JEE MAIN - Chemistry (2023 - 25th January Morning Shift - No. 9)
The radius of the $$\mathrm{2^{nd}}$$ orbit of $$\mathrm{Li^{2+}}$$ is $$x$$. The expected radius of the $$\mathrm{3^{rd}}$$ orbit of $$\mathrm{Be^{3+}}$$ is
$$\frac{16}{27}x$$
$$\frac{4}{9}x$$
$$\frac{9}{4}x$$
$$\frac{27}{16}x$$
Explanation
$r_{\mathrm{Li}^{+}}=r_{0} \times \frac{2^{2}}{3}=x \Rightarrow r_{0}=\frac{3 x}{4}$
$\mathrm{r}_{\mathrm{Be}^{3+}}=\mathrm{r}_{0} \times \frac{3^{2}}{4}$
$r_{\mathrm{Be}^{3+}}=\frac{3 \mathrm{x}}{4} \times \frac{3^{2}}{4}=\frac{27 \mathrm{x}}{16}$
$\mathrm{r}_{\mathrm{Be}^{3+}}=\mathrm{r}_{0} \times \frac{3^{2}}{4}$
$r_{\mathrm{Be}^{3+}}=\frac{3 \mathrm{x}}{4} \times \frac{3^{2}}{4}=\frac{27 \mathrm{x}}{16}$
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