90 questions on this topic of Atomic Structure (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.
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Answer: (A) (√3 / 2) (h / 2π)
For an electron, the spin quantum number s is always 1/2. Therefore, S = √[(1/2)(1/2 + 1)] (h / 2π) = √[3/4] (h / 2π) = (√3 / 2) (h / 2π).
Answer: (C) 10
In a d⁶ subshell adhering to Hund's rule, 5 electrons have parallel spin (say spin-up) and 1 electron has opposite spin (spin-down). The number of exchange pairs among parallel-spin electrons is given by n(n - 1)/2. For n = 5, exchanges = 5 × 4 / 2 = 10. For n = 1, exchanges = 0. Total exchanges = 10.
Answer: (D) Chromium (Z = 24)
The total number of electrons is 2 + 8 + 13 + 1 = 24, corresponding to Z = 24, which is Chromium (Cr). Distributing electrons into subshells gives 1s² (K: 2), 2s² 2p⁶ (L: 8), 3s² 3p⁶ 3d⁵ (M: 2 + 6 + 5 = 13), and 4s¹ (N: 1), which reflects the anomalous half-filled d-subshell configuration of Cr.
Answer: (A) 4f
According to the (n + l) rule: 6s has n + l = 6 + 0 = 6. The next orbitals in increasing energy order are 4f (4 + 3 = 7), 5d (5 + 2 = 7), and 6p (6 + 1 = 7). Among these with n + l = 7, the 4f subshell has the lowest n (n = 4) and is filled immediately following 6s (in the idealized Aufbau sequence, beginning at lanthanum/cerium).
Answer: (C) 3
For n = 3 and l = 1 (the 3p subshell), there are three orbitals (m_l = -1, 0, +1). According to Pauli's exclusion principle, each orbital can hold at most one electron with a given spin (e.g., m_s = +1/2). Hence, the maximum number of electrons with the same spin is 3.
Answer: (C) 3
For the first shell (n = 1), l = 0, giving a single 1s orbital. If each orbital can hold 3 electrons (one for each allowed value of m_s: -1/2, 0, +1/2), the 1s subshell will be completely filled with 3 electrons. Hence, the atomic number of the first noble gas would be 3.
Answer: (A) Both Statement I and Statement II are correct
Statement I is the direct standard definition of Pauli's exclusion principle. Statement II is correct because a subshell with azimuthal quantum number l has (2l + 1) orbitals, and since each orbital can hold at most 2 electrons (with opposite spins), the capacity is 2(2l + 1).
Answer: (C) Statement I is incorrect but Statement II is correct
Statement I is incorrect because the magnetic orbital quantum number (mₗ), not the spin quantum number (mₛ), designates the spatial orientation of the orbital. Statement II is correct because Pauli's exclusion principle requires electrons in the same orbital (identical n, l, mₗ) to have opposite spin values (mₛ = +1/2 and -1/2).
Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A)
In single-electron systems like H, He⁺, and Li²⁺, there is no electron-electron repulsion, meaning screening does not occur. As a result, orbital energy depends solely on the principal quantum number n, making 3s, 3p, and 3d degenerate.
Answer: (C) All three have identical energy
In single-electron systems like the hydrogen atom, the energy of an orbital depends solely on the principal quantum number (n). Therefore, 3s, 3p, and 3d orbitals are degenerate and possess the exact same energy.
Answer: (A) (D) > (C) > (B) > (A)
For 4f, n + l = 4 + 3 = 7. For 6s, n + l = 6 + 0 = 6 (n = 6). For 5p, n + l = 5 + 1 = 6 (n = 5). For 4d, n + l = 4 + 2 = 6 (n = 4). When n + l is the same, orbital energy increases with increasing principal quantum number n. Therefore, the order of decreasing energy is 4f (7) > 6s (6, n=6) > 5p (6, n=5) > 4d (6, n=4), which corresponds to (D) > (C) > (B) > (A).
Answer: (A) (↑ ) (↑ ) (↑ )
Hund's rule of maximum multiplicity states that pairing of electrons in degenerate orbitals does not occur until each orbital contains one electron, and these single electrons must possess parallel spins to maximize total spin and minimize electron-electron repulsion via exchange energy. The arrangement (↑)(↑)(↑) has maximum multiplicity (2S + 1 = 4) and hence the lowest energy.