Question

In ${BI}_3$ molecule distance between two $I$ atoms is found to be $3.54\mathring{A}$. Also, ${BI}_3$ has ${sp}^2$-hybridized boron atom. If radius of covalently bonded $I$ atom is $1.33$, the covalent radius of born will be $X\mathring{A}$. Find value of $100X$ ?

Answer 1

The B-I bond length is $X+1.33A^0$
The I-I bond length is 3.54 $A^0$.
$2\times (X+1.33)\times sin60=3.54$.
$X=0.71$
$100X=71$.