Assign reason for the following:

A. is a stronger reducing agent than

B. Sulphur shows more tendency for catenation than Oxygen.

C. Reducing character increases from HF to HI.

a) First, we will calculate the oxidation state of P in H₃PO₄

Let oxidation state of P is x

Oxidation state of H is 1 and oxygen is (-2)

⇒ 1 × 3 + x + (-2) × 4 = 0

⇒ 3 + x-8 = 0

⇒ x = 5

⇒ Thus, in H₃PO₄, the oxidation state of phosphorus is +5

Now, we will calculate the oxidation state of P in H₃PO₂

Let oxidation state of P is x

Oxidation state of H is 1 and oxygen is (-2)

⇒ 1 × 3 + x + (-2) × 2 = 0

⇒ 3 + x-4 = 0

⇒ x = 1

⇒ Thus, in H₃PO₂, the oxidation state of phosphorus is +1

⇒ H₃PO₂ is a stronger reducing agent than the H₃PO₄ because in H₃PO_2, P is in +1 oxidation state whereas H₃PO_4, P is in +5 oxidation state.

⇒ Hence, H₃PO₂ can be oxidized to higher oxidation state while in H₃PO_4, P is present in its highest oxidation state.

b) Sulphur shows more tendency for catenation than Oxygen because:

⇒ S-S bonds are stronger than O-O bond.

⇒ Due to small size, the lone pairs of electrons on the oxygen atoms repel the bond pair of O-O bond to a greater extent than the lone pairs on the Sulphur atoms in S-S bond.

⇒ As a result, S-S bond is much stronger than O-O bond.

⇒ Hence, Sulphur has a much stronger tendency for catenation than oxygen.

c) Reducing character depends upon the bond dissociation energy. Thus, greater the bond dissociation energy, more stable is the halogen acid and hence weaker is the reducing agent.

⇒ Since, the bond dissociation energies of the halogen acids increase in the order:

HF < HCl < HBr < HI