SrO has a NaCl structure. If the lattice cons | vedclass

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(A) In a $NaCl$ structure,the edge length $a$ is related to the ionic radii by the formula $a = 2(r_{Sr^{2+}} + r_{O^{2-}})$.Given $a = 0.514 \ nm$ an

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$1.44 	imes 10^{-10} \ m$

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(A) In a $NaCl$ structure,the edge length $a$ is related to the ionic radii by the formula $a = 2(r_{Sr^{2+}} + r_{O^{2-}})$.Given $a = 0.514 \ nm$ and $r_{Sr^{2+}} = 0.113 \ nm$.Substituting the values: $0.514 = 2(0.113 + r_{O^{2-}})$.$0.257 = 0.113 + r_{O^{2-}}$.$r_{O^{2-}} = 0.257 - 0.113 = 0.144 \ nm$.Converting to meters: $0.144 \ nm = 0.144 	imes 10^{-9} \ m = 1.44 	imes 10^{-10} \ m$.

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

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