Bond angles in PB r_(3)(101.5^(@)),(PB l_(3)(100^(@)) and PF_(3)(97^(@)) decrease with increase in electronegativities of the surrounding atoms, however, bond angles in BF_(3),BC l_(3) andB Br_(3) do not change with change in electronegativities of the surrounding atoms. explain with reason.

Bond angles in PB r_(3)(101.5^(@)),(PB l_(3)(100^(@)) and PF_(3)(97^(@

1.	Bond angles in PB r_(3)(101.5^(@)),(PB l_(3)(100^(@)) and PF_(3)(97^(@)) decrease with increase in electronegativities of the surrounding atoms, however, bond angles in BF_(3),BC l_(3) andB Br_(3) do not change with change in electronegativities of the surrounding atoms. explain with reason.
Answer» Solution : (i) `PX_(3)` has a trigonal pyramidal geometry with increase in electronegativity of the surrounding halonge (X) atoms, bond pairs are oriented more towards halogen atoms, resulting in decrease in bond PAIR-bond pair repulsion. hence X-P-X bond angles decrease in the order: `PB r_(3)(101.5^(@)) gt PB l_(3)(100^(@)) gt PF_(3)(97^(@))` (ii) `BX_(3)` has a trigonal planar geometry, where 3 halogen atoms are located at 3 corners of an equilateral TRIANGLE. with increse in electronegativity of halogen atom, bond pair tend to concentrate more towards the halogen atoms resulting in decrease in bond pair-bond pair repulsion. since all 3 halogen atoms lie on the same plane, forming 3 equivalent B-X BONDS, there is no change in the X-B-X bond ANGLE `(120^(@))`.