MCQ
The correct order of bond angle
- A$PF_3 < PCl_3 < PBr_3 > PI_3$
- ✓$PF_3 < PCl_3 < PBr_3 < PI_3$
- C$PF_3 > PCl_3 > PBr_3 > PI_3$
- D$PF_3 > PCl_3 < PBr_3 > PI_3$
If $\mathrm{C.A}$. $=$ same Hybridisation same
$L.P.$ $=$ same Bond angle $\propto$ size of side atom
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$A(g)\,\xrightarrow{{{K_1}\, = \,2\, \times \,{{10}^{ - 3}}\,\,{S^{ - 1}}}}2B\,(g)$
$A(g)\,\xrightarrow{{{K_2}\, = \,1\, \times \,{{10}^{ - 3}}\,\,{S^{ - 1}}}}C\,(g)$
