Question

calculate attractive and repulsive forces in ionic bond of kbr. atomic radii: $r_{k+}=0.138$ nm $r_{br-}=0.196$ nm

Explanation:

Step1: Recall ionic - bond force formula

The attractive force in an ionic bond is given by Coulomb's law $F_{attractive}=k\frac{q_1q_2}{r^2}$, where $k = 8.99\times10^{9}\ N\cdot m^{2}/C^{2}$, $q_1$ and $q_2$ are the charges of the ions, and $r$ is the distance between the ions. For $K^+$ and $Br^-$, $q_1 = + 1e$ and $q_2=-1e$ ($e = 1.6\times10^{-19}\ C$), and $r=r_{K^+}+r_{Br^-}$. The repulsive force is due to the overlap of electron - clouds and can be modeled by a power - law function $F_{repulsive}=A/r^n$, where $A$ is a constant and $n$ is typically in the range of 8 - 12. First, calculate the distance between the ions.
$r=r_{K^+}+r_{Br^-}=0.138\ nm + 0.196\ nm=0.334\ nm = 0.334\times10^{-9}\ m$

Step2: Calculate attractive force

$F_{attractive}=k\frac{q_1q_2}{r^2}$, substituting $k = 8.99\times10^{9}\ N\cdot m^{2}/C^{2}$, $q_1 = 1.6\times10^{-19}\ C$, $q_2=-1.6\times10^{-19}\ C$ and $r = 0.334\times10^{-9}\ m$
\[

$$\begin{align*} F_{attractive}&=8.99\times 10^{9}\times\frac{(1.6\times 10^{-19})\times(- 1.6\times 10^{-19})}{(0.334\times10^{-9})^2}\\ &=8.99\times10^{9}\times\frac{- 2.56\times10^{-38}}{1.11556\times10^{-19}}\\ &\approx - 2.06\times10^{-9}\ N \end{align*}$$

\]
The negative sign indicates an attractive force. The repulsive force calculation is more complex as it requires knowledge of the constants $A$ and $n$. But if we just consider the general form, at the equilibrium distance in the ionic bond, the net force $F_{net}=F_{attractive}+F_{repulsive}=0$. So $F_{repulsive}=-F_{attractive}\approx2.06\times10^{-9}\ N$ at equilibrium.

Answer:

The attractive force $F_{attractive}\approx - 2.06\times10^{-9}\ N$ and the repulsive force at equilibrium $F_{repulsive}\approx2.06\times10^{-9}\ N$