A solid compound XY has $N a C l$ structure. If the radius of the cation is $100 p m$ , the radius of the anion ( $Y^{+}$ ) will be :

271.5 p m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

322.5 p m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

241.5 p m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

220.5 p m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $241.5 p m$
$N a C l$ has fcc arrangement in which $C l^{-}$ ions occupy the corners and face centres of a cube while $N a^{+}$ ions are present at the edge centres and body centre.

Since XY has NaCl structure we have,

0.414 ≤ $\frac{r^{+}}{r^{-}}$ ≤ 0.732

Given, $r^{+} = 100 p m$

Therefore, $r^{-}$ = $\frac{r^{+}}{0.414}$ = $\frac{100 p m}{0.414}$

⇒ $r^{-} = 241.5 p m$

Hence, option C is correct.

Suggest Corrections

Similar questions

N a C l

(Y^{-})

A B

N a C l

100 p m

B^{-}

(Y^{-})

X Y

N a C l

(X^{+})

100

(Y^{-})

Join BYJU'S Learning Program