Question

The atomic radii of Mg2+ and F- ions are 0.072 and 0.133 nm, respectively.(a) Calculate the force of attraction between these two ions at their equilibrium inter-ionic separation (i.e., when the ions just touch each other).

Answer 1

`1.09527* 10^(-8)\ N`

Step-by-step explanation:

`q_1` = Mg ion =
`+2q`

`q_2` = F ion =
`-q`

q = Charge of electron =
`1.6* 10^(-19)\ C`

r = Distance between ions =
`0.072+0.133\ nm`

k = Coulomb constant =
`8.99* 10^(9)\ Nm^2/C^2`

Electrical force is given by

`F=-(kq_1q_2)/(r^2)\\\Rightarrow F=-(8.99* 10^9* 2* 1.6* 10^(-19)* -1* 1.6* 10^(-19))/([(0.072+0.133)* 10^(-9)]^2)\\\Rightarrow F=1.09527* 10^(-8)\ N`

The attractive force is
`1.09527* 10^(-8)\ N`