NEETChemistryStructure of Atom
Match List-I with List-II for a multi-electron atom: List-I (Quantum numbers) List-II (Relative energy) (A) n=4, l=1 (I) Highest energy (B) n=4, l=2 (II) Second highest energy (C) n=3, l=2 (III) Third highest energy (D) n=5, l=0 (IV) Lowest energy Choose the correct answer from the options given below:
Options
- A(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
- B(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
- C(A)-(III), (B)-(I), (C)-(II), (D)-(IV)
- D(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
Correct answer
A. (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Step-by-step solution
The energy of an orbital in a multi-electron atom is determined by the (n+l) rule. First, calculate the (n+l) value for each given state: (A) n=4, l=1 4p orbital; (n+l) = 4 + 1 = 5 (B) n=4, l=2 4d orbital; (n+l) = 4 + 2 = 6 (C) n=3, l=2 3d orbital; (n+l) = 3 + 2 = 5 (D) n=5, l=0 5s orbital; (n+l) = 5 + 0 = 5 The orbital with the highest (n+l) value has the highest energy. Thus, (B) has the highest energy (I). For orbitals with the same (n+l) value, the one with the higher principal quantum number ( n ) has higher e