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Q.

The correct decreasing order of energy, for the orbitals having, following set of quantum numbers :

(A) n = 3, l = 0, m = 0

(B) n = 4, l = 0, m = 0

(C) n = 3, l = 1, m = 0

(D) n = 3, l = 2, m = 1

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a

(B) > (D) > (C) > (A)

b

(C) > (B) > (D) > (A)

c

(D) > (B) > (C) > (A)

d

(B) > (C) > (D) > (A)

answer is A.

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Detailed Solution

detailed_solution_thumbnail

(A) n+l=3+0=3

(B) n+l=4+0=4

(C) n+l=3+1=4

(D) n+l=3+2=5

Higher n+l value, higher the energy & if same n+l value, then higher n value, higher the energy.

Thus : D > B > C > A.

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