Q2 / 75 · JEE Main 2026 (05 April Shift 1) · Chemistry

Arrange the following atomic orbitals of multi electron atoms in order of increasing energy.
A. $n = 3, l = 2, m = +1$
B. $n = 4, l = 0, m = 0$
C. $n = 6, l = 1, m = 0$
D. $n = 5, l = 1, m = +1$
E. $n = 2, l = 1, m = +1$
Choose the correct answer from the options given below:

  • A
    $C < D < B < A < E$
  • B
    $B < A < E < C < D$
  • C
    $E < C < D < B < A$
  • Correct answer
    D
    $E < B < A < D < C$

Solution

The energy of an atomic orbital in a multi-electron atom is determined by the $(n + l)$ rule. Orbitals with a lower value of $(n + l)$ have lower energy. If two orbitals have the same $(n + l)$ value, the one with the lower $n$ value has lower energy.

Calculating the $(n + l)$ value for each given orbital:
A. $n = 3, l = 2 \Rightarrow n + l = 5$
B. $n = 4, l = 0 \Rightarrow n + l = 4$
C. $n = 6, l = 1 \Rightarrow n + l = 7$
D. $n = 5, l = 1 \Rightarrow n + l = 6$
E. $n = 2, l = 1 \Rightarrow n + l = 3$

Arranging the orbitals in increasing order of their $(n + l)$ values, we get:
$3 < 4 < 5 < 6 < 7$

This corresponds to the order:
$E < B < A < D < C$

Answer: $E < B < A < D < C$

Hence, the correct answer is option: 'D'.

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