What is the ground state electron configuration for #"O"^(2-)# ion?

1 Answer
Oct 26, 2016

Based on the periodic table, #"O"# is atomic number #8#, which means it has #8# electrons. The first few atomic orbitals are #1s#, #2s#, and #2p#.

Each orbital can hold #2# electrons maximum, and there are #2l+1# of each type of orbital (#s,p,d,f,g,...#), where #l = 0# corresponds to an #s# orbital, #l = 1# means #p# orbital, and so on.

So, the configuration for neutral #"O"# atom is:

#1s^2 2s^2 2p^4#

where the #2p# orbitals hold the #4# electrons like this:

#ul(uarr darr) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr))#

When oxygen gains two electrons, it acquires a charge of #2-#. Therefore, the two electrons, which go into the highest-energy atomic orbitals, give you a new configuration of

#color(blue)(1s^2 2s^2 2p^6)#

for #"O"^(2-)#, which is the same electron configuration as #"Ne"#.

That is known as a noble-gas-like configuration, and in general it is stable if oxygen is like that in a compound (having #8# electrons in its #2s# and #2p# orbitals combined).