Question

Select that pair(s) in which bond angle of first molecule is less than second one.

• A. ${Br{O}_3}^{-}$, ${Cl{O}_3}^{-}$
• B. $As{I}_3$, $Sb{I}_3$
• C. $SbB{r}_3$, $Sb{I}_3$
• D. $N{F}_3$, $N{H}_3$

Answer 1

Answer: (A), (C), (D)

${Br{O}_3}^{-}$, ${Cl{O}_3}^{-}$
$SbB{r}_3$, $Sb{I}_3$
$N{F}_3$, $N{H}_3$

(A) The bond angle decreases with decrease in electronegativities and increases with increase in size of the central atom. Hence, the correct order of $O-X-O$ bond angle is ${Cl{O}_3}^{-}>{Br{O}_3}^{-}$
The approximate values of bond angles of the molecules are as follows:
(B) $As{I}_3={100.2}^0;Sb{I}_3={99}^0$
(C) $SbB{r}_3={95}^0;Sb{l}_3={99}^0$

Since, $I$ is very less electronegative, the electron cloud between $Sb$ and $I$ will be more dispersed towards $Sb$ and hence, the bond angle in this case, will be more than $SbB{r}_3$.
(D) $N{H}_3={106.6}^0;N{F}_3={102.2}^0$

$F$ being more electronegative than $N$ in the $N-F$ bond, will draw the electron cloud towards itself. Thus, there will be lesser electronic repulsions around the central atom $N$, and hence, the bond angle will be less.