QJEEAdv18Che149: JEE Advanced 2018 Chemistry - Class 12 JEEAdv18Che149 Solution: Consider an ionic solid with structure. Construct a new…

The unit cell of initial structure of ionic solid MX looks like

In NaCl type of solids cations (Na⁺) occupy the octahedral voids while anions (Cl ^$-$ ) occupy the face centre positions.

However, as per the demand of problem the position of cations and anions are swapped.

We also know that (for 1 unit cell)

(A) Total number of atoms at FCC = 4

(B) Total number of octahedral voids = 4 (as no. of atoms at FCC = No. of octahedral voids)

Now taking the conditions one by one

(i) If we remove all the anions except the central one than number of left anions.

= 4 $$-$ 3 = 1

(ii) If we replace all the face centred cations by anions than effective number of cations will be = 4 $-$ 3 = 1 Likewise effective number of anions will be = 1 + 3 = 4

(iii) If we remove all the corner cations then effective number of cations will be 1 $-$ 1 = 0

(iv) If we replace central anion with cation then effective number of cations will be 0 + 1 = 1 Likewise effective number of anions will be 4 $-$ 1 = 3

Thus, as the final outcome, total number of cations present in Z after fulfilling all the four sequential instructions = 1 Likewise, total number of anions = 3

Hence, the value of ${{Number\,of\,anions} \over {Number\,of\,cations}} = {3 \over 1} = 3$