Match the following Lists.(a → edge length, r+arrow radius of cation, r arrow arrow radius of anion) ( List-I List-II (P) NaCl type crystal (1) 3 a=4( r_ + +r_ - ) (Q) ZnS type crystal (2) 3 a=2( r_ + +r_ - ) (R) CsCl

Correct answer: D. P arrow 4 ; Q arrow 1 ; R arrow 2 ; S arrow 3 In NaCl crystal Cl ions form fcc unit cell and Na^ + ions occupy all theoctahedral voids. a=2 (r_ + +r_ - ) In ZnS crystal S^ 2- ions form fcc unit cell and Zn^ 2+ ions occupy alternate tetrahedral voids 3 4 a= (r_ + +r_ - ) In CsCl crystal, Cl^ - ions form simple cubic unit cell and Cs^ + ions occupy the centre of each cube 3 a=2 (r_ + +r_ - ) In A _ 3 ~Bcrystal, anions B^ -3 ions form simple cubic unit cell and cations A^ + occupy square planar voids. 2 a=2 ( r_ + +r_ - )

JEE Advanced

Exam: JEE Advanced
Subject: Chemistry
Chapter: Solid State

JEE Advanced · Chemistry · Solid State

Match the following Lists.(a → edge length, r+arrow radius of...

Match the following Lists.(a → edge length,

\mathrm{r}+\rightarrow

radius of cation,

\mathrm{r} \rightarrow \rightarrow

radius of anion) (

List-I	List-II
(P) NaCl type crystal	(1) $\sqrt{3}$ a=4\left( $r_{+}+r_{-}\right)</td></tr><tr><td>(Q) ZnS type crystal</td><td>(2)$ \sqrt{3} $a=2\left($ r_{+}+r_{-}\right)
(R) CsCl type crystal	(3) $\sqrt{2}$ a=2\left( $r_{+}+r_{-}\right)</td></tr><tr><td>(S) A_3 B type crystal in which anion B<br>forms simple cubic unit cell and cation A<br>occupies all the square planar voids</td><td>(4) a=2\left($ r_{+}+r_{-}\right)

)

Options

A $\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 3$
B $\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 4$ ; $\mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 2$
C $\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 4 ; \mathrm{S} \rightarrow 2$
D $\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 3$

Correct answer

D. $\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 3$

Step-by-step solution

In NaCl crystal Cl ions form fcc unit cell and

\mathrm{Na}^{+}

ions occupy all theoctahedral voids.

\therefore \quad a=2 \left(r_{+}+r_{-}\right)

In ZnS crystal

\mathrm{S}^{2-}

ions form fcc unit cell and

\mathrm{Zn}^{2+}

ions occupy alternate tetrahedral voids

\therefore \quad

\frac{\sqrt{3}}{4}

a= \left(r_{+}+r_{-}\right)

In CsCl crystal,

\mathrm{Cl}^{-}

ions form simple cubic unit cell and

\mathrm{Cs}^{+}

ions occupy the centre of each cube

\therefore \quad \sqrt{3}a=2 \left(r_{+}+r_{-}\right)

\mathrm{A}}_{3}

\mathrm{~B}crystal, anions

\mathrm{B}^{-3}

ions form simple cubic unit cell and cations

\mathrm{A}^{+}

occupy square planar voids.

\therefore \quad \sqrt{2}a=2 \left(

r_{+}+r_{-}\right)

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