Question

For principal quantum number $n=4$, the total number of orbitals having $l=3$ is:

• A. $3$
• B. $7$
• C. $5$
• D. $9$

Answer 1

The correct option is B $7$

For $n=4$ and $l=3$, the orbital is $4f$.

Number of values of $m=$ no. of orbitals$=(2l+1)=7$.

Hence, the correct option is $B$