JEE Challenger
More from Solid State
ChemMedAdv 2015 • Paper 1

Fraction of Octahedral and Tetrahedral Holes in Spinel Lattice

If the unit cell of a mineral consists of a cubic close-packed (ccpccp) array of oxide anions, with an mm fraction of octahedral holes occupied by aluminium ions and an nn fraction of tetrahedral holes occupied by magnesium ions, then mm and nn, respectively, are

Options

A

12,18\frac{1}{2}, \frac{1}{8}

Correct
B

1,141, \frac{1}{4}

C

12,12\frac{1}{2}, \frac{1}{2}

D

14,18\frac{1}{4}, \frac{1}{8}

Topics & Concepts

Solid StateStoichiometry

Step-by-Step Solution

To find the fractions mm and nn, we analyze the structure and electrical neutrality of the mineral's unit cell:

  1. Number of Atoms and Voids in a ccpccp Unit Cell:

    • In a cubic close-packed (ccpccp) arrangement formed by oxide ions (O2−\text{O}^{2-}), the effective number of oxide ions per unit cell is: NO2−=4N_{\text{O}^{2-}} = 4
    • The total number of octahedral holes (voids) is equal to the number of lattice points: Noctahedral=4N_{\text{octahedral}} = 4
    • The total number of tetrahedral holes (voids) is twice the number of lattice points: Ntetrahedral=2×4=8N_{\text{tetrahedral}} = 2 \times 4 = 8

  2. Occupancy of Voids by Cations:

    • Aluminium ions (Al3+\text{Al}^{3+}) occupy an mm fraction of the octahedral voids: NAl3+=4mN_{\text{Al}^{3+}} = 4m
    • Magnesium ions (Mg2+\text{Mg}^{2+}) occupy an nn fraction of the tetrahedral voids: NMg2+=8nN_{\text{Mg}^{2+}} = 8n

  3. Condition for Electrical Neutrality: The mineral is an electrically neutral ionic compound (spinel, MgAl2O4\text{MgAl}_2\text{O}_4): Total positive charge=Total negative charge\text{Total positive charge} = \text{Total negative charge} (+2)×NMg2++(+3)×NAl3+=(+2)×NO2−(+2) \times N_{\text{Mg}^{2+}} + (+3) \times N_{\text{Al}^{3+}} = (+2) \times N_{\text{O}^{2-}} 2(8n)+3(4m)=2×42(8n) + 3(4m) = 2 \times 4 16n+12m=816n + 12m = 8 Dividing the entire equation by 44: 3m+4n=23m + 4n = 2

  4. Testing the Given Options:

    • For Option (A): m=12m = \frac{1}{2}, n=18n = \frac{1}{8} 3(12)+4(18)=32+12=2(Satisfied)3\left(\frac{1}{2}\right) + 4\left(\frac{1}{8}\right) = \frac{3}{2} + \frac{1}{2} = 2 \quad \text{(Satisfied)}

      • Number of O2−=4\text{O}^{2-} = 4
      • Number of Al3+=4×12=2\text{Al}^{3+} = 4 \times \frac{1}{2} = 2
      • Number of Mg2+=8×18=1\text{Mg}^{2+} = 8 \times \frac{1}{8} = 1
      • This gives the standard spinel formula MgAl2O4\text{MgAl}_2\text{O}_4.

    • For Option (B): m=1m = 1, n=14  ⟹  3(1)+4(14)=4≠2n = \frac{1}{4} \implies 3(1) + 4\left(\frac{1}{4}\right) = 4 \neq 2

    • For Option (C): m=12m = \frac{1}{2}, n=12  ⟹  3(12)+4(12)=72≠2n = \frac{1}{2} \implies 3\left(\frac{1}{2}\right) + 4\left(\frac{1}{2}\right) = \frac{7}{2} \neq 2

    • For Option (D): m=14m = \frac{1}{4}, n=18  ⟹  3(14)+4(18)=54≠2n = \frac{1}{8} \implies 3\left(\frac{1}{4}\right) + 4\left(\frac{1}{8}\right) = \frac{5}{4} \neq 2

Therefore, the values of mm and nn are 12\frac{1}{2} and 18\frac{1}{8}, respectively.

Correct Answer: A

PreviousNext