Question

The correct order of bond angles is:

• A. $P{F}_3<P{Cl}_3<P{Br}_3<P{I}_3$
• B. $P{F}_3<P{Br}_3<P{Cl}_3<P{I}_3$
• C. $P{I}_3<P{Br}_3<P{Cl}_3<P{F}_3$
• D. $P{F}_3>P{Cl}_3<P{Br}_3<P{I}_3$

Answer 1

Answer: (D)

$P{F}_3>P{Cl}_3<P{Br}_3<P{I}_3$

The correct order of bond angles is $P{F}_3>P{Cl}_3<P{Br}_3<P{I}_3$.

Bond angle is directly proportional to electronegativity of terminal atom. $P{Cl}_3<P{Br}_3<P{I}_3$, the bond angle order is explained in terms of increasing electronegativity of halogens, whereas, $P{F}_3>P{Cl}_3$, bond angle order is explained in terms of $p\pi -d\pi$ bonding in $P{F}_3$.

All in group 7B and their atomic radii are increasing as you go down the period. $F$ is the smallest and most electronegative at the same time, so it attracts more electrons and thus smaller angle. If one compare it to $I$, which is the biggest among the given atoms, $I$ will make greater angle because of its bigger size.