Question

An ionic bond is established between a positive ion $A^{+}$ and negative ion $B^{-}$. How many times strength of the ionic bond is affected by doubling the charge on $A^{+}$ and making the radius halved?

Answer 1

$F$(strength of ionic bond) $=k\frac{z_{1}z_{2}}{r^{2}}=k\frac{z_{1}(A)z_{2}(B)}{r^{2}}$
$z_{1}$ is doubled to $2z_{1}$ and $r$ is halved to$(\frac{r}{2})$
Then $F' =k\frac{2Z_{1} (A)z_{2}(B)}{(\frac{r}{2})^{2}}=8k\frac{z_{1}(A)z_{2}(B)}{r^{2}}=8F$
Thus, ionic strength becomes $8$ times.