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The decreasing order oof bond angles in ...

The decreasing order oof bond angles in `BF_(2), NH_(3), PF_(3)` and `I_(3)^(-)` is:

A

`I_(3)^(-) gtNH_(3) gtPF_(3) gtBF_(3)`

B

`BF_(3) gt I_(3)^(-)gt PF_(3) gt NH_(3)`

C

`BF_(3) gt NH_(3) gtPF_(3) gt I_(3)^(-)`

D

`I_(3)^(-) gt BF_(3) gtNH_(3) gt PF_(3)`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the decreasing order of bond angles in BF3, NH3, PF3, and I3^(-), we need to analyze the molecular geometry and the factors affecting bond angles in each of these compounds. ### Step 1: Analyze BF3 - **Molecular Geometry**: BF3 has a central boron atom with three fluorine atoms bonded to it. Boron has three valence electrons and forms three sigma bonds with fluorine. - **Steric Number**: The steric number is 3 (3 bonding pairs, 0 lone pairs). - **Hybridization**: The hybridization is sp². - **Bond Angle**: The ideal bond angle for sp² hybridization is 120 degrees. Since there are no lone pairs, the bond angle remains 120 degrees. ### Step 2: Analyze NH3 - **Molecular Geometry**: NH3 has a central nitrogen atom bonded to three hydrogen atoms and has one lone pair of electrons. - **Steric Number**: The steric number is 4 (3 bonding pairs, 1 lone pair). - **Hybridization**: The hybridization is sp³. - **Bond Angle**: The ideal bond angle for sp³ hybridization is 109.5 degrees. However, due to the presence of the lone pair, the bond angle is reduced to approximately 107 degrees. ### Step 3: Analyze PF3 - **Molecular Geometry**: PF3 has a central phosphorus atom bonded to three fluorine atoms and has one lone pair of electrons. - **Steric Number**: The steric number is 4 (3 bonding pairs, 1 lone pair). - **Hybridization**: The hybridization is sp³. - **Bond Angle**: Similar to NH3, the ideal bond angle is 109.5 degrees, but due to the lone pair, the bond angle will be slightly less than 109.5 degrees. However, since phosphorus is less electronegative than nitrogen, the bond angle in PF3 is slightly larger than in NH3, approximately around 108 degrees. ### Step 4: Analyze I3^(-) - **Molecular Geometry**: I3^(-) has a linear structure. It consists of three iodine atoms, with the central iodine atom bonded to two other iodine atoms and has three lone pairs. - **Steric Number**: The steric number is 5 (2 bonding pairs, 3 lone pairs). - **Hybridization**: The hybridization is sp³d. - **Bond Angle**: The bond angle in a linear molecule is 180 degrees. ### Step 5: Compare Bond Angles Now, we can summarize the bond angles: - BF3: 120 degrees - I3^(-): 180 degrees - PF3: approximately 108 degrees - NH3: approximately 107 degrees ### Conclusion The decreasing order of bond angles is: **I3^(-) > BF3 > PF3 > NH3**
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