Chemistry - Atomic Structure Question with Solution | TestHub

Radius of orbit $\left(\mathrm{r}\right) = 0.529 \times \frac{{\mathrm{n}}^2}{\mathrm{Z}}$

Where, n = principal quantum number of orbit

Z = Atomic number

$({\mathrm{r}}_{{\mathrm{Li}}^{2+}}{)}_{\mathrm{n}=2} = 0.529 \times \frac{(2{)}^2}{3} = \mathrm{x} -----(1)$

${\left({\mathrm{r}}_{{\mathrm{Be}}^{3+}}\right)}_{\mathrm{n}=3} = 0.529 \times \frac{(3{)}^2}{4} -----\left(2\right)$

Divide equation (1) by equation (2)

$\frac{{\left({\mathrm{r}}_{{\mathrm{Li}}^{2+}}\right)}_{\mathrm{n}=2}}{{\left({\mathrm{r}}_{{\mathrm{Be}}^{3+}}\right)}_{\mathrm{n}=3}} = \frac{0.529 \times {\displaystyle \frac{(2{)}^2}{3}}}{0.529 \times {\displaystyle \frac{(3{)}^2}{4}}}$

$\Rightarrow \frac{\mathrm{x}}{({\mathrm{r}}_{{\mathrm{Be}}^{3+}}{)}_{\mathrm{n}=3}} = \frac{16}{27}$

$\therefore {\left({\mathrm{r}}_{{\mathrm{Be}}^{3+}}\right)}_{\mathrm{n}=3} = \frac{27\mathrm{x}}{16}$