Question

Which one of the following sets of quantum numbers is not possible for an electron in the ground state of an atom with atomic number $19$?

• A. $n=2,l=0,m=0$
• B. $n=2,l=1,m=0$
• C. $n=3,l=1,m=-1$
• D. $n=3,l=2,m=\pm 2$
• E. $n=4,l=0,m=0$

Answer 1

The correct option is D $n=3,l=2,m=\pm 2$

The following set of quantum numbers is not possible for an electron in the ground state of an atom with atomic number 19. $n=3,l=2,m=\pm 2$. It corresponds to 3d orbital. The electronic configuration of the atom is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^1$
Aufbau principle:
In the ground state of the atoms, the orbitals are filled with electrons in order of increasing energy.
The 4s sub-energy level is at lower energy than the 3d sub-energy level

For 3d, $n=3$ and $l=2$. Hence, $(n+l)=3+2=5$
For 4s, $n=4$ and $l=0$. Hence, $(n+l)=4+0=4$

As the value of $(n+l)$ for $4s$ orbital is less than that of $3d$ orbital, $4s$ orbital is filled before $3d$ orbital.