How to find outer election configuration of an element?

How do you to find outer election configuration of an element? Such as for group Li, Al, Cl? I'd like to know the answers, but I also want to know how to do it to figure out the rest of the work.

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1. Okay! Time to learn! Grab your Periodic Table(or get where you can see one) and lets take a look! The rows(horizontally aligned elements) are called "Periods". H and He are in Period 1; Li, Be, B, C, N, O, F and Ne are in Period 2; K, Ca, Sc, Ti... Br, and Kr are in Period 4; etc. The columns(vertically aligned) are called "Groups", and are identified by a Roman numeral followed by A or B. An atom's outer energy level, or shell, is called its valence level. That's what your looking for. In an electron configuration, the number (1-7) is the energy level, the lower-case letter (s,p,d,f or g)is the sublevel, and the superscript number (that's the little ones higher up than the others)is the number of electrons in that sublevel. It should look kinda like this: 3s^2 3p^5 chlorine(Cl)'s electron configuration Note: On message boards, "^" means the following number(s) are superscript. 3 is the energy level, 3s and 3p are sublevels. Get it so far? The 3s sublevel of Cl contains 2 electrons and the 3p contains 5. Now you're probably thinking, "So how do I figure it out?" It's Easy! Just look at the Group Number(e.g., VIA). This doesn't work for the "B" groups, but someone else can explain that later. The Group Number represents how many electrons are in each atom's outer level. The outer(valence) sublevels are always s and p. Also take note that the s sublevel has maximum of 2 electrons, and p has a max of 6. So the electron configuration is the Period Number followed by "s" and then a superscript of the number of electrons in the s sublevel. Then, if there's more than two, put a space followed by the Period Number again, then "p" and a superscript of the number of electrons in the p sublevel. That could be confusing so I'll give some examples: Beryllium(Be) is in Period 2, Group IIA, so it's electron configuration is "2s^2". If there's more than two electrons in its outer shell, the sum of s and p's superscripts must equal the Group Number. Like this: Boron(B) is in Period 2, Group IIIA, so it's electron configuration is "2s^2 2p^1". Phosphorus(P) is in Period 3, Group VA, so it's electron configuration is "3s^2 3p^3". Just in case, here are some more examples: Al 3s^2 3p^1 K 4s^1 Ne 2s^2 2p^6 Still confused? X=Period Number; A Group IA element has "Xs^1" A Group IIA element has "Xs^2" A Group IIIA element has "Xs^2 Xp^1 A Group IVA element has "Xs^2 Xp^2 A Group VA element has "Xs^2 Xp^3 and so on... A Group VIIIA element has "Xs^2 Xp^6" and its outer shell is completely full. So... I hope you learned a lot from this little lesson! Thank you for letting me help!