Hybridisation is something you typically deduce from the geometry of the compound you are observing. It is primarily a mathematical concept.
That said, to form the diatomic molecules you asked about, no hybridisation is needed. In fact, you get much better results from assuming unhybridised atoms and then combining their atomic orbitals to molecular ones. Except for hydrogen (which only has 1σ and 2σ), the molecular orbitals of all these diatomic compounds ($\ce{N2, O2, F2, Cl2, Br2, I2}$) are basically identical save the energy levels of the MOs and their populations. Thus, these would all count as unhybridised compounds.
Also, there is no nead to invoke hybridisation to explain the bonding situation in $\ce{HCl}$ either. Instead agan consider a σ bond formed between hydrogen’s 1s orbital and chlorine’s 3p. All of chlorine’s other orbitals will sit there happily filled and unhybridised.