Step-by-step Derivation:

1. Analyze the Given Arrangement: The figure shows a central cation A⁺ surrounded by six anions X⁻. This is a classic example of an octahedral coordination geometry, where the cation occupies an octahedral void formed by the anions.

2. Identify the Geometric Condition: The figure depicts a situation where the anions (X⁻) are in contact with each other, and the central cation (A⁺) is just large enough to fit in the void and touch all the surrounding anions. This is the limiting condition for stable octahedral coordination.

3. Set up the Geometric Model: Let's consider a 2D cross-section of the octahedron that passes through the center of the A⁺ ion and the centers of four coplanar X⁻ ions. In this plane:

   • The centers of the four X⁻ ions form a square.
   • The center of the A⁺ ion is at the center of this square.
   • Let $r_A⁺$ be the radius of the cation A⁺ and $r_X⁻$ be the radius of the anion X⁻.

4. Relate Radii to Geometric Distances:

   • Since the adjacent anions X⁻ are touching, the distance between their centers is $2 * r_X⁻$. This distance is the side length of the square, let's call it s. So, $s = 2 * r_X⁻$.
   • The cation A⁺ at the center touches the anions at the corners. The distance from the center of A⁺ to the center of any of these X⁻ ions is $r_A⁺ + r_X⁻$.
   • In the square geometry, the distance from the center to a corner is half the length of the diagonal.

5. Apply Pythagorean Theorem:

   • The length of the diagonal (d) of the square with side length s is given by d = s√2.
   • Substituting $s = 2 * r_X⁻$, we get $d = (2 * r_X⁻)√2 = 2√2 * r_X⁻$.
   • The distance from the center to a corner is d/2, which is $(2√2 * r_X⁻) / 2 = √2 * r_X⁻$.

6. Formulate and Solve the Equation:

   • We now have two expressions for the distance between the center of A⁺ and the center of X⁻:
     • From sum of radii: $r_A⁺ + r_X⁻$
     • From geometry: $√2 * r_X⁻$
   • Equating these two expressions: $r_A⁺ + r_X⁻ = \sqrt{2} \cdot r_X⁻$
   • Solving for $r_A⁺$: $r_A⁺ = \sqrt{2} \cdot r_X⁻ - r_X⁻$ $r_A⁺ = (\sqrt{2} - 1) \cdot r_X⁻$ This equation relates the cation radius to the anion radius for the limiting case of octahedral coordination. The ratio $r_A⁺ / r_X⁻ = √2 - 1 ≈ 0.414$ is the minimum radius ratio for stable octahedral coordination.

7. Calculate the Numerical Value:

   • We are given $r_X⁻ = 250$ pm.
   • Using the value √2 ≈ 1.414: $r_A⁺ = (1.414 - 1) \times 250 \text{ pm}$ $r_A⁺ = 0.414 \times 250 \text{ pm}$ $r_A⁺ = 103.5 \text{ pm}$

8. Compare with Options:

   • The calculated value is 103.5 pm.
   • Looking at the options: A: 104 pm B: 125 pm C: 183 pm D: 57 pm
   • The closest option to our calculated value is 104 pm.

Conclusion:

Based on the geometric derivation for the limiting radius ratio in an octahedral void, the radius of the cation A⁺ is calculated to be 103.5 pm, which corresponds to option A.