An element X has six electrons in its outer shell and forms a diatomic molecule. (i) Why would that be so? (ii) What kind of bond would it form? (iii) Draw the structure of the molecule it would form. (iv) A certain other element Y has two electrons in its second shell. Draw the structure of the molecule that X would form with Y.
Element X has six electrons in its outer shell and requires two more electrons to complete its octet and become stable. By forming a diatomic molecule, two atoms of X share two electrons each, fulfilling their need for two additional electrons.
Since the atoms share electrons to complete their octet, they form a covalent bond. Specifically, because two pairs of electrons are shared between the two atoms, they are held together by a double bond.
The structure of the diatomic molecule of X can be represented as X=X, where the two lines indicate a double bond formed by the sharing of two pairs of electrons.
Element Y has two electrons in its second shell (less than four), so it tends to donate its valence electrons to achieve stability. Element X (with more than four valence electrons) tends to gain two electrons. Y donates its two electrons to X, forming an ionic bond. The structure shows Y as a Y²⁺ ion and X as an X²⁻ ion, represented as Y²⁺ X²⁻.
Explanation
The question maps directly to the formation of an oxygen molecule (element X) and an ionic compound like magnesium oxide (element Y acting as a metal). For part (i) and (ii), the textbook explains that an atom with six valence electrons needs two more to complete its octet, leading to the sharing of two electron pairs (a double covalent bond) in a diatomic molecule (O=O). For part (iv), the context explicitly states that atoms with less than four valence electrons generally donate them, while those with more than four gain them. Thus, Y (2 valence electrons) donates to X (6 valence electrons), resulting in an ionic bond rather than a covalent molecule.