(i) Cl₂ from HCl: HCl can be oxidised to chlorine by the number of oxidising agent for example
By the action of any oxidising agent (whose oxidation potential is greater than Cl₂) on HCl
MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
2KMnO₄ + 16HCl → 2KCl + MnCl₂ + 8H₂O + 5Cl₂

(ii) HCl from Cl₂: By the direct combination of elements i.e. hydrogen and chlorine in presence of sunlight.

H₂ + Cl₂→2HCl

PCl₅ can act as an oxidizing agent not as the reducing agent.

PCl₅ can act as an oxidizing agent. The highest oxidation state that P can show is +5. In PCl₅, phosphorus is in its highest state (+5). However, it can decrease its oxidation state and act as an oxidizing agent .

(i) PCl₅ is hydrolysed to give the oxoacid (H₃,PO₄) in the + 5 oxidation state.
PCl₅(s) + 4H₂O→ H₃PO₄ (aq) + 5HCl(aq)

(ii)    In the gas phase, PCl₅ dissociates into trihalide and the chlorine.
${\mathrm{PCl}}_5 \rightleftharpoons {\mathrm{PCl}}_3\left(\mathrm{g}\right) + {\mathrm{Cl}}_2\left(\mathrm{g}\right)$ 

(i) Electronic configuration: They all have six electrons in the outermost shell and have ns² np⁴ general electronic configuration.
(ii) Oxidation state: The outer configuration of all these elements is ns² np⁴. Therefore, they complete their octet either by gaining two electrons or by sharing two electrons. Two types of oxidation states are shown by these elements.
(a) Negative oxidation state: Except the compound OF₂ oxygen shows-2 oxidation state in all its compounds. Due to hgh electronegativity, it forms O²' ion in most of the metal oxides.
The electronegativities of S, Se, Te are low hence their compounds even with most electropositive elements are not more than 50% ionic. Hence S^2–', Se^2–' and Te^2–' are less probable. Being a metal Po does not form Po²⁺ ion at all.
(b) Positive oxidation state: Oxygen does not show positive oxidation state except OF₂(O = + 2). With the increase in atomic number of electro negativity is decreasd in this group, hence the tendency to show the positive oxidation states will increase. S, Se, Te, Po show + 4, +6 oxidation state in addition to + 2.
(iii) Hydride formation: All the elements O, S, Se, Te and Po form M₂M type hydrides (where M = O, S, Se, Te and Po)