For Helium atom we know there are spin singlet and spin triplet state corresponding to $S=0$ and $S=1$. But what if the electrons are more than two and what does singlet mean for orbital degree of freedom? For many-electron system, does $L=0$ mean an orbital singlet state, or does $L=2,4\ldots$ also mean an orbital singlet? And what's the relation between orbital singlet and symmetry of the orbital part wave function?
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The terms singlet, doublet, triplet, quartet are names given to part of the Term Symbol in $LS$ coupling. It specifies the value of $S$, but in an encoded form where the left superscript is the numerical value of $2S+1$: $^1S$ denotes $S=0$. It's entirely historical: the classification of levels into terms and their naming was an empirical product of spectroscopy which long pre-dated quantum mechanics. Very old textbooks such as White Atomic Spectra, discuss the pre-quantum spectroscopic evidence in great detail before giving the quantum interpretation.
So to answer your specific questions: yes you can have singlet whenever $S=0$ is a possible result of adding up all the spins. This requires an even number of spins. And it is nothing to do with the value of L, the orbital angular momentum.
