Question

In $NaF$ or $Mg{{F}_{3}}$, which is formed more easily?
(A) $Mg{{F}_{3}}$
(B) $NaF$
(C) Both are easily formed
(D) Can’t say

Answer 1

Hint: Both the compounds are formed by the electron transfer from sodium to fluorine and it’s an ionic compound. The compound with higher ionic nature will be more stable and it will be formed easily. Thus by analyzing the ionic nature of both compounds we will be able to come to a conclusion.

Complete step by step answer:
  -Let us in detail try and understand about ionic bonding. They result from the transfer of electrons from a metal atom to a non-metal atom. An anion is formed when a nonmetal gains an electron and cations are formed when a metal loses its valence electrons.
 - A cation is positively charged ion whereas the anion is negatively charged. Since these two ions have opposite charge, ionic bond is formed as a result of the electrostatic force between them.
 - In the formation of sodium fluoride ($NaF$), the sodium atom loses its single valence electron to the fluorine atom which has the exact space to accept that donated electron. The oppositely charged sodium cation and fluorine anion are attracted through electrostatic force of attraction and hence an ionic bond is formed.
- As we know the $N{{a}^{+}}$ ion is larger than the magnesium ion and it also has the lower charge and the fluoride anion ${{F}^{-}}$ is common in both the compounds. Hence $NaF$ is more ionic than $Mg{{F}_{3}}$ and more stable also. Thus the compound $NaF$ is formed more easily than $Mg{{F}_{3}}$.
So, the correct answer is “Option B”.

Note: It should be noted that covalent bonds differ from ionic bonds. The electrons are shared between the two atoms in the covalent bonds. Also, all the ionic bonds have some covalent character and the larger the difference in electronegativity between the two atoms, the greater will be the ionic character of the interaction.