A) \[n=2,\,L{{i}^{2+}}\]
B) \[n=2,\,B{{e}^{3+}}\]
C) \[n=2,\,H{{e}^{+}}\]
D) \[n=3,\,L{{i}^{2+}}\]
Correct Answer: B
Solution :
\[r=\frac{0.529{{n}^{2}}}{Z}{{A}^{o}}\] \[Z=1\] for \[H\] \[{{r}_{H}}=\frac{0.529\,{{n}^{2}}}{1}{{A}^{o}}\] \[Z=4\] for \[B{{e}^{3+}}\] \[{{r}_{Be}}^{3+}=\frac{0.529\,{{n}^{2}}}{4}{{A}^{o}}\] Given, \[{{r}_{H}}={{r}_{Be}}^{3+}\] Hence, it is equal, when \[n=2\] for \[B{{e}^{3+}}\] Thus \[{{r}_{2}}(B{{e}^{3+}})={{r}_{{{1}_{(H)}}}}\]
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