JEE Chemistry 2023 Question with Solution

Given: The radius of the $2^{\text{nd}}$ orbit of $\text{Li}^{2+}$ is $x$.

Find: The radius of the $3^{\text{rd}}$ orbit of $\text{Be}^{3+}$.

Use the formula for hydrogen-like species:

$$r_n = k\cdot \frac{n^2}{Z}$$

For $\text{Li}^{2+}$, $Z=3$ and $n=2$:

$$r_2 = k\cdot \frac{2^2}{3} = \frac{4k}{3}$$

Since this radius is given as $x$, we have $x=\frac{4k}{3}$.

For $\text{Be}^{3+}$, $Z=4$ and $n=3$:

$$r_3 = k\cdot \frac{3^2}{4} = \frac{9k}{4}$$

Now take the ratio:

$$\frac{r_3}{r_2} = \frac{\frac{9k}{4}}{\frac{4k}{3}} = \frac{27}{16}$$

So,

$$r_3 = \frac{27}{16}x$$

Therefore, the radius of the $3^{\text{rd}}$ orbit of $\text{Be}^{3+}$ is $\frac{27}{16}x$. The solution states that the correct option is D, but the computed value matches option C.

Given: Two hydrogen-like ions, $\text{Li}^{2+}$ and $\text{Be}^{3+}$.

Find: The orbit radius of $\text{Be}^{3+}$ in terms of $x$.

For hydrogen-like species, radius is proportional to:

$$r \propto \frac{n^2}{Z}$$

Hence,

$$\frac{r_{\text{Be}^{3+}}}{r_{\text{Li}^{2+}}} = \frac{\frac{3^2}{4}}{\frac{2^2}{3}} = \frac{\frac{9}{4}}{\frac{4}{3}} = \frac{27}{16}$$

Since $r_{\text{Li}^{2+}}=x$, it follows that

$$r_{\text{Be}^{3+}} = \frac{27}{16}x$$

Therefore, the correct value is $\frac{27}{16}x$, which corresponds to option C.