Need for hybridization
In general, you should never assume a need for hybridisation unless you have a bonding situation (such as a tetrahedron or related tetravalent structures) which cannot be explained without hybridisation.
As soon as you have a lone pair on any atom, it is more favourable to have this lone pair in an unhybridised s orbital rather than in a hybridised $\mathrm{sp}^n$ orbital. This can be seen for example in phosphane, hydrogen sulphide, etc. in which practically unhybridised central atoms are possible and thus the bond angles show pure p orbitals taking part in the $\ce{A-H}$ bonds.
If you consider very small central atoms (e.g. nitrogen in ammonia) or very large ligands (e.g. chlorine in $\ce{PCl3}$), then having a perfectly unhybridised central atom — meaning a bonding angle of $90^\circ$ — would introduce too much steric strain between the ligands. Thus, the overall system is more favourable if the central atom’s bonding orbitals are partially hybridised, i.e. if the bond angles are enlargened slightly to reduce steric stress. The final bond angle and thus the final central atom’s configuration is a carefully balanced compromise between a hybridisation with a p contribution as high as possible and the ligands separated maximally.

By the way, chlorine is practically never hybridised and it is most certainly unhybridised in $\ce{ClF3}$. Rather, this molecule should be understood as having:

a single $\ce{Cl-F}$ σ bond using a p orbital both of chlorine and of fluorine

a four-electron-three-centre bond using a p orbital of each of the three atoms that form the linear $\ce{F\bond{...}Cl\bond{...}F}$ fragment (i.e. a p orbital of chlorine). This bond can be understood with the following resonance structures:
$$\ce{F-\overset{+}{Cl}\bond{...}\overset{-}{F} &lt;-&gt; \overset{-}{F}\bond{...}\overset{+}{Cl}-F}$$

one chlorine lone pair in a p orbital perpendicular to all $\ce{C-F}$ bonds

one chlorine lone pair in an s-type orbital.