Question

Which of the following compounds has the smallest bond angle $(X-A-X)$ in each series respectively?
$\left(A\right)OS{F}_2$ $OSC{l}_2$ $OSB{r}_2$
$\left(B\right)SbC{l}_3$ $SbB{r}_3$ $Sb{I}_3$
$\left(C\right)P{I}_3$ $As{I}_3$ $Sb{I}_3$

• A. $OS{F}_2,SbC{l}_3$ and $P{I}_3$
• B. $OSB{r}_2,Sb{I}_3$ and $P{I}_3$
• C. $OS{F}_2,Sb{I}_3$ and $P{I}_3$
• D. $OS{F}_2,SbC{l}_3$ and $Sb{I}_3$

Answer 1

The correct option is D $OS{F}_2,SbC{l}_3$ and $Sb{I}_3$

Order of decreasing bond angle.


Since $F$ has the highest electronegativity among $Cl,Br$ and $F$, the electron density on the $S-F$ bond is more towards the flourine atom. So, bond pair- bond pair repulsion is least in case of $OS{F}_2$ and the bond angle $\angle FSF$ is also smallest.



Since $Cl$ has the highest electronegativity among $Cl,Br$ and $I$, the electron density on the $Sb-Cl$ bond is more towards the chlorine atom. So, bond pair-bond pair repulsion is least in case of $SbC{l}_3$ and the bond angle is smallest in $SbC{l}_3$.


Inert pair effect becomes more predominant as we go down the group. The $s$-electrons thus become more tightly held (more penetrating) and therefore, the energy gap between $s$ and $p$ orbital increases. Thus, the tendency of $s$ orbital to participate in hybridization decreases. Hence, down the group the $s-$character decreases in $A-I$ bond and lesser is the
$s-$character, smaller is the bond angle.