Solid State - Result Question 29

29. Consider an ionic solid $M X$ with $NaCl$ structure. Construct a new structure $(Z)$ whose unit cell is constructed from the unit cell of $M X$ following the sequential instruction given below. Neglect the charge balance.

(2018 Adv.)

(a) Remove all the anions $(X)$ except the central one

(b) Replace all the face centered cations $(M)$ by anions $(X)$

(c) Remove all the corner cations $(M)$

(d) Replace the central anion $(X)$ with cation $(M)$

The value of $\left(\dfrac{\text { Number of anions }}{\text { Number of cations }}\right)$ in $Z$ is _______

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Answer:

Correct Answer: 29. $(3)$

Solution:

  1. The unit cell of initial structure of ionic solid $M X$ looks like

<img src=“https://temp-public-img-folder.s3.amazonaws.com/sathee.prutor.images/sathee_image/cropped_2024_01_16_b4fdca9f34924034e8d8g-140_jpg_height_345_width_582_top_left_y_2112_top_left_x_1302.jpg"width="300">

In $NaCl$ type of solids cations $\left(Na^{+}\right)$occupy the octahedral voids while anions $\left(Cl^{-}\right)$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 $\dfrac{\text { Number of anions }}{\text { Number of cations }}=\dfrac{3}{1}=3$