Question

Radius of the following species follows the order
$H^{+},H,H{e}^{+}\text{\&}H{e}^{+2}$

• A. $H^{+}>H{e}^{+}>H>H{e}^{2+}$
• B. $H{e}^{+}>H>H{e}^{2+}>H^{+}$
• C. $H>H{e}^{+}>H{e}^{2+}>H^{+}$
• D. $H^{+}>H{e}^{2+}>H{e}^{+}>H$

Answer 1

The correct option is C $H>H{e}^{+}>H{e}^{2+}>H^{+}$

Here $H$ and $H{e}^{+}$ both are isoelectronic. Thus radius of $H$ is more compared to $H{e}^{+}$.
But in $H^{+}$ and in $H{e}^{2+}$, electrons are not there and the formula to calculate the radius of nucleus is $r=r_o{A}^{\frac{1}{3}}$, where $A$ is the mass number of the nucleus. Since $H{e}^{2+}$ has a larger mass number thus $H{e}^{2+}$ radius is more compare to $H^{+}$
So the correct order of radius will be:
$H>H{e}^{+}>H{e}^{2+}>H^{+}$