Quote (RzChaos @ Nov 2 2016 09:04pm)
Fill your orbitals out in order until 35 electrons have been used. Count how many are in P orbitals.
1S (2 Electrons), 2S (2 Electrons), 2P (6 Electrons), 3S (2 Electrons), 3P (6 Electrons), 4S (2 Electrons), 3D (10 electrons), 4P (5 Electrons)
2P (6) + 3P (6) + 4P (5) = 17 electrons in P orbitals.
Question 2 is essentially the same, the only trick is to know what the L value represents. L = 0 will be the S Oribtal.
1S (2). 2S (2), 2P (6), 3S (2), 3P (6), 4S (2) = 20 Electrons
Four S orbitals are filled with 2 electrons. = 8 electrons are in orbitals with L = 0
n = 4 means the 4th "Shell".
The 4th shell can contain: 4S, 4P, 4D, and 4F orbitals. These can contain 2, 6, 10, and 14 electrons for a total of 32 electrons in the fourth shell.
Ne has an atomic number of 10, so it has 10 electrons. You can then list out the orbitals and their electrons as we did in questions 1 and 2: 1S(2), 2S(2), 2P(6)...and realize that at 10 electrons, all the oribtals are completely filled, for no unpaired electrons.
P has an atomic number of 15, so it has 15 electrons. You can do this the same way as we did with Neon, or you can look at the group that P falls into (16), and recognize that elements in group 16 will have 3 electrons in their P orbitals.
The question now is, with three electrons in the P-Orbitals, how many are unpaired, which requires us to know the rules of filling orbitals. One of the rules essentially states that electrons would rather be in an orbital of their own, so they will only double up in a orbital if the others are filled. So, with three electrons and three P orbitals, they will fill so that each orbital has 1 electron, and you will have three unpaired.
tyvm for the help