Use these answers for questions 4 - 7.

Question

(A) 1s2 2s22p5 3s23p5
(B) 1s2 2s22p6 3s23p6
(C) 1s2 2s22p62d10 3s23p6
(D) 1s2 2s22p6 3s23p63d5
(E) 1s2 2s22p6 3s23p63d3 4s2
4. An impossible electronic configuration

5. The ground-state configuration for the atoms of a transition element

6. The ground-state configuration of a negative ion of a halogen

7. The ground-state configuration of a common ion of an alkaline earth element

Electron configurations really confuse me. How come E goes from 3d3 to 4s2? I need help with #6 & 7. I don't understand what they're asking or how to figure them out. This is a very weak area for me!

For #6, all of the halogens are in group 17 (VIIA depending upon the system you are using). Therefore, they have a ns2, np5 configuration (note 7 electrons in the outside shell) where n is 2, 3, 4, 5 etc depending upon the particular halogen we are discussing. They become the -1 ion by adding an electron to the p level to make the outside shell ns2, np6.

For #7, all alkaline earths are in group 2 (or IIA) and have a ns2 configuration. If they lose those two electrons, their electronic configuration becomes that of the noble gas two elements below it; for example, ₁₂Mg is 1s2 2s2 2p6 3s2 (12 total). Losing those two electrons makes it 1s2 2s2 2p6 (total 10) which makes it like Ne (12-10=2 elements below Mg).

Answer 1

Therefore, the answers to #6 and #7 are (B) and (E) respectively.