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  • Help me answer: - Chemical Bonding and Molecular Structure - JEE Main-2

The decreasing order of bond angles in BF3, NH3, PF3 and I_{3}^{-}is :

  • Option 1)

    I_{3}^{-} > NH_{3}> PF_{3}> BF_{3}

     

     

     

  • Option 2)

    I_{3}^{-} >BF_{3}> NH_{3}> PF_{3}

  • Option 3)

    BF_{3} >I_{3}^{-}}>PF_{3}>NH_{3}

  • Option 4)

    BF_{3} >NH_{3}>PF_{3}>I_{3}^{-}}

 

Answers (1)

As we learned 

 

S - P Hybridisation -

When one S - and one P - orbital belonging to the same main shell of an atom are mixed together to form two new SP - orbitals.

- wherein

shapes

           Diagonal or Linear

 

 

sp2 Hybridisation -

When one s - and two p - orbitals of the same shell of an atom mix to form three new equivalent orbitals.The hybridised orbital is called sp2 orbital.

- wherein

Shape is Trigonal planar 

 

 

 

 

sp3 hybridisation -

when one s - and three p - orbital belonging to the same shell of an atom mix together to form four new equivalent orbitals is called sp3 orbital.

- wherein

Shape Tetrahedral 

 

 

 

 

                           Bond\: angle=120^{\circ}

 

 

       Bond\: angle=107^{\circ}

 

   when size of central atom increases bond angle decreases 

                                                    B.A of NH3   >    PH3

 

 

        Bond\: angle=180^{\circ}

PF3 < NH3 < BF3 < I_{3}^{-}

 

 

 


Option 1)

I_{3}^{-} > NH_{3}> PF_{3}> BF_{3}

 

 

 

Option 2)

I_{3}^{-} >BF_{3}> NH_{3}> PF_{3}

Option 3)

BF_{3} >I_{3}^{-}}>PF_{3}>NH_{3}

Option 4)

BF_{3} >NH_{3}>PF_{3}>I_{3}^{-}}

Posted by

Vakul

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