Answer
425.1k+ views
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.
Complete answer:
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.
Now, as there are two tetrahedral sites available in every Face centred cubic close packed structure for every calcium ion, the fluoride ions occupy all the tetrahedral sites, and the stoichiometry of the compound becomes 1:2.
Each fluoride ion is surrounded by four calcium ions whereas each calcium ions is surrounded by eight fluoride ions which are placed towards the corners of a cube. The coordination of the compound is 8:4.
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.
Hence, the correct answer will be option A which is, the coordination numbers of calcium and fluoride ions in calcium fluoride are 8 and 4 respectively.
So, the correct answer is “Option A”.
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.
Complete answer:
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.
Now, as there are two tetrahedral sites available in every Face centred cubic close packed structure for every calcium ion, the fluoride ions occupy all the tetrahedral sites, and the stoichiometry of the compound becomes 1:2.
Each fluoride ion is surrounded by four calcium ions whereas each calcium ions is surrounded by eight fluoride ions which are placed towards the corners of a cube. The coordination of the compound is 8:4.
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.
Hence, the correct answer will be option A which is, the coordination numbers of calcium and fluoride ions in calcium fluoride are 8 and 4 respectively.
So, the correct answer is “Option A”.
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.
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