In compounds of type $E C l_{3}$ , wher $E = B, P, A s o r B i,$ the angles $C l - E - C l$ for different $E$ are in the order:

B > P = A s = B i

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B > P > A s > B i

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B < P = A s = B i

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B < P < A s < B i

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Solution

The correct option is B $B > P > A s > B i$
In $B C l_{3},$ $H = \frac{1}{2} (3 + 3 + 0 - 0) = 3; s p^{2}$ hybridization $(b o n d a n g l e = 120^{\circ})$ similarly $P C l_{3}, A s C l_{3}, a n d B i C l_{3}$ are found to have $s p^{3}$ hybrized central atom with one lone pair of electrons on the central atom. The $b o n d a n g l e \leq 109^{\circ} 28^{'}$ , since the central atoms belong to the same group, the bond angle of the chlorides decreases as we go down the group. Thus, the order of bond angle is, $B C l_{3} > P C l_{3} > A s C l_{3} > B i C l_{3}$ .

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