Question

In an atom the order of increasing energy of electrons with quantum numbers
(i) $n=4,l=1$
(ii) $n=4,l=0$
(iii) $n=3,l=2$ and
(iv) $n=3,l=1$ is

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

Answer 1

Answer: (D)

The order of increase of energy can be calculated from $(n+l)$ rule. If two orbitals have same value of $(n+l)$, the orbital with lower value of $n$ will be filled first.

(i) For $n=4,l=1,(n+l)=4+1=5$

(ii) For $n=4,l=0,(n+l)=4+0=4$

(iii) For $n=3,l=2,(n+l)=3+2=5$

(iv) For $n=3,l=1,(n+l)=3+1=4$

Therefore correct order is

$(iv)<(ii)<(iii)<(i)$.