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Question

For which of the following, the radius will be same as for hydrogen atom having $$n=1$$?


A
He+,n=2
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B
Li2+,n=2
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C
Be3+,n=2
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D
Li2+,n=3
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Solution

The correct option is B $$Be^{3+}, n=2$$
$$r_n\propto \dfrac{n^2}{Z}$$

$$r_n = 0.529\times \dfrac{n^2}{Z} A^0$$  

For $$H$$-atom, 
n = 1
Z = 1
$$r_{(H)} =0.529A^0$$.

and for $$Be^{3+}$$, 
Z = 4
n = 2
$$r_{(Be^{3+})} = 0.0529\times \dfrac{2^2}{4}A^0$$
$$r_{(Be^{3+})} = 0.0529A^0$$

Therefore, for the second orbit of $$Be^{+3}$$ radius is same as that for 1st orbit of hydrogen.

Hence, option C is correct.

Chemistry

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