In compounds of type ECl_3,where E = B, P, As | vedclass

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(B) The correct order is $BCl_3 > PCl_3 > AsCl_3 > BiCl_3$. $BCl_3$ has a trigonal planar geometry with a bond angle of $120^{\circ}$ due to $sp^2$ hy

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$B > P > As > Bi$

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(B) The correct order is $BCl_3 > PCl_3 > AsCl_3 > BiCl_3$. $BCl_3$ has a trigonal planar geometry with a bond angle of $120^{\circ}$ due to $sp^2$ hybridization. In the case of $PCl_3$,$AsCl_3$,and $BiCl_3$,the central atom is $sp^3$ hybridized with one lone pair,resulting in a pyramidal geometry. As we move down the group from $P$ to $Bi$,the electronegativity of the central atom decreases,causing the bond pairs to move further away from the central atom. Consequently,the bond angle decreases as we move down the group.

Molecule	Geometry
$(a) SnCl_2$	$(i)$ Angular (or) Bent
$(b) XeF_4$	$(ii)$ See-saw
$(c) ClF_3$	$(iii)$ Square pyramidal
$(d) IF_5$	$(iv)$ $T$-shape
-	$(v)$ Square planar

In compounds of type $ECl_3$,where $E = B, P, As$ or $Bi$,the angles $\angle Cl-E-Cl$ for different $E$ are in the order

Similar Questions

Match the following?
Molecule Geometry
$(a) SnCl_2$ $(i)$ Angular (or) Bent
$(b) XeF_4$ $(ii)$ See-saw
$(c) ClF_3$ $(iii)$ Square pyramidal
$(d) IF_5$ $(iv)$ $T$-shape
- $(v)$ Square planar

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