The radius of the `2 ^(nd)` orbit of `B ^(4+)` ion is

Options

A. 4.23 Å
B. 0.2340 Å
C. 0.4232 Å
D. 0.3241 Å

Answer

C

Explanation

`"Radius of the nth orbit"(r_n)=0.529xxn^2/ZdotA`
`"For "B^(4+)ion,(Z)=5,"second orbit"(n)=2`
`r_2(B^(4+))=(0.529xx(2)^2)/5dotA=(0.529xx4)/5dotA`
`r_2(B^(4+))=0.432dotA`