JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Solid State

The arrangement of

X^{-}

ions around

A^{+}

ion in solid AX is given in the figure (not drawn to scale). If the radius of

X^{-}

250 p m

, the radius of

A^{+}

Answer: (A)

Solution

The figure shows an octahedral arrangement of

X^{-}

ions around

A^{+}

. For an octahedral void, the radius ratio is:

\frac{r _{+}}{r _{-}} = 0.414

Given

r_{-} = 250

pm:

r_{+} = 0.414 \times 250 = 103.5 \approx 104 pm

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Single correct option