Question

What is the correct orbital designation of an electron with the quantum number,
$n=4,l=3,m=-2,s=\frac{1}{2}?$

• A. $3s$
• B. $4f$
• C. $5p$
• D. $6s$

Answer 1

Answer: (B)

The value of n decides the shell of electrons.
The value of $l$ decides the orbital; $l=0$ is $s,l=1$ is d and $l=3$ is $f$
Given, $n=4,l=3,m=-2,s=-\frac{1}{2}$
$\therefore$ It is $4f$ orbital.