To determine the bond angles, we need to find the hybridisation and geometry of each species using VSEPR theory:

• $\text{BCl}_3$: The central boron atom has 3 valence electrons and forms 3 $\sigma$ bonds with chlorine atoms, leaving 0 lone pairs. Its steric number is 3, corresponding to $sp^2$ hybridisation. The geometry is trigonal planar, which results in bond angles of exactly $120^\circ$.
• $\text{PH}_3$: Phosphorus has 5 valence electrons. It forms 3 $\sigma$ bonds and has 1 lone pair. Due to Drago's rule, hybridisation is negligible, and the bond angle is close to $90^\circ$ ($\sim 93.5^\circ$).
• $\text{NCl}_3$: Nitrogen has 5 valence electrons. It forms 3 $\sigma$ bonds and has 1 lone pair ($sp^3$ hybridised). The geometry is trigonal pyramidal, and due to lone pair-bond pair repulsion, the bond angle is compressed to less than $109.5^\circ$ ($\sim 107^\circ$).
• $\text{ClF}_3$: Chlorine has 7 valence electrons. It forms 3 $\sigma$ bonds and has 2 lone pairs ($sp^3d$ hybridised). The geometry is T-shaped, with bond angles slightly less than $90^\circ$ due to strong lone pair repulsions. Therefore, $\text{BCl}_3$ is the only species with a bond angle of $120^\circ$.