A solid compound `XY` has `NaCl` structure. If the radius of the cation is `100` pm, the radius of the anion `(Y^(-))` will be

Options

A. `165.7p m`
B. `275.1p m`
C. `322.5p m`
D. `241.5p m`

Answer

4

Explanation

`NaCl` has fcc arrangement in which `Cl^(-)` ions occupy the corners and face centres of a cube while `Na^(+)` ions are present at the edge centres and body centre. Since `XY` has `NaCl` structure we have
`0.414 le (r_(+))/(r_(-)) le 0.732`
Therefore
`r_(-)=(r_(+))/(0.414)=(100p m)/(0.414)`
`=241.5p m`