Question

What is the full ground state electron configuration of ${{O}^{+}}~$?

Answer 1

We know that to write the electronic configuration of oxygen atom in its neutral state. Then add two electrons to the outermost orbital of the configuration. The key to using this method is to identify the noble gas closest to the desired element that is at a lower energy (has a lower atomic number if I'm loosely speaking).

Complete answer:
For writing the electronic configuration of any element, we have to know its atomic number. The atomic number of an element is the number of protons present inside the nucleus of its atom or the number of electrons present inside a neutral atom of that element. On the periodic table, the element oxygen occupies the position at the crossing of the second period and sixteenth group. The atomic number of oxygen is $8.$ Typically, an atom of $O$ has $8$ electrons, so based on the electron configuration system that would be $1{{s}^{2}}2{{s}^{2}}2{{p}^{4}},\text{ }or~2+2+4=8e-.$
However, ${{O}^{+}}~$ means you've lost an electron, hence the positive charge. Thus, you subtract an electron:
$8-1=7{{e}^{-}}$
$\therefore 1{{s}^{2}}2{{s}^{2}}2{{p}^{4-1}}\Rightarrow 1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}$
Oxygen is the second most electronegative element in the periodic table after fluorine. The oxygen element is relatively stable in its anionic state than other elements because of its higher electronegativity and small size. The small size of oxygen is due to its position in the periodic table. As you know, the size of an atom increases as we go down the group and across a period.

Note:
Remember that even if more electronegative elements are quite stable in their anionic forms, they cannot exist for much time if they are not supported by external agents. An example of such an external agent is water. The water molecules dissociate and surround these anions to stabilise the excess negative charge.