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

Answer: 8

$F$(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 $F^{\prime }=k\frac{2Z_1(A)z_2(B)}{(\frac{r}{2}{)}^2}=8k\frac{z_1(A)z_2(B)}{r^2}=8F$
Thus, ionic strength becomes $8$ times.