Discussion Forum

The radius of the $2^{nd}$  orbit of $B^{4+}$ ion is 

Answer:

Explanation:

Radius of the nth orbit (r_n)= $0.529 \times \frac{n^{2}}{Z}$  Å

 For $B^{4+}$ion, (Z) =5, second  orbit (n)=2

 $r_{2}(B^{4+})= \frac{0.529 \times (2)^{2}}{5}$ Å= $\frac{0.529 \times 4}{5}$Å

 $r_{2}(B^{4+})$= 0.4232 Å