Q. 165.0( 1 Vote )

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Answer :

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 H3PO2, P is in +1 oxidation state whereas H3PO4, P is in +5 oxidation state.


Hence, H3PO2 can be oxidized to higher oxidation state while in H3PO4, 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


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