Question

The correct order of bond angle
(1) $P{F_3} < PC{l_3} < PB{r_3} < P{I_3}$
(2) $P{I_3} < PC{l_3} < PB{r_3} < P{F_3}$
(3) $P{F_3} < PB{r_3} < PC{l_3} < P{I_3}$
(4) $P{F_3} < PC{l_3} < P{I_3} < PB{r_3}$

Answer 1

Hint: The electronegativity difference between the bonded atoms decide the existence of electrons on the respective atoms. The close the electrons exist around the central atom the more is the repulsion between them. Phosphorus is an element in the periodic table with atomic number\[15\]. The electronic configuration of phosphorus is $P:1{s^2}2{s^2}2{p^6}3{s^2}3{p^3}$

Complete step by step answer:
The outer shell of phosphorus is\[3\]. It has a total of five electrons in the outermost shell with two electrons in \[3s\] and three electrons in \[3p\] shell. Out of the five electrons three are bonded to three halide atoms and two electrons remain non-bonded. The central atom phosphorus is \[s{p^3}\] hybridized.
Now the given set of compounds varies in the halide partner. From fluorine to iodine electronegativity decreases down the group. As the size of the atom increases the electrons become diffuse in the space around the nucleus of each atom. So the electron pull of the electrons present in the outermost shell decreases as they are shielded by electrons present on inner shells.
As the electronegativity decreases from fluorine to iodine then the shared pair of electrons between phosphorus and halogen will reside on phosphorus more in iodine than on fluorine. Higher the electron density of phosphorus more will be the lone pair-bond pair repulsion. The greater the repulsion the more is the bond angle.
Thus the bond angle of \[P{F_3}\] is lowest and the bond angle of \[P{I_3}\] is highest.

Hence, the correct order of bond angle is \[P{F_3} < PC{l_3} < PB{r_3} < P{I_3}\], i.e. option (1) is the correct answer. .

Note: In the Stock and Pohlands method, the yield of borazine is in very less quantity because of the formation of many by-products. Borazine can also be prepared by heating $BC{l_3}$ and $N{H_4}Cl$ at $140^\circ C$ and also by heating mixture of $LiB{H_4}$ and $N{H_4}Cl$ at $230^\circ C$.