Question

Most ionic compound among the following:
(a) Sodium fluoride
(b) Sodium Chloride
(c) Sodium Bromide
(d) Sodium Iodide

Answer 1

Ionic bonds require an electron donor, often a metal and an electron acceptor, often a non – metal. Ionic bonding is observed because metals have few electrons in their outermost shell. By losing these electrons, these metals can achieve noble gas configuration and satisfy the outer rule.

Complete step by step answer:
In this given question, we can see that sodium is common in the given compounds. Therefore, the bother atoms will decide which compound will be the most ionic one.
Sodium will form the most ionic compound with higher electronegative atoms for the smaller anion. As fluorine is highly electronegative than other atoms and it has the smallest size too, sodium fluoride is more ionic than other molecules.
Fluorine, as compound to other elements, is more electronegative because it requires only one electron to complete an octet and attain stability. Also, the distance at which the valence electrons is the minimum as compared with other elements having 7 electrons in the outermost orbit.
Bromine, codium, chlorine and fluorine all belong to group 17 of the periodic table and called halogens. Closer the electrons are to the nucleus of an atom, the attraction between them increases.
Therefore, order of electronegativity will be
$Fluorine > Chlorine > Bromine > Iodine$ with fluorine being the most electronegative.

Note: 1. Electronegativity is the ability of an element to attract the bonded electrons towards itself. It depends upon the atomic number but more importantly on the distance at which electrons reside.
2. Electronegativity increases as we move left to right in a period.
Electronegativity increases as we move down in a group.