Question

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

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

Answer 1

Answer: (B)

(iv) < (ii) < (iii) < (i)

Answer 2

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