a) Nitrogen : $1s^2$$2s^2$$2p^3$ b) Chlorine: $1s^2$$2s^2$$2p^6$$3s^2$$3p^5$ or [Ne]$3s^2$$3p^5$ c) Oxygen: $1s^2$$2s^2$$2p^4$ d) Neon: $1s^2$$2s^2$$2p^6$
Work Step by Step
a) Nitrogen has 7 electrons, and is in group 5A, period 2. Therefore, it occupies the n=2 energy level and has 5 valence electrons in its outermost shell (sublevel). First, you fill the orbitals in energy level n=1 with the maximum 2 electrons ($1s^2$), then you fill the n=2 energy level until you have reached 7 total electrons ($2s^2$$2p3$). Because a p sublevel can hold up to 6 electrons and a s sublevel can hold a maximum of 2 electrons, 1s has 2 electrons, 2s has two electrons, and 2p has 3 electrons. b) Chlorine has 17 electrons, and falls in the n=3 energy level (period 3), with 7 valence electrons (Group 7A). You first fill the n=1 energy level ($1s^2$), then the n=2 energy level ($2s^2$$2p^6$), then the n=3 energy level ($3s^2$$3p^5$) until you have a total of 17 electrons. You can abbreviate this configuration by using the inert Noble gas in the period before it in brackets and working your configuration out from the first s sublevel after it ([Ne]$3s^2$$3p^5$). You have 3 energy levels because Cl falls in the 3rd period and 7 valence electrons because Cl is in Group 7A. c) Oxygen: 2nd period, Group 6A. Energy level is n=2, and there are 6 valence electrons; thus, we obtain: $1s^2$$2s^2$$2p^4$, for a total of 8 electrons. d) Neon: It is a noble gas, which is inert, so all orbitals and energy levels will be full. It is in period 2 and Group 8A, which means the energy level is n=2, and there are 8 valence electrons. Thus, the configuration is $1s^2$$2s^2$$2p^6$, for a total of 10 electrons.