Question

What is the electronic configuration of Na and Cl before the formation of NaCl?
a.) $Na:[Ne]3{{s}^{1}};Cl:[Ne]3{{s}^{2}}3{{p}^{5}}$
b.) $Na:[Ne]2{{s}^{2}};Cl:[Ne]3{{s}^{2}}3{{p}^{6}}$
c.) $Na:[Ne]2{{s}^{2}};Cl:[Ne]3{{s}^{2}}3{{p}^{4}}$
d.) $Na:[Ne]2{{s}^{1}};Cl:[Ne]3{{s}^{2}}3{{p}^{6}}$

Answer 1

Hint: NaCl is an ionic compound. Therefore, one element loses electrons and the other gains an electron to form subsequent cation and anion. These ions are then bound by strong electrostatic forces of attraction.

Complete step by step answer:
NaCl is an ionic compound, i.e. it is made up of ions – cation and anion. Sodium loses an electron, it is a cation, whereas Chlorine is an anion as it gains an electron.

& Na\to N{{a}^{+}}+{{e}^{-}} \\\ & Cl+{{e}^{-}}\to C{{l}^{-}} \\\ & N{{a}^{+}}+C{{l}^{-}}\to NaCl \\\ \end{aligned}$$ Sodium (Na) – is a group 1 element, i.e. an alkali metal. It has one electron in its outermost orbital. It tends to lose an electron to gain the valence shell configuration like that of Neon. Its configuration is – $$[Ne]3{{s}^{1}}$$. Chlorine (Cl) – is a group 17 element, i.e. a halogen. It has seven electrons in its outermost orbital. It tends to gain an electron to gain the valence shell configuration like that of Argon. Its configuration is – $$[Ne]3{{s}^{2}}3{{p}^{5}}$$. Therefore, the answer is – option (a). Additional Information: Covalent compounds are the type of compounds that are formed by the sharing of electrons. No ions are formed in these compounds. Note: Ionic compounds have the following characteristics – They are bonded by strong electrostatic forces They have high melting and boiling point They don’t conduct electricity as such but they do conduct electricity in aqueous solution. They dissociate in a polar solvent.