Based on the ionic crystal lattice above, identify the substance based on the knowledge that the ions are relatively equal in size.

K F

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M g C l_{2}

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N a C l

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C a F_{2}

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Solution

The correct option is D $C a F_{2}$
Let the red atoms be $A$ and blue atoms be $B$

$Z_{e f f e c t i v e}$ of $A = 6 \times \frac{1}{2} + 8 \times \frac{1}{8} = 4$
$Z_{e f f e c t i v e}$ of $B = 8$ (No. of tetrahedral voids)

Molecular formula: $A_{4} B_{8} : A B_{2}$
$⟶$ In $M g C l$ $r_{M g^{+ 2}}$ $<<$ $r_{C l^{-}}$
In $C a F_{2}$ $r_{C a^{+ 2}}$ $\sim$ $r_{F^{-}}$
So the ionic crystal lattice is of $C a F_{2}$ .

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Based on the ionic crystal lattice above, identify the substance based on the knowledge that the ions are relatively equal in size.

The correct option is D CaF2Let the red atoms be A and blue atoms be BZeffective of A=6×12+8×18=4Zeffective of B=8(No. of tetrahedral voids)Molecular formula:A4B8:AB2⟶ In MgCl rMg+2 << rCl−In CaF2 rCa+2 ∼ rF−So the ionic crystal lattice is of CaF2.