Q1 / 90 · JEE Main 2012 (26 May Online) · Chemistry

The following sets of quantum numbers represent four electrons in an atom. (i) $n=4, l=1$ (ii) $n=4, l=0$ (iii) $n=3, l=2$ (iv) $n=3, l=1$ The sequence representing increasing order of energy, is

  • A
    (iii) < (i) < (iv) < (ii)
  • Correct answer
    B
    (iv) < (ii) < (iii) < (i)
  • C
    (i) < (iii) < (ii) < (iv)
  • D
    (ii) < (iv) < (i) < (iii)

Solution

(i) $4 p$ (ii) $4 \mathrm{~s}$ (iii) $3 d$ (iv) $3 p$ According to Bohr Bury's $(n+\ell)$ rule, increasing order of energy will be (iv) < (ii) $ < $ (iii) $ < $ (i). Note: If the two orbitals have same value of $(n+\ell)$ then the orbital with lower value of $n$ will be filled first.

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

Relevant Questions