$\text{ Decreasing order of bond angle of (I) }{\mathrm{NF}}_3,\text{ (II) }{\mathrm{PH}}_3\text{ (III) }$${\text{ AsF }}_3\text{ is }$

I. ${\mathrm{NCl}}_3>{\mathrm{NF}}_3$

II. $\text{ Bond angle of }{\mathrm{NCl}}_3<{\mathrm{NF}}_3\text{. }$

III. $\text{ Bond angle of }{\mathrm{NH}}_3>{\mathrm{NF}}_3\text{. }$

IV.$\text{ Bond angle of }{\mathrm{NH}}_3<{\mathrm{NF}}_3\text{. }$

$\left[\text{ Bond angle for }{\mathrm{AB}}_x\text{ molecule }\right]\propto \frac{1}{\mathrm{EN}\text{ of atom }\mathrm{B}}$

$\text{ EN of }\mathrm{Cl}<\mathrm{F}\text{. So, bond angle of }{\mathrm{NH}}_3>{\mathrm{NF}}_3\text{. }$

Hence I is correct. EN of H < F. So, bond angle of NH₃ > NF₃. Hence III is also correct.

Thus, decreasing order of bond angle is:

NH₃ > NCl₃ > NF₃. Since EN of H < Cl.