51 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: (C) 3
The valence configuration of phosphorus is 3s² 3p³. The three electrons in the 3p subshell have parallel spins. The number of exchange pairs is given by n(n - 1)/2 = 3(2)/2 = 3.
Answer: (B) Statement I is correct but Statement II is incorrect
Statement I is correct because Cr acquires the more stable half-filled d⁵ configuration. Statement II is incorrect because half-filled subshells possess maximum (larger) exchange energy, not smaller exchange energy, which accounts for their extra stability.
Answer: (C) 10
In a d⁶ system, by Hund's rule, 5 electrons have parallel spins (spin up) and 1 electron has opposite spin (spin down). Exchange only occurs between electrons with parallel spins. Thus, number of exchange pairs = ⁵C₂ + ¹C₂ = 10 + 0 = 10.
Answer: (A) 13
In a 3d⁸ configuration, Hund's rule gives 5 electrons with spin +1/2 and 3 electrons with spin -1/2. Exchange interactions occur only between electrons with parallel spins. Exchanges among spin-up electrons = 5(4)/2 = 10; exchanges among spin-down electrons = 3(2)/2 = 3. Total exchange interactions = 10 + 3 = 13.
Answer: (D) Nitrogen has a stable half-filled 2p subshell requiring extra energy to remove an electron
Nitrogen has the configuration 1s² 2s² 2p³, where the 2p subshell is exactly half-filled and possesses extra stability. Oxygen has 1s² 2s² 2p⁴, in which removal of one electron yields a stable half-filled 2p³ configuration, making it easier to ionize than nitrogen.
Answer: (A) 4
In the high-spin d⁵ state, all 5 electrons have parallel spins, yielding 5(4)/2 = 10 exchange interactions. When one electron flips spin, there are 4 electrons with spin +1/2 and 1 electron with spin -1/2. The exchange pairs become 4(3)/2 + 0 = 6. The number of exchange interactions lost is 10 - 6 = 4.
Answer: (C) Zn²⁺ (Z = 30)
Cu⁺ has 28 electrons with the configuration [Ar] 3d¹⁰ 4s⁰. Zn²⁺ (Z = 30) loses its two 4s electrons to also yield [Ar] 3d¹⁰ 4s⁰. Therefore, Cu⁺ and Zn²⁺ are isoelectronic with a stable, completely filled d¹⁰ subshell.
Answer: (B) [Ar] 3d¹⁰ 4s¹
A completely filled 3d subshell is extra stable, so one 4s electron moves to 3d, giving 3d¹⁰ 4s¹ instead of 3d⁹ 4s².
NCERT Class 11
Answer: (A) 21
In the 4f⁷ configuration, all 7 electrons have parallel spins according to Hund's rule. The number of exchange pairs is ⁷C₂ = (7 × 6) / 2 = 21.
Answer: (D) Silver (Z = 47)
Silver (Z = 47) possesses the ground-state electronic configuration [Kr] 4d¹⁰ 5s¹ instead of [Kr] 4d⁹ 5s². This shift of an electron from the 5s to the 4d orbital occurs to achieve the extra stability associated with a completely filled 4d¹⁰ subshell.
Answer: (A) [Ar] 4s¹
By the (n + l) rule, 4s (n + l = 4) has lower energy than 3d (n + l = 5), so the 19th electron enters 4s.
NCERT Class 11
Answer: (D) Fe²⁺ (Z = 26)
The ground-state configurations of the 3d subshells are: V²⁺ is 3d³ (3 unpaired electrons), Ni²⁺ is 3d⁸ (2 unpaired electrons), Ti⁺ is 3d² 4s¹ (2 unpaired electrons in 3d), and Fe²⁺ is 3d⁶ (4 unpaired electrons in 3d). Thus, Fe²⁺ has the maximum number of unpaired electrons in the 3d subshell.