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 :

• A. \((B)>(A)>(C)>(E)>(D)\)
• B. \((E)>(C)<(D)<(A)<(B)\)
• C. \((E)>(C)>(A)>(D)>(B)\)
• D. \((C)<(E)<(D)<(A)<(B)\)

Answer 1

The Correct Option is D

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: \((C) < (E) < (D) < (A) < (B)\).