A) \[H{{e}^{+}}(n=2)\]
B) \[L{{i}^{2+}}(n=2)\]
C) \[L{{i}^{2+}}(n=3)\]
D) \[B{{e}^{3+}}(n=2)\]
Correct Answer: D
Solution :
\[{{r}_{H}}=0.529\frac{{{n}^{2}}}{z}{\AA}\] For hydrogen ; \[n=1\] and \[z=1\]therefore \[{{r}_{H}}=0.529{\AA}\] For \[B{{e}^{3+}}:\,Z=4\] and \[n=2\] Therefore \[{{r}_{B{{e}^{3+}}}}=\frac{0.529\times {{2}^{2}}}{4}=0.529{\AA}\].
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