Redox Reactions And Electrochemistry Ques 753

Question: Match List I with List II and select the correct answer using the code given below the lists:
List I List II
Oxidising agent (p) Disproportionation
$ Mn_3O_4 $ (q) Redox reaction
$ C_6H_6 $ (r) Decreases oxidation number
$ 2C{{u}^{+}}\xrightarrow{{}}C{{u}^{2+}} $ $ +Cu^{0} $ (s) Fractional oxidation number
(E) $ H_2O_2+O_3\xrightarrow{{}} $ $ H_2O+2O_2 $ (t) Oxidation number is -1

Codes:

Options:

A) A $ \to $ q, B $ \to $ p, C $ \to $ t, D $ \to $ s, E $ \to $ r

B) A $ \to $ t, B $ \to $ s, C $ \to $ r, D $ \to $ q, E $ \to $ P

C) A $ \to $ r, B $ \to $ t, C $ \to $ s, D $ \to $ p, E $ \to $ q

D) A $ \to $ r, B $ \to $ s, C $ \to $ t, D $ \to $ p, E $ \to $ q

Show Answer

Answer:

Correct Answer: D

Solution:

[d] [a] Oxidising agent, which undergo in reduction process and decreases its oxidation number. [b] It is $ MnO,Mn_2O_3 $ and O.S. $ =\frac{+8}{3} $ [c] $ C_6H_66x+6=0 $ $ x=-1 $ [d] Disproportionate reaction. (e) Redox reaction.



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