A solid AB has NaCl type structure with edge length 580. 4pm. The radius of A^ + is 100pm. What is the radius of B^ - in pm?

Correct answer: A. 190. 2 NaCl has fcc structure and thus, r _ c + r _ a = (a)/(2) 100 + r _ a = (580. 4)/(2) = 290. 2 100 + r _ a = 290. 2 So r _ a = 290. 2 - 100 = 190. 2pm

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Chemistry
Chapter: Solid State

NTA Abhyas 2020 · Chemistry · Solid State

A solid AB has NaCl type structure with edge length 580. 4pm. The...

A solid AB has NaCl type structure with edge length

580. 4pm.

The radius of

A^{+}

100pm.

What is the radius of

B^{-}

in pm?

Options

A $190. 2$
B $540. 13$
C $525$
D $78. 12$

Correct answer

A. $190. 2$

Step-by-step solution

NaCl

has fcc structure and thus,

{r}_{c} + {r}_{a} = \dfrac{a}{2}

100 + {r}_{a} = \dfrac{580. 4}{2} = 290. 2

100 + {r}_{a} = 290. 2

{r}_{a} = 290. 2 - 100 = 190. 2pm

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