The distance between Na⁺ and Cl⁻ ions in solid NaCl of density 43.1 g cm⁻³ is _______________ × 10⁻¹⁰ m. (Nearest Integer)
(Given : N A = 6.02 × 10²³ mol⁻¹)

Solution

Chemistry ma’am · 4 steps

Get the edge length from density first, then remember Na⁺ and Cl⁻ touch along half of that edge.

1. Know the cell. NaCl has a rock-salt (fcc) structure with Z = 4 formula units per unit cell. The molar mass is M = 23 + 35.5 = 58.5 g mol⁻¹.

   a³ = (Z M)/(ρ N_A) = (4 × 58.5)/(43.1 × 6.02 × 10²³) cm³

2. Find the edge. The numerator is 234 and the denominator is 2.595 × 10²⁵, so a³ = 9.02 × 10⁻²⁴ cm³. Taking the cube root gives a = 2.08 × 10⁻⁸ cm = 2.08 × 10⁻¹⁰ m.
3. Nearest neighbours. Along each cube edge the order is Cl⁻, Na⁺, Cl⁻, so a Na⁺ ion sits halfway along the edge. The Na⁺ to Cl⁻ distance is a/2 = 1.04 × 10⁻¹⁰ m.
4. So. The interionic distance is about 1.04 × 10⁻¹⁰ m, which rounds to 1.

The common slip: 2. Reporting the edge length a instead of a/2 gives about 2.08 × 10⁻¹⁰ m, which rounds to 2.

The idea to keep. For rock-salt crystals, use ρ = ZM/(a³ N_A) with Z = 4, then halve a: cation and anion touch along the cube edge, so r⁺ + r⁻ = a/2.

Answer from the NTA final answer key. Solution by PrepWiser.