Question

In a lithium ion $L{i}^{2+}$, an orbit has a diameter of about $3.18\stackrel{\circ }{A}$. What is the maximum number of electrons that can be accommodated in this orbit?

Answer 1

Diameter $=3.18\stackrel{\circ }{A}$

Radius of the orbit $=\frac{3.18}{2}=1.59\stackrel{\circ }{A}$
$\Rightarrow 1.59=0.53\times \frac{n^2}{Z}$
Where n = the orbit and Z = 3 as a lithium ion is given
$\to n^2=\frac{4.77}{0.53}=9$
$\Rightarrow n=3$
Therefore, the maximum number of atoms that can be accommodated in this orbit will be $2n^2=18$