Question

Which of the following is not associated with an ionic bond?

• A. $\mathrm{NaCl}$
• B. ${\mathrm{MgCl}}_2$
• C. ${\mathrm{AlCl}}_3$
• D. ${\mathrm{CaCl}}_2$

Answer 1

The correct option is C

${\mathrm{AlCl}}_3$

Explanation for the correct option:

Covalent bonds are the bonds, that are formed by the sharing of electrons between the same and different kinds of atoms. Covalent bonds are the result of the electrostatic force of attraction.
(C) Aluminium chloride $({\mathrm{AlCl}}_3)$: Aluminum has 6 electrons in its outer shell even after making three bonds. Hence $\mathrm{Aluminium}\left(\mathrm{Al}\right)$ atom in Aluminium chloride is electron deficient as its octet is incomplete. To complete its octet Aluminium will share its electron with Chlorine $\left(\mathrm{Cl}\right)$ and Aluminium forms covalent bonds with Chlorine atoms.
Explanation for the incorrect options:

The bond formed by the transfer of electron(s) is called an ionic bond. In ionic bond formation, one atom loses an electron and the other atom accepts it.

(A) Sodium chloride $\left(\mathrm{NaCl}\right):$ In Sodium chloride, Sodium ($\mathrm{Na}$) gives one of its electrons to form ${\mathrm{Na}}^{+}$ and Chlorine accepts that one electron from Sodium and forms ${\mathrm{Cl}}^{-}$.

(B) Magnesium chloride $\left({\mathrm{MgCl}}_2\right)$ : During the formation of Magnesium chloride, Magnesium $\left(\mathrm{Mg}\right)$ loses 2 electrons and forms ${\mathrm{Mg}}^{2+}$ and Chlorine accepts those two electrons and forms $2{\mathrm{Cl}}^{-}$.

(D) Calcium chloride $\left({\mathrm{CaCl}}_2\right)$ : During the formation of Calcium chloride, Calcium $\left(\mathrm{Ca}\right)$ loses 2 electrons and Chlorine accepts those two electrons and forms $2{\mathrm{Cl}}^{-}$.

Thus, option (C) i.e., Aluminium chloride is the correct option.