Question

The electrons represented by following set of quantum numbers in decreasing order of energy is:

a) $n=4,l=0,m=0,s=+\frac{1}{2}$
b) $n=3,l=1,m=1,s=-\frac{1}{2}$
c) $n=3,l=2,m=0,s=+\frac{1}{2}$

d) $n=3,l=0,m=0,s=-\frac{1}{2}$

• A. a > b > c > d
• B. c > a > b >d
• C. c > b > a > d
• D. a > c > b > d

Answer 1

The correct option is B c > a > b >d

The energy order of the respective orbital can be compared by their $n+l$ values. In case of equal n+l values of orbital higher energy corresponds to the orbital whose n value is greater.

In (a), the electron is present in 4s-orbital $\Rightarrow n+l=4$

In (b), the electron is present in 3p-orbital $\Rightarrow n+l=4$

In (c), the electron is present in 3d-orbital $\Rightarrow n+l=5$

In (d), the electron is present in 3s-orbital $\Rightarrow n+l=3$

Therefore, the decreasing order of energies of orbitals is: $3d>4s>3p>3s$.