I am studying LS coupling and term symbols. In my textbook, there is an exercise:
Why is it impossible for a $2\ ^{2}\text{D}_{3/2}$ state to exist?
The answer says, the total orbital angular momentum quantum number must less than the principal quantum number. But in my opinion, considering the electron configuration, $1s^{2}2s^{2}2p^{2}$, if the two electrons in $2p$, the outer subshell, have quantum numbers $(1, 1/2)$ and $(1, -1/2)$ respectively which are in the term of $(m_{l}, m_{s})$, $m_{l}$ is the magnetic quantum number, and $m_{s}$ is the spin magnetic quantum number, then the total orbital angular quantum number is $1+1=2$ which is equal to its principal quantum number. This example is conflict against the answer.
Which is wrong, my example or the answer in the textbook?
Best Answer
Your example is wrong. You have two active electrons in the $p$ shell, and their total spin must couple either to $S=0$ or $S=1$, which correspond to singlet ($2S+1=1$) or triplet ($2S+1=3$) states. The target state you've been given is a doublet state (indicated by the $2S+1=2$ superscript), so you've already missed the mark.
More generally, if you want a doublet state (with $S=1/2$), then you need an odd number of electrons, since even numbers of electrons always have integer-valued total spin.
This then puts you into trouble, because having $n=2$ limits you to having only $p$ electrons with $\ell=1$ contributing to the orbital angular momentum, and if you have an odd number of such electrons, then you're restricted to an odd-integer value for $L$. This then completely eliminates the possibility of any $2 \ {}^2\mathrm{D}_J$ state, whatever the $J$.
(If any of the above is unfamiliar, then it's almost certainly because of an incomplete preparation in the quantum-mechanical procedure for adding angular momenta. This is a large and complex topic, and you should take it from the ground up.)
As for your more general question,
No, this is not the case (at least, for excited states). With a half-filled shell, say, on atomic nitrogen, it's perfectly possible to achieve $\rm F$ states with $L=3$, by taking the parallel configuration for the three individual orbital angular momenta.