The decreasing order of bond angles in $BF_{3} , NH_{3} , PF_{3}$ and $I_{3}{^{-}}$ is

Question

How many of the following compounds have zero dipole moment ?
$NH_3, H_2O, HF, CO_2, SO_2, BF_3, CH_4$

Answer 1

To determine the decreasing order of bond angles in the molecules $ BF_3 $, $ NH_3 $, $ PF_3 $, and $ I_3^{-} $, it's essential to consider the molecular geometry and the factors affecting bond angles.

$ BF_3 $ (Boron Trifluoride):
- This molecule has a trigonal planar geometry because boron is bonded to three fluorine atoms with no lone pairs on the central atom.
- The bond angle is exactly $ 120^\circ $.
$ NH_3 $ (Ammonia):
- This molecule has a trigonal pyramidal geometry due to the presence of one lone pair on the nitrogen atom.
- The lone pair-bond pair repulsion reduces the bond angle from the tetrahedral $ 109.5^\circ $ to approximately $ 107^\circ $.
$ PF_3 $ (Phosphorus Trifluoride):
- Similar to $ NH_3 $, this molecule also has a trigonal pyramidal geometry, but the presence of larger fluorine atoms results in bond angle strain.
- The bond angle is slightly less than that of $ NH_3 $, around $ 96^\circ $.
$ I_3^{-} $ (Triiodide Ion):
- This is a linear molecule because the central iodine has three lone pairs, effectively arranging the two terminal iodines at $ 180^\circ $.

From the above analysis, the order of bond angles is highest for the linear molecule $ I_3^{-} $ followed by $ BF_3 $, and then by $ NH_3 $ and $ PF_3 $, respectively.

The correct order is: I_3{^{-}} > BF_3 > NH_3 > PF_3.

Answer 2

To determine the decreasing order of bond angles in the molecules $ BF_3 $, $ NH_3 $, $ PF_3 $, and $ I_3^{-} $, it's essential to consider the molecular geometry and the factors affecting bond angles.

$ BF_3 $ (Boron Trifluoride):
- This molecule has a trigonal planar geometry because boron is bonded to three fluorine atoms with no lone pairs on the central atom.
- The bond angle is exactly $ 120^\circ $.
$ NH_3 $ (Ammonia):
- This molecule has a trigonal pyramidal geometry due to the presence of one lone pair on the nitrogen atom.
- The lone pair-bond pair repulsion reduces the bond angle from the tetrahedral $ 109.5^\circ $ to approximately $ 107^\circ $.
$ PF_3 $ (Phosphorus Trifluoride):
- Similar to $ NH_3 $, this molecule also has a trigonal pyramidal geometry, but the presence of larger fluorine atoms results in bond angle strain.
- The bond angle is slightly less than that of $ NH_3 $, around $ 96^\circ $.
$ I_3^{-} $ (Triiodide Ion):
- This is a linear molecule because the central iodine has three lone pairs, effectively arranging the two terminal iodines at $ 180^\circ $.

From the above analysis, the order of bond angles is highest for the linear molecule $ I_3^{-} $ followed by $ BF_3 $, and then by $ NH_3 $ and $ PF_3 $, respectively.

The correct order is: I_3{^{-}} > BF_3 > NH_3 > PF_3.

The decreasing order of bond angles in $BF_{3} , NH_{3} , PF_{3}$ and $I_{3}{^{-}}$ is

The Correct Option is B

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