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

Answer:

$1.09527\times 10^{-8}\ N$

Explanation:

$q_1$ = Mg ion = $+2q$

$q_2$ = F ion = $-q$

q = Charge of electron = $1.6\times 10^{-19}\ C$

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

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Electrical force is given by

$F=-\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=-\dfrac{8.99\times 10^9\times 2\times 1.6\times 10^{-19}\times -1\times 1.6\times 10^{-19}}{[(0.072+0.133)\times 10^{-9}]^2}\\\Rightarrow F=1.09527\times 10^{-8}\ N$

The attractive force is $1.09527\times 10^{-8}\ N$