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.

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

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