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If the radius of $\mathrm{H}$ is $0.53 Ã…$, then what will be the radius of ${ }_{3} \mathrm{Li}^{2+}$?
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The correct answer is:
$0.17 Ã…$
Radius of orbit $=\frac{\mathrm{n}^{2} \mathrm{a}_{\mathrm{o}}}{\mathrm{z}}\left(\mathrm{a}_{\mathrm{o}}=0.529 \mathrm{~A}\right)$
Radius of $\mathrm{H}=\frac{(1)^{2} \times 0.529 \mathrm{~A}}{1}=0.53 \mathrm{~A}$
Thus, the radius of ${ }_{3} \mathrm{Li}^{2+}$ will be $=\frac{(1)^{2} \times 0.529}{3}=0.17 Ã…$
Radius of $\mathrm{H}=\frac{(1)^{2} \times 0.529 \mathrm{~A}}{1}=0.53 \mathrm{~A}$
Thus, the radius of ${ }_{3} \mathrm{Li}^{2+}$ will be $=\frac{(1)^{2} \times 0.529}{3}=0.17 Ã…$
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