Answer:
The attraction force between the cation and anion is [tex]- 5.479\times 10^{- 11} N[/tex]
Solution:
According to the question:
Separation distance between the center of the charges, d = 2.9 nm = [tex]2.9\times 10^{- 9} m[/tex]
Also, valence is the no. of electrons required to reach the stable configuration of octet in the outer most orbit or shell of an atom.
Thus
+ 2 means +2e = Q
-1 means - 1e = Q'
where
e = electronic charge = [tex]1.6\times 10^{- 19} C[/tex]
The Coulombian Force is given as:
[tex]F_{c} = \frac{1}{4\pi\epsilon_{0}}\frac{QQ'}{d^{2}}[/tex]
where
[tex]\frac{1}{4\pi\epsilon_{0}} = 9\times 10^{9}[/tex]
Now,
[tex]F_{c} = 9\times 10^{9}\times \frac{2(1.6\times 10^{- 19}\times -1(1.6\times 10^{- 19}))}{(2.9\times 10^{- 9})^{2}}[/tex]
[tex]F_{c} = - 5.479\times 10^{- 11} N[/tex]
Here, the negative sign is indicative of the attractive nature of force between two oppositely charged particles.
