REDOX REACTION QUIZ-12
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Chemistry is important because everything you do is chemistry! Even your body is made of chemicals. Chemical reactions occur when you breathe, eat, or just sit there reading. All matter is made of chemicals, so the importance of chemistry is that it's the study of everything..
Q1. Which of the following reactions does not involve either oxidation or reduction?
Solution
(c). Since oxidation state of Cr in both reactant and product is same.
(c). Since oxidation state of Cr in both reactant and product is same.
Q2.Which of the following is not a disproportionation reaction?
1. NH4NO3 ---> N2O+H2O
2. P4 ---> PH3+HPO2^(⊖)
3. PCl5--->PCl_3+Cl2
4. IO3^( ⊖)+I^⊝⟶I2
1. NH4NO3 ---> N2O+H2O
2. P4 ---> PH3+HPO2^(⊖)
3. PCl5--->PCl_3+Cl2
4. IO3^( ⊖)+I^⊝⟶I2
Solution
(b). Reduction II is disproportionation, while I, III, and IV are not
(b). Reduction II is disproportionation, while I, III, and IV are not
Q3. 
Solution
(c). Isocyanides are oxidised to isocyanates. So the answer is (c)
(c). Isocyanides are oxidised to isocyanates. So the answer is (c)
Q4. In the compound YBa2Cu3O7 which shows superconductivity, what is the oxidation state of Cu? Assume that the rare earth element yttrium is in its usual +3 oxidation state
Solution
(a). (YBa2Cu3O7) 3+(2×2)+3x-14=0 ∴ x=+7/3
(a). (YBa2Cu3O7) 3+(2×2)+3x-14=0 ∴ x=+7/3
Q5.An alkene on ozonolysis yields only ethanal. There is an isomer of the alkene which on ozonolysis yields:
Solution
option-(a)
option-(a)
Q6. The final product obtained in the oxidation of t-butyl benzene with Na2Cr2O7+H2SO4 is:
Solution
(d). Due to (+I) effect of three (Me) groups in t-butyl group, the benzene is activated and unstable, so the oxidation of the benzene ring takes place. The oxidation of the side chain containing t-butyl group does not take place because it does not have the benzylic H atom
(d). Due to (+I) effect of three (Me) groups in t-butyl group, the benzene is activated and unstable, so the oxidation of the benzene ring takes place. The oxidation of the side chain containing t-butyl group does not take place because it does not have the benzylic H atom
Q7.The number of moles of KMnO_4 reduced by 1 mol of KI in alkaline medium is
Solution
(b). In a basic medium, 3e^-+MnO4^(⊖)⟶MnO_2 I^⊖⟶IO3^(⊖)+6e^- Equivalent of MnO_4^⊖=Equivalent ofI^⊖ (n=3)(n=6) 1 Eq=1 Eq 1/3 mol=1/6 mol 6/3 mol ofMnO4^( ⊖)=1 mol of I^⊝ ∴2 mol of MnO4^( ⊖)=1 mol ofI^⊖
(b). In a basic medium, 3e^-+MnO4^(⊖)⟶MnO_2 I^⊖⟶IO3^(⊖)+6e^- Equivalent of MnO_4^⊖=Equivalent ofI^⊖ (n=3)(n=6) 1 Eq=1 Eq 1/3 mol=1/6 mol 6/3 mol ofMnO4^( ⊖)=1 mol of I^⊝ ∴2 mol of MnO4^( ⊖)=1 mol ofI^⊖
Q8.To an acidic solution of an anion, a few drops of KMnO4 solution are added. Which of the following, if present, will not decolourise the KMnO4 solution?
Solution
(a). Oxidation state of C in CO3^(2-)is +4, which is maximum. So, it will not be oxidised
(a). Oxidation state of C in CO3^(2-)is +4, which is maximum. So, it will not be oxidised
Q9.Oxidation states of the metal in the minerals haematite and magnetite, respectively, are
Solution
(d). Haematite is Fe2O3, in which oxidation number of iron is III. Magnetite is Fe3O4 which is infact a mixed oxide (FeO.Fe2O3.), hence iron is present in both II and III oxidation state.
(d). Haematite is Fe2O3, in which oxidation number of iron is III. Magnetite is Fe3O4 which is infact a mixed oxide (FeO.Fe2O3.), hence iron is present in both II and III oxidation state.
Q10. Which single reagent can be used in the following conversions?
Solution
(c). H2 + Pd + C (charcoal) selectively reduces (C≡C) to (C=C) bond, benzyl alcohol to toluene, azides to amines, and (-NO2) to (-NH2) group. So the answer is (c)
(c). H2 + Pd + C (charcoal) selectively reduces (C≡C) to (C=C) bond, benzyl alcohol to toluene, azides to amines, and (-NO2) to (-NH2) group. So the answer is (c)
