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Question:
Compare the energies of following sets of quantum numbers for multielectron system.
(A) n = 4, 1 = 1
(B) n= 4, l = 2
(C) n = 3, l = 1
(D) n = 3, l = 2
(E) n = 4, 1 = 0
Choose the correct answer from the options given below :
Compare the energies of following sets of quantum numbers for multielectron system.
(A) n = 4, 1 = 1
(B) n= 4, l = 2
(C) n = 3, l = 1
(D) n = 3, l = 2
(E) n = 4, 1 = 0
Choose the correct answer from the options given below :
(A) n = 4, 1 = 1
(B) n= 4, l = 2
(C) n = 3, l = 1
(D) n = 3, l = 2
(E) n = 4, 1 = 0
Choose the correct answer from the options given below :
Updated On: Nov 23, 2024
\((B)>(A)>(C)>(E)>(D)\)
\((E)>(C)<(D)<(A)<(B)\)
\((E)>(C)>(A)>(D)>(B)\)
\((C)<(E)<(D)<(A)<(B)\)
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The Correct Option is D
Solution and Explanation
In multielectron systems, the energy of an electron in an orbital depends on both the principal quantum number (\(n\)) and the azimuthal quantum number (\(l\)). The energy increases as the value of \(n + l\) increases. For orbitals with the same \(n + l\), the one with the lower \(n\) has lower energy.
- (A) \(n + l = 4 + 1 = 5\)
- (B) \(n + l = 4 + 2 = 6\)
- (C) \(n + l = 3 + 1 = 4\)
- (D) \(n + l = 3 + 2 = 5\)
- (E) \(n + l = 4 + 0 = 4\)
Order by \(n + l\):
\((C) = (E) < (D) < (A) < (B)\)
For orbitals with the same \(n + l\), compare \(n\):
\((C) < (E)\), as \(n = 3\) for (C) and \(n = 4\) for (E).
Final Answer: \((C) < (E) < (D) < (A) < (B)\).
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