Question

Calculate the force of attraction between a cation with a valence of +2 and an anion with a valence of -1, the centers of which are separated by a distance of 2.9 nm.

Answer 1

The attraction force between the cation and anion is
`- 5.479* 10^(- 11) N`

Solution:

According to the question:

Separation distance between the center of the charges, d = 2.9 nm =
`2.9* 10^(- 9) m`

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 =
`1.6* 10^(- 19) C`

The Coulombian Force is given as:

`F_(c) = (1)/(4\pi\epsilon_(0))(QQ')/(d^(2))`

where

`(1)/(4\pi\epsilon_(0)) = 9* 10^(9)`

Now,

`F_(c) = 9* 10^(9)* (2(1.6* 10^(- 19)* -1(1.6* 10^(- 19))))/((2.9* 10^(- 9))^(2))`

`F_(c) = - 5.479* 10^(- 11) N`

Here, the negative sign is indicative of the attractive nature of force between two oppositely charged particles.