The atomic radii of a divalent cation and a monovalent anion are 0.62 nm and 0.121 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).

The force of attraction between these two ions at their equilibrium interionic separation is .

Explanation:

Given:

The divalent cation and monovalent anion with atomic radii of 0.62 nm and 0.121 nm, respectively.

To find:

Force of attraction between two ions at their equilibrium interionic separation.

Solution:

The charge on divalent cation =

The charge on monovalent anion =

Atomic radius of divalent cation =

Atomic radius of monovalent anion =

Charge on an electron =

The inter ionic seperation between ions = d

The permittivity of vacuum =

The force of attraction between ions is given by:

(negative sign indicates attractive force)

The force of attraction between these two ions at their equilibrium interionic separation is .

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Explanation:

The formula for the force exerted between two ions is  

where the Coulomb constant

k = 8.988 × 10⁹ N·m·C⁻²

Data:

z₁ = +2; r₁ = 0.62    nm

z₂ =  -1;  r₂ = 0.121 nm

Calculations:

r = r₁ + r₂ = 0.62 + 0.121 = 0.741 nm