Question

In calcium fluoride ($Ca{{F}_{2}}$) structure, the coordination numbers of calcium and fluoride ions are:a.) 8 and 4b.) 6 and 8c.) 4 and 4d.) 4 and 8

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Hint: The $C{{a}^{2+}}$and ${{F}^{-}}$ in $Ca{{F}_{2}}$​ crystal makes a fluorite type of structure. The $C{{a}^{2+}}$ in $Ca{{F}_{2}}$ crystal makes a face centered cubic close packed lattice and ${{F}^{-}}$ ions occupy the tetrahedral voids at face-centred cubic lattice points and in tetrahedral voids. So, we make the structure and calculate the number ions surrounding each type of ion in the compound.

In $Ca{{F}_{2}}$​, the calcium ions form the CCP arrangement, that is, these ions occupy all the corner positions and also the centre of each face of the cube.
Hence, as each $C{{a}^{2+}}$ ion in surrounded by 8 ${{F}^{-}}$ ions and each ${{F}^{-}}$ ion is surrounded by 4 $C{{a}^{2+}}$ ions, the coordination number of $C{{a}^{2+}}$ and ${{F}^{-}}$ ions in $Ca{{F}_{2}}$ structure will be 8 and 4 ions respectively.
Note: The face centred cubic close packed structure has atoms located at each of the corners and the centres of all the cubic faces in the lattice. Each of the corner atoms is the corner of another cube so the corner atoms are shared among eight unit cells and the atom located on the face of the cube is shared by 2 cubes. This gives a representation of a compound having FCC type of structure. For example, $Ca{{F}_{2}}$, $NaCl$, etc.