1. Tardigrade
  2. Question
  3. Chemistry
  4. The SrO has NaCl structure. If the lattice constant a =0.514 nm and ionic radius of Sr 2+=0.113 nm. The radius of O 2- ion is :-

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Q. The $SrO$ has $NaCl$ structure. If the lattice constant $a =0.514\, nm$ and ionic radius of $Sr ^{2+}=0.113\, nm$. The radius of $O ^{2-}$ ion is :-

Solution:

$2 r _{ Sr ^{+}}+2 r _{ O ^{-2}}=0.514$ nm
$\therefore r _{ O ^{-2}}=0.144$ nm
$=1.44 \times 10^{-10} \,m$