Ionic potential $(\phi )$ of electropositive elements will be highest in which of the following compounds?
(A) $CsCl$
(B) $MgC{l_2}$
(C) $Al{F_3}$
(D) $S{F_6}$

Hint:As we know that the ionic potential is basically defined as the ratio of the charge to size of any ion. We also know that while moving down the group ionic radius of an atom increases.

Complete step-by-step answer:We all know that the ionic potential gives us the information about the strength of an ion which will be electrostatically attracted or repelled by the ions with opposite charge. It is determined as the ratio as shown below:
$\phi = \dfrac{{charge}}{{ionic\;radius}}$
We can see that the ionic radius is inversely proportional to the ionic potential, so we can say that the ions with smaller radii will have more ionic potential and vice versa. However , ionic potential gives a sense of how strongly or weakly the ion will be electrostatically attracted to ions of opposite charge.
Ionic potential is very useful in understanding the behaviour of hard cations . Cations of low ionic potential like $N{a^ + }$ are typically soluble because they make weak bonds with ${O^{2 - }}$.
Now, we know that the Sulphur hexafluoride ($S{F_6}$) possesses the smallest ionic radii, followed by Aluminium fluoride, Magnesium chloride and lastly Cesium chloride.
Thus the order of ionic potential becomes: $S{F_6} > Al{F_3} > MgC{l_2} > CsCl$

Therefore, the correct answer is (D).

Note:Remember that the ionic potential is also used in determining the polarising power of a cation. Also, it tells us about the charge density of the ion, if the denser is the charge, stronger will be the bond which is formed by that ion with the ions having opposite charge.