The electrons identified by quantum numbers n and l :(a) n = 4, l = 1 (b) n = 4, l = 0 (c) n = 3, l = 2 (d) n = 3, l = 1 Can be placed in order of inc - Chemistry MCQ AIEEE 2012 2026 | ExamTest.live

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The electrons identified by quantum numbers n and l :(a) n = 4, l = 1 (b) n = 4, l = 0 (c) n = 3, l = 2 (d) n = 3, l = 1 Can be placed in order of increasing energy as:

Solution & Step-by-step Explanation

According to the (n + l) rule:(a) n + l = 4 + 1 = 5 (4 p)(b) n + l = 4 + 0 = 4 (4 s)(c) n + l = 3 + 2 = 5 (3 d)(d) n + l = 3 + 1 = 4 (3 p)Energy order:For different (n + l), lower (n + l) has lower energy.So, {d, b} < {c, a} .For same (n + l), lower n has lower energy.Between d and b : n_{d} = 3, n_{b} = 4 ⟹ d < b .Between c and a : n_{c} = 3, n_{a} = 4 ⟹ c < a .Total order: (d) < (b) < (c) < (a).

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The electrons identified by quantum numbers n and l :(a) n = 4, l = 1 (b) n = 4, l = 0 (c) n = 3, l = 2 (d) n = 3, l = 1 Can be placed in order of increasing energy as:

(c) < (d) < (b) < (a)

(d) < (b) < (c) < (a)

(b) < (d) < (a) < (c)

(a) < (c) < (b) < (d)

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