(i) Write the electron-dot structures for sodium, oxygen and magnesium.
(ii) Show the formation of `Na_(2)O` and `M_(g)O` by the transfer of electrons.
(iii) What are the ions present in these compounds?

Explanation

(i) Sodim atom has 1 electorn in its outermost shell, so the electron dot structure of sodium is Na (1 dot on the symbol Na). Oxygen atom has 6 electrons in the outermost shell, so the electron-dot structure of oxygen is `underset(..)overset(..)(O: )` (6 dots on the symbol O).
(ii). The formation of sodium oxide `(Na_(2)O)` can be explained as follows: A sodium atom has 1 outermost electron to donate but an oxygen atom requires 2 electrons to achieve the 8-electrons structure. So, two sodim atoms will combine with one oxygen atom to form sodium oxide compound.
In the formation of sodium oxide, two sodium atoms transfer their 2 outermost electrons to an oxygen atom. by losing 2 electrons the two sodium atoms form two sodium ions `(2Na^(+))`. And by gaining 2 electrons,
the oxygen atom forms an oxide ion `(O^(2-))`.
The oppositely charged sodium ions and oxide ion are held together by stong electrostatic forces of attraction to form the ionic sodium oxide compound `2Na^(+)O^(2-)` or `Na_(2)O`.
(iii). The ions present in sodium oxide compound `(Na_(2)O)` are: sodium ions `(2Na^(+))` and oxide ion `(O^(-))`.