Question

Assuming the additivity of covalent radii in the $C-I$ bond, what would be the iodine-iodine distance in each of the diiodobenzene? Assume that the ring is regular hexagon and that each $C-I$ bond lies on a line through the centre of the hexagon. (Take $C-C$ bond-length $=1.40\mathring{A}$ radius of iodine atom $=1.33\mathring{A}$, radius of carbon atom $=0.77\mathring{A}$)
In $o-,m-$ and $p-$ isomers, bond angles are respectively ${60}^o,{120}^o$ and ${180}^o$. Hence, the distance between two-atoms is:

• A. $3.5\mathring{A}$
• B. $4.0\mathring{A}$
• C. $7.0\mathring{A}$
• D. $8.0\mathring{A}$

Answer 1

The correct option is C $7.0\mathring{A}$

(a) $I-$atom are at $C$ and $D$ positions. $CD=?$
Also, $△OCD$ is equilateal triangle.
$CD=OC=OA+AC$
$=1.40+1.33+0.77$
$=3.50\mathring{A}$
($AC$ is the sum of radii of carbon and iodine and $OA=AB$)
(b) $CD=2DE$
But $\frac{DE}{OD}=sin{60}^o$
$DE=ODsin{60}^o$
$=(OB+BD)sin{60}^o$
$=(1.40+0.77+1.33)sin{60}^o$
$=3.50\times \frac{\sqrt{3}}{2}=3.03\mathring{A}$
$\therefore CD=6.06\mathring{A}$
($BD=AC$ as in (a))
(c) $OC=OD=3.50\mathring{A}$
$CD=2\times 3.50=7.0\mathring{A}$