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A cubic unit cell has manganese (Mn) ions at all corner positions and fluoride (F⁻) ions at the centre of every edge. Given that the radius of Mn ion is 66 pm and the radius of F⁻ ion is 134 pm, what is the shortest distance between any two fluoride ions?

1. 4.00 angstrom
2. 2.00 angstrom
3. 2.83 angstrom
4. 3.83 angstrom

Correct answer: 2.83 angstrom

Solution

Along an edge of the cube the contact condition gives r(Mn) + r(F⁻) = a/2, so a = 2 * (66 + 134) pm = 400 pm = 4 angstrom. F⁻ ions sit at edge-centres. The shortest F⁻-F⁻ distance connects two F⁻ ions on adjacent edges sharing a common corner vertex; this distance equals sqrt((a/2)² + (a/2)²) = a / sqrt(2) = 4 / sqrt(2) = 2.83 angstrom.

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