Question

What are the possible values of $l$ and $m_l$ for an atomic orbital $4f$?

• A. $n=4$, $l=0$, $1$, $2$, $3$, $m_l=-2$, $-1$, $0$, $+1$, $+2$
• B. $n=4$, $l=3$, $m_l=-3$, $-2$, $-1$, $0$, $+1$, $+2$, $+3$
• C. $n=4$, $l=2$ , $m_l=-2$, $-1$, $0$, $+1$, $+2$, $+3$
• D. $n=4$, $l=0$, $1$ , $m_l=-1$, $0$, $+1$

Answer 1

Answer: (B)

For $n^{\text{th }}$ orbital possible values of azimuthal quantum number, 1 are from 0 to $(n-1)$. Total of $n$ values.
In $4f$ orbital, $n=4$ and $f$-orbital has $1=3$
Magnetic quantum number $m_1$ have values from $-1$ to $+1$ and total of $21+1$ values.
For $1=3$ total; values of $m_1$ are 7 and are:
For $1=3,m_1=-3-2,-1,0,1,2,3$