Question

The distance between any two $I$ atoms in $B{I}_3$ is found to be $3.46\stackrel{o}{A}$. What is the covalent radius of boron. If covalent radius of $I$ atom is $1.33\stackrel{o}{A}$? (Assume all bonds are single bonds in $B{I}_3$)
(Given, $\sqrt{3}=1.73,\sqrt{3}.99=1.99$)

• A. $0.66\stackrel{o}{A}$
• B. $1.77\stackrel{o}{A}$
• C. $1.2\stackrel{o}{A}$
• D. $1.32\stackrel{o}{A}$

Answer 1

The correct option is A $0.66\stackrel{o}{A}$

The hybridisation of $B{I}_3$ is $s{p}^2$
So, it has trigonal planar shape.
$\therefore \angle BIB={120}^o$
Given,
The distance between two $I$ atoms is$3.46A^o$.
Covalent radius of $I$ atom is $=1.33\stackrel{o}{A}$

Since,
$B-I$ bond length are same, thus $\angle BII={30}^o$

If we draw a perpendicular $Bx$, then length of $Ix$ would be $1.73\stackrel{o}{A}$
Now,
$tan{30}^o=\frac{Bx}{Ix}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{Bx}{1.73}$
$\Rightarrow Bx=\frac{1.73}{1.73}=1$
Again,
$B{I}^2=B{x}^2+I{x}^2$
$=1^2+(1.73{)}^2$
$=3.99$
$\Rightarrow BI=1.99$
Let, the covalent radius of $B=Z$
$\therefore$
According to the question,
$1.33+Z=1.99$
$\Rightarrow Z=0.66\stackrel{o}{A}$