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Q.

The radius of the 2^nd orbit of Li²⁺ is . The expected radius of the 3^rd orbit of Be³⁺ is

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a

$\frac{9}{4} x$

b

$\frac{4}{9} x$

c

$\frac{27}{16} x$

d

$\frac{16}{27} x$

answer is D.

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Detailed Solution

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$\begin{array}{l} r_{n} = \frac{0.529 n^{2}}{z} \\ \frac{r_{{Li}^{+ 2}}}{r_{{Be}^{+ 3}}} = \frac{n_{{Li}^{+ 2}}^{2}}{n_{{Be}^{+ 3}}^{2}} \cdot \frac{z_{{Be}^{+ 3}}}{z_{{Le}^{+ 2}}} \\ \frac{x}{r_{{Be}^{+ 3}}} = \frac{4}{9} \times \frac{4}{3} \\ r_{{Be}^{+ 3}} = \frac{27}{16} x \end{array}$
Option 4 is correct

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The radius of the 2nd orbit of Li2+ is . The expected radius of the 3rd orbit of Be3+ is