Question

Consider the following iodides :
$P{I}_{3}As{I}_{3}Sb{I}_3$
${102}^{\circ }{100.2}^{\circ }{99}^{\circ }$
The bond angle is maximum in $P{I}_3$ which is :

• A. due to small size of phosphorus
• B. due to more bond pair-bond pair repulsion in $P{I}_3$
• C. due to less electronegativity of P
• D. none of these

Answer 1

The correct option is B due to more bond pair-bond pair repulsion in $P{I}_3$

As the electronegativity of the halogen increases, the halogen exerts a stronger pull on electron pairs it shares with the central atom. This effect reduces the concentration of electrons near the central atom, decreasing somewhat the repulsion between the bonding pairs near the central atom, and allows the lone pair to have more impact in compressing the halogen central atom halogen angles. Consequently, the molecules with the most electronegative outer atoms, $P{F}_3$ and $OS{F}_2$, have the smallest angles.