Question

When sodium reacts with fluorine:
A. each fluorine atom loses one electron
B. each sodium atom gains one electron
C. a covalent bond is formed
D. the compound formed is a good conductor of electricity in the molten state.

Answer 1

Hint: Chemical bonding is of different types. Some of them are ionic bonding, covalent bonding, hydrogen bonding, metal bonding, coordinate bond, coordinate covalent bond etc. Generally, ionic bonds are formed between metals and nonmetals. Strength of the bond depends upon the molecules or atoms involved in the process of bond formation.

Complete step by step solution:
Strength of the bond depends upon the molecules or atoms involved in the process of bond formation. Major types of chemical bonding are ionic bonding, covalent bonding, hydrogen bonding and metal bonding.
When an electron leaves one atom and enters into orbit around another, it is said to be an ionic bond. In this bond, oppositely charged ions attract each other. They are formed between metals and nonmetals. Covalent bond is formed when there is a mutual sharing of electrons between two atoms.
Sodium is a metal and fluorine is a non-metal. So the bond between them may be an ionic bond. Sodium has one valence electron and fluorine is highly electronegative. Sodium needs to lose electron and fluorine needs to lose electrons to form a chemical bond. Thus sodium donates its valence electron to fluorine, forming a cation, and fluorine accepts the electron, forming an anion. Thus they distribute the difference in the charge between them.
So we can say that when sodium and fluorine reacts, sodium fluoride is formed which is a good conductor of electricity in the molten state.

Hence the correct option is D.

Note: In the solid state, there are no free ions. Thus it cannot conduct electricity in the solid state. But when it is melted or mixed in water, it forms sodium ions and fluoride ions. They become mobile and have the ability to move freely.