(A) SO₂ is reducing agent and TeO₂ is oxidising agent
(R) From SO₂ to TeO₂ acidic nature of dioxides increases.

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(A) is true but (R) is false

Both (A) and (R) are true and (R) is the correct explanation of (A)

Both (A) and (R) are false

Both (A) and (R) are true and (R) is not the correct explanation of (A)

answer is C.

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Detailed Solution

(A) SO₂ is a reducing agent because it tends to undergo oxidation and lose electrons, causing another substance to undergo reduction and gain electrons. For example, in the reaction between SO₂ and H₂O₂, SO₂ reduces H₂O₂ to H₂O and itself oxidizes to H₂SO₄.

TeO₂, on the other hand, is an oxidizing agent because it tends to undergo reduction and gain electrons, causing another substance to undergo oxidation and lose electrons. For example, in the reaction between TeO₂ and H₂S, TeO₂ oxidizes H₂S to S and itself reduces to Te.

(R) The acidic nature of dioxides generally decreases from left to right across the periodic table. This is because as we move from left to right, the electronegativity of the elements increases, and the dioxides become more acidic. However, this is not the case with SO₂ and TeO₂. SO₂ is acidic, but TeO₂ is amphoteric (can act as both an acid and a base) or weakly acidic, depending on the conditions. Therefore, statement (R) is false.

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