The coordination number of cation and anion i | vedclass

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(C) In $CaF_2$ (fluorite structure),$Ca^{2+}$ ions form an $FCC$ lattice,and $F^{-}$ ions occupy all $8$ tetrahedral voids. The coordination number $(

Vedclass · Accepted Answer

$8 : 4$ and $4 : 8$

Vedclass · Answer

(C) In $CaF_2$ (fluorite structure),$Ca^{2+}$ ions form an $FCC$ lattice,and $F^{-}$ ions occupy all $8$ tetrahedral voids. The coordination number $(C.N.)$ of $Ca^{2+}$ is $8$ and that of $F^{-}$ is $4$. Thus,the $C.N.$ ratio for fluorite is $8:4$. In $Na_2O$ (anti-fluorite structure),the positions of cations and anions are interchanged compared to $CaF_2$. Here,$O^{2-}$ ions form an $FCC$ lattice,and $Na^{+}$ ions occupy all $8$ tetrahedral voids. The coordination number $(C.N.)$ of $O^{2-}$ is $8$ and that of $Na^{+}$ is $4$. Therefore,the $C.N.$ ratio for anti-fluorite is $4:8$. The correct answer is $8:4$ and $4:8$.

The coordination number of cation and anion in fluorite $CaF_2$ and anti-fluorite $Na_2O$ are respectively

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