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Hi I'm not sure what you mean by thoroughly but n denotes the period on the periodic table so in this case if n= 2 we are looking at period 2 so all of the elements from Lithium to Neon, l can be any value up to the n-1 so if we have n-2 the maximum value of l=1 but also includes 0. A general rule that may help you is if l=0 we are looking at the s subshell if l=1 we are looking at the p subshell and if l=2 we are looking at the d subshell now since we know the period is two and we are looking at the p subshell there are 6 possible elements with those quantum numbers from Boron to Neon and we would only be able to refine it a little more if we had values of Ml and Ms. Hope that helps!
A way that makes it a lot easier for me is thinking about certain orbitals as a multiple of n and p. For example you said n=2 and l =1. From class we know that when l=1 we are talking about p orbitals. n*p in this case would give us the 2p orbital. from the periodic table there are six elements that have this orbital partially filled (5) and then the noble gas Neon which has it fully filled. I'm a little bit confused on your question though, could you elaborate a little?
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