MCQ
In calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion $\ce{(Ca^{2+})}$ and fluoride ion $(F^-)$ are:
  • $8$ and $4$
  • B
    $4$ and $8$
  • C
    $4$ and $2$
  • D
    $6$ and $6$

Answer

Correct option: A.
$8$ and $4$
Calcium fluoride crystallizes in a Face$-$Centered Cubic unit cell $(\text{FCC})$ having an edge length of $5.463$ Angstroms.
The cell is displayed here with calcium cations $($in blue$)$ defining $\text{FCC}$ lattice sites, and fluoride anions $($in green$)$ occupying all tetrahedral sites.
Fluoride anions have $4$ neighbors of opposite charge arranged at vertices of an tetrahedron. Calcium cations have $\text{EIGHT}$ neighbors of opposite charge arrange at corners that outline a smaller cube.
So the $Ca:\ F$ coordination ratio is $8 : 4$ or $2 : 1.$
In $\ce{4CaF_2}$​ the the co ordination number of $\ce{Ca^{+2}}$ is $\ce{8 F^-}$ is $4.$

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