Question

In $Ca{{F}_{2}}$ unit cell, the coordination number of ${{F}^{-}}$ ion is?

Answer 1

Hint: To find the coordination number of the cations and the anions in a crystal lattice, we should know the type of structure of the crystal lattice formed by the ions. A basic idea of the crystal lattice can be obtained from the formula of the compound also.


Complete step by step solution:
-Lattice is a 3-D arrangement of the particles which is repeated to give the complete structure of the compound formed. Its smallest portion is called a unit cell which has 3 edges and 3 angles between the edges.
-Coordination number is the number of the nearest neighbouring atoms associated to a particular atom present inside the lattice. Both the cations and the anions present in the crystal lattice have a particular coordination number which may or may not be the same.
-$Ca{{F}_{2}}$ crystal lattice is of the type $A{{B}_{2}}$ . The calcium ions form the face-centered cubic lattice which is also called FCC lattice. The fluoride ions occupy the position of the tetrahedral voids present in the lattice of the calcium ions.
-The effective number of $C{{a}^{2+}}$ ions is 4 and the effective number of ${{F}^{-}}$ ions is 8. Thus the formula of the compound formed is $C{{a}_{4}}{{F}_{8}}=Ca{{F}_{2}}$ .
-The effective number of the formula unit cells which is denoted by Z for this crystal lattice is 4. This is helpful in calculating the density of the lattice which is given by
Density = $\dfrac{ZxM}{{{N}_{A}}x{{a}^{3}}}$
Where M=molecular weight of the compound and ${{N}_{A}}$ is avogadro’s number.
-For calcium ions, the nearest neighbours are the fluoride ions present in the tetrahedral voids and the total number of such ions are 8 and so the coordination number of calcium ions is 8. They are present at a distance of $\dfrac{\sqrt{3}}{4}$ from the calcium ions.
-For fluoride ions, the nearest neighbors are the calcium ions which are present in the crystal FCC lattice. There are 4 such ions and so the coordination number of fluoride ions is 4 which is present at the distance of $\dfrac{\sqrt{3}}{4}$from the fluoride ions.

Therefore, the coordination number of ${{F}^{-}}$ ion is 4 in $Ca{{F}_{2}}$ unit cell.


Note: The ratio of the fluoride ions and the calcium ions lies in the range of 0.225 and 0.414. This tells us about the type of void occupied by the lattice and thus gives us the coordination number. $Ba{{F}_{2}},SrC{{l}_{2}},BaC{{l}_{2}}$ also form this type of lattice.