Consider an ionic solid MX with NaCl structure. Construct a new structure ( Z ) whose unit cell is constructed from the unit cell of MX following the sequential instructions given below. Neglect the charge balance. (i) Remove all the anions ( X ) except the central one (ii) Replace all the face centered cations (M) by anions ( ( X ) (iii) Remove all the corner cations (M) (iv) Replace the central anion (X) with cation (M) The value of (frac number of anions number of cations ) in Z is .

Chemistry · Solid State

JEE Advanced 2018 — Paper 1 — Question 21

Consider an ionic solid MX with NaCl structure. Construct a new structure ( $\mathbf{Z}$ ) whose unit cell is constructed from the unit cell of MX following the sequential instructions given below. Neglect the charge balance. (i) Remove all the anions ( $\mathbf{X}$ ) except the central one (ii) Replace all the face centered cations (M) by anions ( $\mathbf{( X )}$ (iii) Remove all the corner cations (M) (iv) Replace the central anion (X) with cation (M) The value of $\left(\frac{\text { number of anions }}{\text { number of cations }}\right)$ in $\mathbf{Z}$ is $\qquad$ .

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{X}^{-}=$ Octahedral void

$\mathrm{M}^{+}=$ FCC point

$\begin{array}{ll} & \mathrm{M}^{+} \quad \mathrm{X}^{-} \\ \text {(i) } & 4 \quad 4-3=1 \end{array}$

(ii) $4-6 \times \frac{1}{2} \quad 1+6 \times \frac{1}{2}$

(iii) $1-1 \quad 3+1=4$

(iv) $0+1 \quad 4-1=3$

Hence $\frac{\text { anion }}{\text { cation }}=\frac{3}{1}=3$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2018
Paper: Paper 1
Subject: Chemistry
Chapter: Solid State
Topic: Packing in Ionic Solids - all types