The atomic radii of Mg2+ and F-1 ions 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?
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(a) To calculate the force of attraction between Mg2+ and F-1 ions, we can use Coulomb's law. Coulomb's law states that the force of attraction or repulsion between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The formula is given by:
F = (k * |q1 * q2|) / r^2
Where:
F is the force of attraction/repulsion,
k is the electrostatic constant (9 × 10^9 N m^2 C^-2),
q1 and q2 are the charges of the ions,
r is the distance between the ions.
In this case, the charge of Mg2+ ion (q1) is +2e, and the charge of F-1 ion (q2) is -1e, where e is the elementary charge (1.6 × 10^-19 C).
The equilibrium interionic separation (r) is the sum of the atomic radii of the ions.
r = atomic radius of Mg2+ + atomic radius of F-1
r = 0.072 nm + 0.133 nm
Now we can substitute these values into the formula to calculate the force of attraction:
F = (9 × 10^9 N m^2 C^-2) * (|2e * -1e|) / (0.205 nm)^2
Converting the distance to meters:
F = (9 × 10^9 N m^2 C^-2) * (|2e * -1e|) / (0.205 × 10^-9 m)^2
Calculating the force:
F = (9 × 10^9 N m^2 C^-2) * (2e * 1e) / (0.205 × 10^-9 m)^2
= (9 × 10^9 N m^2 C^-2) * 2e^2 / (0.042025 × 10^-18 m^2)
= 9 × 10^9 N m^2 C^-2 * 2 * (1.6 × 10^-19 C)^2 / (0.042025 × 10^-18 m^2)
≈ 1.729 N
Therefore, the force of attraction between Mg2+ and F-1 ions at their equilibrium interionic separation is approximately 1.729 N.
(b) If the ions just touch one another, the distance between them is effectively zero. At this separation distance, the force of repulsion between the ions is also calculated using Coulomb's law:
F = (k * |q1 * q2|) / r^2
Using the same charges as before and considering the distance between them as zero:
F = (9 × 10^9 N m^2 C^-2) * (|2e * -1e|) / (0 nm)^2
= (9 × 10^9 N m^2 C^-2) * 2e^2 / 0
= ∞
Therefore, the force of repulsion between Mg2+ and F-1 ions at their equilibrium interionic separation is infinite.
To calculate the force of attraction between the Mg2+ and F-1 ions, you can use Coulomb's Law. Coulomb's Law states that the force of attraction or repulsion between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
(a) To find the force of attraction between the two ions, you need to calculate the charges on the ions and determine their equilibrium interionic separation.
The charge of an Mg2+ ion is +2, and the charge of an F-1 ion is -1. The equilibrium interionic separation is the sum of the atomic radii of the two ions. So,
Equilibrium interionic separation = r(Mg2+) + r(F-1)
= 0.072 nm + 0.133 nm
= 0.205 nm
Note: The atomic radii are given in nanometers (nm).
Now, plug in the values into Coulomb's Law equation:
Force of attraction = (charge1 * charge2) / (4πε₀ * r^2)
where:
charge1 = charge on Mg2+ ion = +2
charge2 = charge on F-1 ion = -1
r = equilibrium interionic separation = 0.205 nm
ε₀ = permittivity of free space = 8.854 x 10^-12 C^2/Nm^2 (constant)
Substituting the values, we get:
Force of attraction = (2 * -1) / (4π * 8.854 x 10^-12 * (0.205 x 10^-9)^2)
Now, calculate this expression to find the force of attraction between the two ions.
(b) To find the force of repulsion at the equilibrium interionic separation, you'll use the same formula but with the charges having like polarities. In this case, both charges are negative, so the force will be repulsive.
Force of repulsion = (charge1 * charge2) / (4πε₀ * r^2)
where:
charge1 = charge on F-1 ion = -1
charge2 = charge on F-1 ion = -1
r = equilibrium interionic separation = 0.205 nm
ε₀ = permittivity of free space = 8.854 x 10^-12 C^2/Nm^2 (constant)
Substituting the values into the equation, you'll get the force of repulsion between the two ions.