• anonymous
Give the electron configuration for the calcium ion
  • Stacey Warren - Expert
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
  • chestercat
I got my questions answered at in under 10 minutes. Go to now for free help!
  • pottersheep
  • abb0t
Well, everytime that you move from the left to the right, you jump orbitals. S orbitals can only hold 2 electrons. Those, such as Mg (magnesium) has an electron configuration of: \[1s^2 2s^2 2p^6 3s^2\] When you move to elements such as bromine, carbon, nitrogen, and those below it, they are in the P-orbital. The p-orbitals can hold up to SIX electrons. Since it is in the farthest, 6th position. For instance, Neon [Ne] has an electron configuration: \[1s^2 2s^2 2p^6\] however, carbon has: \[1s^2 2s^2 2p^2\] meaning that it's p-orbital only has 2 electrons in it's orbital shell. Now, with this information, try and do it for Ca (calcium). I am not sure if your teacher would like, but you can also use:\[[Ne]3s^2\] for magnesium the [Ne] basically just means represents the electron configuration for everything behind neon, which i wrote before. It's shorrt-hand notation, basically.

Looking for something else?

Not the answer you are looking for? Search for more explanations.