Question

gThe atomic radii of a divalent cation and a monovalent anion are 0.085 nm and 0.125 nm, respectively. (a) Calculate the force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another)./homework-help/electron-subshell-filled-rare-earth-series-elements-periodic-chapter-2-problem-14qp-solution-9781118324578-exc

Answer 1

Answer:

1.05 × 10⁻⁸ N

Explanation:

Using Coulomb's law

F force of attraction between them = k $\frac{q^{2} }{r^{2} }$ = k $\frac{z1z2q^{2} }{r^{2} }$

K coulomb constant = 8.99 × 10⁹ N m²C⁻²

Z₁ = + 2 ( valency electron)

Z₂ = -1 ( valency electron)

r is the distance between then since they just touch one another = 0.085 nm + 0.125 nm = 0.21 nm = 0.21 × 10⁻⁹ m

q charge = 1.602 × 10⁻¹⁹ C of an electron

F = ( 2 × 1 × 8.99 × 10⁹ N m²C⁻² × (1.602 × 10⁻¹⁹ C)² / ( 0.21 × 10⁻⁹ m )² = 1046.34 × 10 ⁻¹¹ = 1.05 × 10⁻⁸ N