# 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) $BF_{3} >NH_{3}>PF_{3}>I_{3}^{-}}$

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 sporbital.

- 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)

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

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