How do valence electrons affect ionic bonding?

Hint: In the formation of ionic bond charge distribution between the two atoms takes place due to the transfer of electrons between the two atoms. The valence electrons are the electrons present in the outermost orbital of the electronic configuration.

Complete step by step answer:
The valence electrons are considered as the electrons present in the outermost electronic configuration of the chemical element which take part in chemical bonding. The chemical bonds formed by the manipulation of the valence electrons are covalent bonds and ionic bonds.
Here in this question we are dealing with ionic bonding. The ionic bonds are the chemical bonds formed by the transfer of valence electrons between the two atoms in order to form the bond. In other words in ionic bonding distribution of charge is seen where one atom forms a cation and one atom forms an anion. The ionic bond is observed between metal atom and non-metal atom.
The ions bonded together by an ionic bond forms an ionic compound.
The atom loses or gains electrons to obey octet rule and forms a stable configuration. Generally metal atoms lose electrons to form a cation and the nonmetal gains electrons to form an anion.
For example: Sodium chloride is an ionic compound formed by sodium cation and chloride anion which are bonded by ionic bonds. The atomic number of sodium is 11 and the electronic configuration of sodium is $[Ne]3{s^1}$. The valence electrons present in sodium is 1, so it loses its one electron to form a $N{a^ + }$ ion of stable configuration. The atomic number of chlorine is 17 and the electronic configuration is $[Ne]3{s^2}3{p^5}$. The valence electron in chlorine is 7 so to obey octet rule (8 electrons) it gains one electron to form $C{l^ - }$.

Note: The ionic compound is neutral in nature so the ions should combine with each other in such a way that they neutralize each other. The ionic bonds are relatively stronger due to the presence of electrostatic force of attraction between the ions.