Question

The NaCl crystal structure consists of alternating Na⁺ and Cl⁻ ions lying next to each other in three dimensions. If the Na⁺ radius is 56.4% of the Cl⁻ radius and the distance between Na⁺ nuclei is 566 pm, what are the radii of the two ions?

Answer 1

Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

Explanation :

As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.

Let us assume that the radius of Cl⁻ be, (x) pm

So, the radius of Na⁺ = $x\times \frac{56.4}{100}=(0.564x)pm$

In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.

Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.

Given:

Distance between Na⁺ nuclei = 566 pm

Thus, the relation will be:

$2\times \text{Radius of }Cl^-+2\times \text{Radius of }Na^+=\text{Distance between }Na^+\text{ nuclei}$

$2\times x+2\times 0.564x=566$

$2x+1.128x=566$

$3.128x=566$

$x=180.9\approx 181pm$

The radius of Cl⁻ ion = (x) pm = 181 pm

The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm

Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.