Question

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

• A. $n=4,l=0,1,2,3$, $m_1=-2,-1,0,+1,+2$
• B. $n=4,l=3$, $m_1=-3,-2,-1,0,+1,+2,+3$
• C. $n=4,l=2$, $m_1=-2,-1,0,+1,+2,+3$
• D. $n=4,l=0,1$, $m_1=-1,0,+1$

Answer 1

The correct option is B $n=4,l=3$, $m_1=-3,-2,-1,0,+1,+2,+3$

For $n^{th}$ orbital possible values of azimuthal quantum number, $l$ are from $0$ to $(n-1)$. Total of $n$ values.

In $4f$ orbital, $n=4$ and f-orbital has $l=3$

Magnetic quantum number $m_l$ have values from $-l$ to $+l$ and total of $2l+1$ values.

For $l=3$ total; values of $m_l$ are $7$ and are:

For $l=3,m_l=-3-2,-1,0,1,2,3$