Use Lewis symbols to determine the formula for the compound that forms between each pair of elements.
a. Sr and Se
b. Ba and Cl
c. Na and S
d. Al and O
Solution Step 1 of 4 Couple a) Strontium (Sr) belongs to II A group, thus it has two electrons in its outer shell Selenium (Se) belongs to VI A group thus it has six electron in its outer shell. Because of Selenium’s high electronegativity respect to Strontium, Selenium can gain electrons from Strontium. In order to reach their stability, Strontium can give two electrons to Selenium, thus leaving the inner shell complete. On the other hand, Selenium can gain two electrons from strontium, completing its outer shell. This is the reason why Sr and O combine together in a 1:1 ratio, thus obtaining the formula SrO. Strontium turns to be a positive, bivalent, ion (because it loses two electrons) Selenium becomes a negative, bivalent ion (because it acquires two electron). Thus the Lewis formula is Step 2 of 4 Couple b) Barium (Ca) belongs to II A group, thus it has two electrons in its outer shell Chlorine (I) belongs to VII A group thus it has seven electrons in its outer shell. Because of the Chlorine’s high electronegativity respect to Barium, Chlorine can gain one electron from Calcium. Unfortunately, when an atom of Chlorine acquires an electron, it completes its outer shell. Calcium, instead, when it loses a single electron doesn’t reach its stability because another electron still remains in the outer shell. That means that Calcium needs to give its other electron to another Chlorine atom. Thus Barium and Chlorine combine together in a 1:2 ratio. The final formula will be BaCl . 2 Barium turns to be a positive, bivalent, ion (because every atom of Barium loses two electrons), Chlorine becomes a negative, monovalent ion (because it acquires one electron). Thus, the Lewis structure of the ionic compound is as follows: