JEE Challenger
More from Solid State
ChemMedAdv 2018 • Paper 1

Ratio of Anions to Cations in Modified Rock Salt Lattice

An ionic solid MX\mathbf{MX} crystallizes in the NaCl\text{NaCl} type structure. A modified unit cell (Z)(\mathbf{Z}) is derived from the unit cell of MX\mathbf{MX} by carrying out the following sequential modifications, neglecting charge neutrality:

(i) Eliminate all anions (X)(\mathbf{X}) except the one located at the body center. (ii) Substitute all face-centered cations (M)(\mathbf{M}) with anions (X)(\mathbf{X}). (iii) Remove all cations (M)(\mathbf{M}) situated at the corners. (iv) Replace the body-centered anion (X)(\mathbf{X}) with a cation (M)(\mathbf{M}).

The ratio (number of anionsnumber of cations)\left( \frac{\text{number of anions}}{\text{number of cations}} \right) in the resulting structure Z\mathbf{Z} is ______.

Official Numerical Answer3

Topics & Concepts

Solid StateStoichiometry

Step-by-Step Solution

To determine the ratio of the number of anions to the number of cations in the modified unit cell Z\mathbf{Z}, we follow the given sequential instructions step-by-step.

1. Initial Structure of NaCl\text{NaCl}-type MX\mathbf{MX} Lattice

From the statements in the problem:

  • Cations (M\mathbf{M}) occupy the positions of a face-centered cubic (fcc\text{fcc}) lattice:

    • 88 corners: Contribution =8×18=1= 8 \times \frac{1}{8} = 1
    • 66 face centers: Contribution =6×12=3= 6 \times \frac{1}{2} = 3
    • Total number of M\mathbf{M} cations initially =1+3=4= 1 + 3 = 4

  • Anions (X\mathbf{X}) occupy the octahedral voids:

    • 1212 edge centers: Contribution =12×14=3= 12 \times \frac{1}{4} = 3
    • 11 body center (central position): Contribution =1×1=1= 1 \times 1 = 1
    • Total number of X\mathbf{X} anions initially =3+1=4= 3 + 1 = 4

2. Sequential Modifications

  • Step (i): Remove all the anions (X\mathbf{X}) except the central one

    • The 1212 edge-centered anions are removed.
    • Only the body-centered anion remains.
    • Number of anions =1×1=1= 1 \times 1 = 1
    • Number of cations =4= 4 (11 from corners, 33 from face centers)

  • Step (ii): Replace all the face-centered cations (M\mathbf{M}) by anions (X\mathbf{X})

    • The 66 face-centered positions are now occupied by anions (X\mathbf{X}).
    • Cations remaining: only at the 88 corners   ⟹  8×18=1\implies 8 \times \frac{1}{8} = 1
    • Anions present: 66 at face centers +1+ 1 at the body center   ⟹  (6×12)+(1×1)=3+1=4\implies \left(6 \times \frac{1}{2}\right) + (1 \times 1) = 3 + 1 = 4

  • Step (iii): Remove all the corner cations (M\mathbf{M})

    • All 88 corner cations are removed.
    • Number of cations remaining =0= 0
    • Number of anions present =4= 4 (33 from face centers, 11 from the body center)

  • Step (iv): Replace the central anion (X\mathbf{X}) with cation (M\mathbf{M})

    • The body-centered position now contains 11 cation (M\mathbf{M}).
    • Number of cations in Z=1×1=1\mathbf{Z} = 1 \times 1 = 1
    • Number of anions in Z=6×12=3\mathbf{Z} = 6 \times \frac{1}{2} = 3 (only the face-centered anions remain)

3. Calculating the Desired Ratio

In the resulting structure Z\mathbf{Z}: Total number of anions=3\text{Total number of anions} = 3 Total number of cations=1\text{Total number of cations} = 1

Therefore, the ratio is: number of anionsnumber of cations=31=3\frac{\text{number of anions}}{\text{number of cations}} = \frac{3}{1} = 3

PreviousNext