A solid AB has NaCl type structure with edge length $580.4$ pm. The radius of $A^{+}$ is $100$ pm. What is the radius (in pm) of $B^{-}$ ?

190.20

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

540.13

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

525.23

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

78.12

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

Open in App

Solution

The correct option is A $190.20$
As we know, NaCl has fcc structure and thus,
$r_{c} + r_{a} = \frac{a}{2}$
$100 + r_{a} = \frac{580.4}{2} = 290.2$ pm
$100 + r_{a} = 290.2$ pm
$r_{a} = 290.2 - 100 = 190.2$ pm

Suggest Corrections

Similar questions

A solid AB has NaCl type structure. If the radius of cation A is 100 pm, then the radius of the anion B will be

A^{+}

B^{-}

A B

N a C l

A^{+}

100 p m

B^{-}

A binary solid $(A^{+} B^{-})$ has a rock salt structure. If the edge length is 400 pm and radius of cation is 75 pm, the radius of anion is

B^{-}

250

Join BYJU'S Learning Program