Question

In compound of the type $EC{l_3}$ , where $E = B,P,As,Bi$ the angles $Cl - E - Cl$ for different $E$ are in the order:
A. $B > P = As = Bi$
B. $B > P > As > Bi$
C. $B < P < As = Bi$
D. $B < P < As < Bi$

Answer 1

The two most important keys for a compound for the determination of bond angle are: In order to determine the bond angles between them firstly we must know the geometry of the compound or the structure of the compound and secondary the hybridization of the compound.

Complete answer:
In the case of $BC{l_3}$ ,
It is $s{p^2}$ hybridized compound and its geometry is trigonal planar. It has a bond angle that is the angle between $Cl - B - Cl$ is ${120^0}$ .
In case of $PC{l_3}$ ,
It is $s{p^3}$ hybridized compound and its geometry is pyramidal. It has a bond angle that is the angle between $Cl - P - Cl$ is ${109^0}$ approximately.
In case of $AsC{l_3}$ ,
It is also $s{p^3}$ hybridized compound and its geometry is pyramidal. It has a bond angle that is the angle between $Cl - As - Cl$ is $> {109^0}$ . Its bond angle is less than that of phosphorus because on going down the group bond angle decreases.
In case of $BiC{l_3}$ ,
It is also $s{p^3}$ hybridized compound and its geometry is pyramidal. It has a bond angle that is the angle between $Cl - Bi - Cl$ is $> > {109^0}$ . Its bond angle is less than that of astatine because on going down the group bond angle decreases.
So the correct order of bond angle is: $BC{l_3} > PC{l_3} > AsC{l_3} > BiC{l_3}$

So, option B is correct.

Note:
For the determination of bond angle VSEPR theory is used. It is also called the Gillespie-Nyholm theory after its two main developers, Ronald Gillespie and Ronald Nyholm.VSEPR theory refers to the valence shell electron pair repulsion theory. This theory gives some consideration which determines the bond angle. Number of lone pairs, hybridization and electronegativity of the central atom is responsible for the bond angle of a compound.