A student is investigating why table salt \( (\mathrm{NaCl}) \) forms when sodium metal \( (\mathrm{Na}) \) reacts with chlorine gas \( \left(\mathrm{Cl}_{2}\right) \). The student learns that this reaction happens because of how atoms gain or lose electrons to become stable. 1. Briefly explain why AND how the transfer of electrons leads to the formation of NaCl . The transfer of clectrons lead to the formertion of NaCl becense 2. Based on periodic trends, provide 1 example of another pair of elements that would react in a similar way. Justify your answer using at least 1 periodic trend and atomic structure.

1. When sodium (\( \mathrm{Na} \)) reacts with chlorine (\( \mathrm{Cl}_2 \)), each sodium atom loses one electron to become a positively charged ion, \( \mathrm{Na}^+ \). At the same time, each chlorine atom gains one electron to become a negatively charged ion, \( \mathrm{Cl}^- \). The electron transfer allows each atom to achieve a stable electron configuration (a full valence shell). The electrostatic attraction between the oppositely charged \( \mathrm{Na}^+ \) and \( \mathrm{Cl}^- \) ions leads to the formation of the ionic compound sodium chloride (\( \mathrm{NaCl} \)). 2. Another example is the reaction between potassium (\( \mathrm{K} \)) and bromine (\( \mathrm{Br}_2 \)). Potassium, like sodium, is an alkali metal in Group 1 and has one electron in its outermost shell, so it readily loses that electron to form \( \mathrm{K}^+ \). Bromine is a halogen in Group 17 and needs one electron to achieve a complete octet, so it gains an electron to form \( \mathrm{Br}^- \). Periodic trends, such as low ionization energy for alkali metals and high electron affinity for halogens, justify this similar behavior. The electrostatic attraction between \( \mathrm{K}^+ \) and \( \mathrm{Br}^- \) ions results in the formation of an ionic compound similar to \( \mathrm{NaCl} \).