A CsCl (cesium chloride) salt crystal is built from the unit cells shown in the figure below. Cl− ions form the corners of a cube and a Cs+ ion is at the centre of the cube. The edge length of the cube, which is called the lattice constant, is 0.4 nm.

(a) What is the magnitude of the net force exerted on the cesium ion by its eight nearest Cl− neighbours? in N

(b) If the Cl− in the lower left corner is removed, what is the magnitude of the net force exerted on the cesium ion at the centre by the seven remaining nearest chlorine ions? in N

a) Presumably zero. All forces should cancel.

b) The four matched corners will again cancel so this leaves only the unmatched corner.the distance between them can be found by pythagorean with one side 4nm the other (a diagonal across the bottom say) is 4sqrt(2) nm. Then use Coulombs law, kq1q2/r(sqr).
A picture might help...just doing this in my head.

To find the magnitude of the net force exerted on the cesium ion (Cs+) by its nearest neighbors, we need to use Coulomb's law, which states that the force between two charged particles is proportional to the product of their charges and inversely proportional to the square of the distance between them.

(a) To calculate the net force on Cs+ due to its eight nearest Cl- neighbors, we need to consider that each Cl- ion exerts an equal force on the Cs+ ion.

Given:
Lattice constant (edge length of the cube) = 0.4 nm

Since there are eight Cl- ions at the corners of the cube surrounding the Cs+ ion, we consider them as eight point charges equally spaced around a circle. The distance from the center Cs+ ion to each Cl- ion is equal to half of the edge length of the cube, which is 0.4 nm / 2 = 0.2 nm.

Now, let's calculate the magnitude of the net force using Coulomb's law:

1. Calculate the charge of each Cl- ion: Since the Cs+ ion has a charge of +1 (due to losing one electron) and Cl- ions have a charge of -1, the charge of each Cl- ion is -1.

2. Calculate the force between the Cs+ and each Cl- ion using Coulomb's law:
F = k * (|q1| * |q2|) / r^2

Substituting the values:
F = (9 × 10^9 N·m²/C²) * (|+1| * |-1|) / (0.2 × 10^-9 m)^2

F = (9 × 10^9 N·m²/C²) * (1) / (0.2 × 10^-9 m)^2

F = (9 × 10^9 N·m²/C²) * (1) / (4 × 10^-20 m²)

F = 9 × 10^9 N·m²/C² * (1 / 4 × 10^-20)

F = 9 × 10^29 N

3. Since there are eight Cl- ions, we multiply the force by 8 to get the net force:
Net Force = 8 * (9 × 10^29 N) = 72 × 10^29 N

Therefore, the magnitude of the net force exerted on the cesium ion by its eight nearest Cl- neighbors is 72 × 10^29 N.

(b) If the Cl- ion in the lower left corner is removed, there will be seven remaining Cl- ions that are nearest neighbors to the Cs+ ion. To find the magnitude of the net force exerted on Cs+ by the seven remaining Cl- ions, we follow a similar process as in part (a):

1. Calculate the charge of each Cl- ion: Same as in part (a), the charge of each Cl- ion is -1.

2. Calculate the force between the Cs+ and each Cl- ion using Coulomb's law:
F = k * (|q1| * |q2|) / r^2

The distance from the center Cs+ ion to each of the seven Cl- ions is still 0.2 nm.

F = (9 × 10^9 N·m²/C²) * (|+1| * |-1|) / (0.2 × 10^-9 m)^2

F = 9 × 10^9 N·m²/C² * (1 / (0.2 × 10^-9 m)^2)

F = 9 × 10^9 N·m²/C² * (1 / 4 × 10^-20)

F = 9 × 10^29 N

3. Since there are seven Cl- ions, we multiply the force by 7 to get the net force:
Net Force = 7 * (9 × 10^29 N) = 63 × 10^29 N

Therefore, the magnitude of the net force exerted on the cesium ion at the center by the seven remaining nearest chlorine ions is 63 × 10^29 N.