Question

Compare $F-\hat{I}-Oand{F}_{axial}-\hat{I}-F_{axial}$ bond angle in $IO{F}_3$ molecule :

• A. $F-\hat{I}-O>F_{axial}-\hat{I}-F_{axial}$
• B. $F_{axial}-\hat{I}-F_{axial}>F-\hat{I}-O$
• C. $F_{axial}-\hat{I}-F_{axial}=F-\hat{I}-O$
• D. None of these

Answer 1

The correct option is B $F_{axial}-\hat{I}-F_{axial}>F-\hat{I}-O$

In case of ${IOF}_3$ the central atom is bonded to 4 other atoms and 1 lone pair (${AX}_{2}E)$ in AXE notation. Such molecules possess a disphenoidal or Seesaw type of molecular geometry.

Compounds with disphenoidal geometry (See-Saw Geometry) have two types of ligands, one pair related by 180degree often called axial ligands. A separate pair of ligands is situated orthogonal to the axial ligands.

In IOF $F_{axial}-I-F_{axial}\left({180}^0\right)>F-I-O\left({120}^0\right)$