Question

During the formation of \[{\text{NaCl}}\] , element \[{\text{X}}\] is oxidised and element \[{\text{Y}}\] is reduced. Then elements \[{\text{X}}\] and \[{\text{Y}}\] are respectively :
(A) \[{\text{Na, C}}{{\text{l}}_2}\]
(B) \[{\text{Na, Cl}}\]
(C) \[{\text{N}}{{\text{a}}^ + }{\text{, Cl}}\]
(D) \[{\text{Cl, Na}}\]

Answer 1

Hint: :Loss of electrons is oxidation and gain of electrons is reduction. Identify the species undergoing oxidation and the species undergoing reduction. Also write the correct chemical formula of different species.

Complete answer:
The compound NaCl is an ionic compound. It contains sodium cations and chloride anions. Sodium cations are represented as \[{\text{N}}{{\text{a}}^ + }\] ions and chloride anions are represented as \[{\text{C}}{{\text{l}}^ - }\] ion. Strong electrostatic forces of attraction are present between sodium cation and chloride anion. Two elements are present in sodium chloride compounds. These two elements are sodium metal and chlorine nonmetal. Sodium metal has highly electropositive and chlorine nonmetal is highly electronegative. When the electronegativity difference between two atoms is high, an ionic bond is formed. The chemical symbol of sodium metal is Na. The chemical symbol of chlorine nonmetal is \[{\text{C}}{{\text{l}}_2}\] .
During formation of sodium chloride molecule, sodium atom loses one electron. Chlorine atom gains this electron. Loss of electrons is oxidation and gain of electrons is reduction. Thus, sodium metal is oxidized to sodium cations whereas chlorine nonmetal is reduced to chloride ions.
\[{\text{Na }} \to {\text{ N}}{{\text{a}}^ + }{\text{ + }}{{\text{e}}^ - } \\
\dfrac{1}{2}{\text{C}}{{\text{l}}_2}{\text{ + }}{{\text{e}}^ - }{\text{ }} \to {\text{ C}}{{\text{l}}^ - } \\
{\text{Na + }}\dfrac{1}{2}{\text{C}}{{\text{l}}_2}{\text{ }} \to {\text{ N}}{{\text{a}}^ + }{\text{ + C}}{{\text{l}}^ - }{\text{ }} \to {\text{ NaCl}} \\\]
Hence, the correct option is the option (B).

Note: During formation of sodium chloride molecule, both sodium cation and chloride anion have completed their octet. Sodium has one valence electron and chlorine has seven valence electrons. Sodium loses one valence electron, so that the penultimate shell containing eight valence electrons becomes the outermost shell. Chlorine gains one electron so that valence shell has now eight electrons.