial node and 2 nodal planes, (II) 1 radial node and 1 nodal plane, (III) 2 radial nodes and no nodal plane, (IV) 1 radial node and no nodal plane. Answer: For any orbital the radial node count is n minus l minus 1 and the number of nodal planes equals l. For 2s, n = 2 and l = 0, so there is one radial node and no nodal plane, which is IV. For 3s, n = 3 and l = 0, giving two radial nodes and no nodal plane, which is III. For 3p, n = 3 and l = 1, giving one radial node and one nodal plane, which is II. For 4d, n = 4 and l = 2, giving one radial node and two nodal planes, which is I. The matching is A-IV, B-III, C-II and D-I.

Q7 (NCERT Exemplar, 3 marks) Which conclusion from the list below cannot be drawn from Rutherford's alpha particle scattering experiment? (i) The atom is mostly empty space. (ii) The atomic radius is about 10^-10 m and the nuclear radius about 10^-15 m. (iii) Electrons travel in circular paths of fixed energy, known as orbits. (iv) The nucleus and the electrons attract each other by electrostatic force. Answer: The experiment supports (i), (ii) and (iv). Most alpha particles passing through shows that the atom is mostly empty. The tiny fraction scattered through large angles fixes the small nuclear size and the large atomic size. The electrostatic attraction between the positive nucleus and the negative electrons is the force that holds the atom together. Statement (iii) is the one that cannot be derived: fixed circular orbits with fixed energy were proposed later by Bohr to explain spectra, and the scattering experiment says nothing about them. The answer is (iii). Source: NCERT Exemplar Class 11 Chemistry, Unit 2, MCQ 1.

Q8 (NCERT Exemplar, 3 marks) Which of the following options does not represent the ground state electronic configuration of an atom? (i) 1s2 2s2 2p6 3s2 3p6 3d8 4s2 (ii) 1s2 2s2 2p6 3s2 3p6 3d9 4s2 (iii) 1s2 2s2 2p6 3s2 3p6 3d10 4s1 (iv) 1s2 2s2 2p6 3s2 3p6 3d5 4s1 Answer: Count the electrons to identify the element and then check the ground state. Option (i) has 28 electrons, which is nickel, and its ground state is 3d8 4s2, so it is correct. Option (iii) has 29 electrons, copper, whose ground state is 3d10 4s1, so it is correct. Option (iv) has 24 electrons, chromium, whose ground state is 3d5 4s1, so it is correct. Option (ii) also has 29 electrons but keeps 3d9 4s2; for copper this arrangement costs energy, and the observed ground state is the fully filled 3d10 4s1. The option that is not a ground state configuration is (ii). Source: NCERT Exemplar Class 11 Chemistry, Unit 2, MCQ 2.

Q9 (NCERT Exemplar, 3 marks) The number of radial nodes for a 3p orbital is: (i) 3 (ii) 4 (iii) 2 (iv) 1. Answer: The number of radial nodes is n minus l minus 1. For a 3p orbital n is 3 and l is 1, so the count is 3 minus 1 minus 1 = 1. The 3p orbital has one radial node and one angular node, a nodal plane, so its total number of nodes is n minus 1 = 2. The correct option is (iv). Source: NCERT Exemplar Class 11 Chemistry, Unit 2, MCQ 8.

Q10 (NCERT Exemplar, 3 marks) Two atoms are isobars if: (i) they have the same atomic number but different mass number. (ii) they have the same number of electrons but different numbers of neutrons. (iii) they have the same number of neutrons but different numbers of electrons. (iv) the sum of protons and neutrons is the same but the number of protons is different. Answer: The mass number A is the sum of protons and neutrons, so two atoms are isobars when their A values are equal while their atomic numbers are different. That is exactly statement (iv). Statement (i) describes isotopes, which share Z but differ in A. Statement (ii) and (iii) describe species with equal electron or neutron counts, which are not the isobar definition. The correct option is (iv). Source: NCERT Exemplar Class 11 Chemistry, Unit 2, MCQ 7.