Question #89d4c

Start by looking for calcium, #"Ca"#, in the Periodic Table. You will find that this element is located in period #4#, group #2# and that it has an atomic number equal to #20#.

This tells you that an atom of calcium contains #20# protons inside the nucleus and #20# electrons surrounding the nucleus.

You can thus say that the electron configuration of an atom of calcium must account for a total of #20# electrons.

As you know, you can use the Aufbau Principle to write the full electron configuration for an atom of calcium.

You will end up with

#"Ca: " 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2#

Now, when calcium loses two electrons, it becomes a cation with an overall #2+# charge. These two electrons will be removed from the fourth energy level, the valence shell of an atom of calcium. More specifically, these two electrons will come from the #42# orbital.

#"Ca: " 1s^2 2s^2 2p^6 3s^2 3p^6 color(red)(cancel(color(black)(4s^2)))#

This implies that the full electron configuration of a calcium action will look like this

#"Ca"^(2+): 1s^2 2s^2 2p^6 3s^2 3p^6#

Finally, to write the abbreviated electron configuration of a calcium cation, you need to look in the Periodic Table for the noble gas that comes immediately before calcium.

This noble gas is neon, #"Ne"#. The full electron configuration of an atom of neon looks like this

#"Ne: " 1s^2 2s^2 2p^6 3s^2 3p^6#

Notice that this is the same electron configuration as the one you have for the calcium cation. To show this, you can write

#"Ca"^(2+): ["Ne"] -># the nogle gas shorthand of a calcium cation

Here #["Ne"]# simply means the full electron configuration of neon.

Notice that you can write the noble gas shorthand of an atom of calcium like this

#"Ca: " ["Ne"] 4s^2#

Once again, the #["Ne"]# simply means the electron configuration of neon, the noble gas that comes immediately before calcium in the Periodic Table.