A particular element (A) has one electron in its third shell. There is another element (B) with six electrons in its second shell. (i) How many electrons does A tend to give or take to become stable? (ii) What kind of ion would it form? (iii) How many electrons does B tend to give or take to become stable? (iv) What kind of ion would it form? (v) If A and B were to combine, what kind of bond would be formed? (vi) What would be the formula for the compound thus formed?
Element A tends to give 1 electron to become stable.
It forms a positive ion.
Element B tends to take 2 electrons to become stable.
It forms a negative ion.
An ionic bond would be formed.
The formula for the compound would be A₂B.
Explanation
Element A has the electronic configuration 2, 8, 1. Since its valence shell has less than four electrons, it will donate its one valence electron to achieve a stable octet, forming a positive ion. Element B has the configuration 2, 6. With more than four valence electrons, it will gain two electrons to complete its octet, forming a negative ion. The transfer of electrons from A to B creates an ionic bond. To balance the charges (A⁺ and B²⁻), two A ions are needed for one B ion, resulting in the formula A₂B.