Question

Compare the energy of orbitals for multielectronic species. \[ \begin{array}{c|c|c|c} & n & l & m \\ \hline (A) & 3 & 0 & 0 \\ (B) & 3 & 1 & -1 \\ (C) & 4 & 2 & 0 \\ (D) & 3 & 2 & 1 \\ \end{array} \]

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

Answer 1

The Correct Option is A

Concept: For multielectron atoms, orbital energy depends on the value of \(n+l\). If two orbitals have the same \(n+l\), the one with higher \(n\) has higher energy. Step 1: Identify orbitals \[ (A): n=3, l=0 \Rightarrow 3s \] \[ (B): n=3, l=1 \Rightarrow 3p \] \[ (C): n=4, l=2 \Rightarrow 4d \] \[ (D): n=3, l=2 \Rightarrow 3d \] Step 2: Apply \(n+l\) rule \[ 3s: n+l=3 \] \[ 3p: n+l=4 \] \[ 3d: n+l=5 \] \[ 4d: n+l=6 \] Step 3: Arrange energies Higher \(n+l\) means higher energy. \[ 4d > 3d > 3p > 3s \] Thus \[ C > D > B > A \]