The face diagonal length of f.c.c. cubic cell | vedclass

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(B) For an $f.c.c.$ unit cell,the face diagonal is given by $\sqrt{2} a = 660 \sqrt{2} \, pm$,which implies the edge length $a = 660 \, pm$.In an $f.c

Vedclass · Accepted Answer

$220$

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(B) For an $f.c.c.$ unit cell,the face diagonal is given by $\sqrt{2} a = 660 \sqrt{2} \, pm$,which implies the edge length $a = 660 \, pm$.In an $f.c.c.$ lattice,the anions form the lattice and the cations occupy the octahedral voids.The relation for the edge length in terms of ionic radii for an octahedral void is $a = 2(r_+ + r_-)$.Substituting the values: $660 = 2(110 + r_-)$.$330 = 110 + r_-$.$r_- = 330 - 110 = 220 \, pm$.The radius ratio is $\frac{r_+}{r_-} = \frac{110}{220} = 0.5$.Since $0.414 < 0.5 < 0.732$,the cation fits perfectly into the octahedral void.

The face diagonal length of $f.c.c.$ cubic cell is $660 \sqrt{2} \, pm$. If the radius of the cation is $110 \, pm$,the radius of the anion is ............ $pm$.

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