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Question: A solid AB has the NaCl structure. If radius of cation\(A^{+}\) is 120 pm, calculate the maximum pos...

A solid AB has the NaCl structure. If radius of cationA+A^{+} is 120 pm, calculate the maximum possible value of the radius of the anion B−B^{-}

A

240 pm

B

280 pm

C

270 pm

D

290 pm

Answer

290 pm

Explanation

Solution

We know that for the NaCl structure

radius of cation/radius of anion = 0.414; rA+rB−=0.414\frac{r_{A^{+}}}{r_{B^{-}}} = 0.414; rB−=rA+0.414=1200.414=290pmr_{B^{-}} = \frac{r_{A^{+}}}{0.414} = \frac{120}{0.414} = 290pm