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Q.

(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

(A) is true but (R) is false

b

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

c

Both (A) and (R) are false

d

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

answer is C.

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Detailed Solution

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