Respuesta :
Answer:
[tex]r_{cation}[/tex] = 0.2162 nm
Explanation:
The force of attraction can be written as:
[tex]F=\frac {K\times |Z_{cation}|e\times |Z_{anion}|e}{r^2}[/tex]
Where,
K is the Coulomb's constant having value 9×10⁹ N. m²/C²
[tex]Z_{cation}[/tex] is the charge on the cation
[tex]Z_{anion}[/tex] is the charge on the anion
e is electronic charge = [tex]1.602\times 10^{-19}\ C[/tex]
r is the distance between the cation and anion and in bond, it is the sum of the ionic radius of cation and anion
So,
Given, F = [tex]1.12\times 10^{-8}\ N[/tex]
[tex]Z_{cation}=Z_{anion}=2[/tex]
Applying in the equation,
[tex]1.12\times 10^{-8}=\frac{9\times 10^9\times 2\times 1.602\times 10^{-19}\times 2\times 1.602\times 10^{-19}}{r^2}[/tex]
[tex]r^2=\frac{9\times \:10^9\times \:2\times \:1.602\times \:10^{-19}\times \:2\times \:1.602\times \:10^{-19}}{1.12\times \:10^{-8}}[/tex]
[tex]r^2=\frac{10^{-29}\cdot \:92.390544}{10^{-8}\cdot \:1.12}[/tex]
[tex]r=0.2872\times 10^{-9}\ m[/tex]
Also, 1 m = [tex]10^{-9}\ nm[/tex]
So, r = 0.2872 nm
Also,
[tex]r=r_{cation}+r_{anion}[/tex]
[tex]r_{cation}[/tex] = 0.071 nm
Thus,
[tex]0.2872\ nm=0.071\ nm+r_{anion}[/tex]
[tex]r_{cation}[/tex] = 0.2162 nm
