Question

Compare the energy of orbitals for multielectronic species : (A) \(n = 3, \ell = 0, m = 0\)
(B) \(n = 3, \ell = 1, m = -1\)
(C) \(n = 4, \ell = 2, m = 0\)
(D) \(n = 3, \ell = 2, m = 1\)

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

Answer 1

The Correct Option is D

Concept:
For multielectron atoms, the relative energy of orbitals depends on the \((n+\ell)\) rule (Madelung rule). According to this rule: \[ \text{Energy} \propto (n+\ell) \]

• Orbital with lower \(n+\ell\) value has lower energy.
• If two orbitals have the same \(n+\ell\) value, the orbital with lower \(n\) has lower energy.

The magnetic quantum number \(m\) does not affect the energy of orbitals in the absence of an external field. Step 1: Calculate \((n+\ell)\) values. For A: \[ n=3,\; \ell=0 \] \[ n+\ell = 3 \] For B: \[ n=3,\; \ell=1 \] \[ n+\ell = 4 \] For C: \[ n=4,\; \ell=2 \] \[ n+\ell = 6 \] For D: \[ n=3,\; \ell=2 \] \[ n+\ell = 5 \] Step 2: Arrange orbitals according to energy. Higher \((n+\ell)\) value corresponds to higher energy. \[ C(6) > D(5) > B(4) > A(3) \] Step 3: Final energy order. \[ \boxed{C > D > B > A} \]