Question

The atomic radii of Mg2+ and Fions are 0.072 and 0.133 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) b) What is the force of repulsion at this same separation distance

Answer 1

The forces between Mg2+ and F- ions are calculated using Coulomb's Law. The attraction force uses the ions' charges and their separation distance, while the repulsive force is not applicable as they have opposite charges.

Step-by-step explanation:

The question involves calculating the force of attraction and the force of repulsion between Mg2+ and F- ions at their equilibrium interionic separation, as well as understanding lattice energy and ionization energy in the context of ionic compounds. To calculate these forces, one would typically use Coulomb's Law, which states that the force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.

For part (a), the equilibrium interionic separation is the sum of the radii of Mg2+ (72 pm) and F- (133 pm), so the separation distance is 205 pm. The force of attraction can be calculated by plugging these values into Coulomb's Law formula:

F = (k * |q1 * q2|) / r^2

Where k is Coulomb's constant, q1 and q2 are the charges, and r is the separation distance. The proportionality constant mentioned in the reference is used as k in the formula.

For part (b), the force of repulsion would not exist between Mg2+ and F- as they are oppositely charged and thus always attract each other. The repulsive force condition would apply if both ions had the same charge.

Answer 2

`\large \boxed{\text{(a) }-1.10 * 10^(-8) \text{ N}; \text{(b) }+1.10 * 10^(-8) \text{ N}}`

Step-by-step explanation:

(a) Attractive force

To calculate the attractive force (F), we can use Coulomb's Law:

`F =k(Q_(1)Q_(2))/(r^(2))`

where

Q₁ and Q₂ are the charges on the ions,

r is the distance between them, and

k = the Coulomb constant

Data:

Q₁ = 2+

Q₂ = 1-

r₊ = 0.072 nm

r₋ = 0.133 nm

k = 8.988 × 10⁹ N· m²C⁻²

Calculations:

Q₁ = (+2) × 1.602 × 10⁻¹⁹ C = +3.204 × 10⁻¹⁹ C

Q₂ = (-1) × 1.602 × 10⁻¹⁹ C = -1.602 × 10⁻¹⁹ C

r = r₊ + r₋ = 0.072 nm + 0.133 nm = 0.205 nm = 0.205 × 10⁻⁹ m

`\begin{array}{rcl}F & = & 8.988 * 10^(9)*(3.204 * 10^(-19) * (-1.602) * 10^(-19))/((0.205 * 10^(-9))^(2))\\\\& = & \mathbf{-1.10 * 10^(-8)} \textbf{ N}\\\end{array}\\\text{The attractive force between the two ions is $\large \boxed{\mathbf{-1.10 * 10^(-8)} \textbf{ N}}$}`

(b) Repulsive force

The equilibrium position is reached when the ions just touch each other. If they come any closer, the nuclear repulsions will outweigh the Coulomb attraction.

Thus, the repulsive force is equal and opposite to the attractive force.

`\text{The repulsive force is $\large \boxed{\mathbf{+1.10 * 10^(-8)} \textbf{ N}}$}`