Question

Sodium reacts with chlorine to form the compound sodium chloride $2Na+C{{l}_{2}}\to 2NaCl$ . How would you describe, in terms of electron arrangement, the types of bonding in a molecule of chlorine?

Answer 1

Hint: Sodium is a very reactive element in the periodic table. It lies in alkali metal series, whereas chlorine is a pale-yellow colored element which releases noxious gas. When these elements combine together, they form sodium chloride.

Complete answer:
Sodium atoms have one electron in their valence shell. To form a stable octet, it does not require much energy to remove an electron and hence it forms $N{{a}^{+}}$ ion, whereas chlorine atoms have seven electrons in their valence shell. To form a stable octet, it requires a high amount of energy to gain one electron and form $C{{l}^{-}}$ ion. The reaction given is:
$2Na+C{{l}_{2}}\to 2NaCl$
It is an ionic reaction, that requires number of steps:
First, sodium is converted from solid state to gaseous state. This reaction is an endothermic reaction. Now, let us see the conversion:
$N{{a}_{(s)}}\to N{{a}_{(g)}}$
Then loss of electrons takes place from sodium atom and forms $N{{a}^{+}}$ ion. The amount of energy consumed is first ionization energy.
$N{{a}_{(g)}}\to N{{a}^{+}}_{(g)}+{{e}^{-}}$
When the bonds of half $C{{l}_{2}}$ molecule breaks, it forms chlorine atoms.
$\dfrac{1}{2}C{{l}_{2(g)}}\to C{{l}_{(g)}}$
Addition of electrons takes place to form $C{{l}^{-}}$ ion. The amount of energy released is equal to the electron affinity.
$C{{l}_{(g)}}+{{e}^{-}}\to C{{l}^{-}}_{(g)}$
When the sodium ion and chloride ion react with each other, they give sodium chloride.
$N{{a}^{+}}+C{{l}^{-}}\to NaCl$
Here, the octet of sodium ion and chloride is completely fulfilled, that is, sodium ion has $2,8$ configuration and chloride ion has$2,8,8$ configuration. In the question, reaction given is:
$2Na+C{{l}_{2}}\to 2NaCl$
Chlorine is a halogen; therefore, it always exists as a diatomic molecule. To balance the chemical reaction, we add coefficient of in front of sodium atom, and therefore the product formed is $2NaCl$
The type of bonding a chlorine molecule shows a covalent bonding which holds the atom tightly

Note:When loss and gain of electrons takes place, it results in the formation of cations and anions. Ions formed have eight electrons in their valence shell. Then atoms gain electrons, it forms anion and when atom loses electrons it forms cation.