Chemistry Examplar Problems (EN)

I. Multiple Choice Questions (Type-I) 1. Which of the following is not an example of redox reaction? (i) CuO + H2 ⎯→ Cu + H2O (ii) FeO + 3CO ⎯→ 2Fe + 3CO 23 2 (iii) 2K + F⎯→ 2KF 2 (iv) BaCl2 + H2SO4 ⎯→ BaSO4 + 2HCl 2. The more positive the value of EV, the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below find out which of the following is the strongest oxidising agent. EV values: Fe3+/Fe2+ = + 0.77; I(s)/I– = + 0.54; 2 Cu2+/Cu = + 0.34; Ag+/Ag = + 0.80V (i) Fe3+ (ii) I(s) 2 (iii) Cu2+ (iv) Ag+ EV 3. values of some redox couples are given below. On the basis of these values choose the correct option. EV values : Br/Br– = + 1.90; Ag+ /Ag(s) = + 0.80 2 Cu2+/Cu(s) = + 0.34; I(s)/I– = + 0.54 2 (i) Cu will reduce Br– (ii) Cu will reduce Ag (iii) Cu will reduce I– (iv) Cu will reduce Br2 4. Using the standard electrode potential, find out the pair between which redox reaction is not feasible. EV values : Fe3+/Fe2+ = + 0.77; I2/I– = + 0.54; Cu2+/Cu = + 0.34; Ag+/Ag = + 0.80 V (i) Fe3+ and I– (ii) Ag+ and Cu Fe3+ (iii) and Cu (iv) Ag and Fe3+ 5. Thiosulphate reacts differently with iodine and bromine in the reactions given below: 2SO2– + I→ SO2– + 2I– 232 46 2– 2– S2O3 + 2Br2 + 5H2O → 2SO4 + 2Br–+ 10 H+ Which of the following statements justifies the above dual behaviour of thiosulphate? (i) Bromine is a stronger oxidant than iodine. (ii) Bromine is a weaker oxidant than iodine. (iii) Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reactions. (iv) Bromine undergoes oxidation and iodine undergoes reduction in these reactions. 6. The oxidation number of an element in a compound is evaluated on the basis of certain rules. Which of the following rules is not correct in this respect? (i) The oxidation number of hydrogen is always +1. (ii) The algebraic sum of all the oxidation numbers in a compound is zero. (iii) An element in the free or the uncombined state bears oxidation number zero. (iv) In all its compounds, the oxidation number of fluorine is – 1. 7. In which of the following compounds, an element exhibits two different oxidation states. (i) NH2OH (ii) NHNO 43 (iii) NH 24 (iv) N3H 8. Which of the following arrangements represent increasing oxidation number of the central atom? (i) CrO–2 , ClO3– , CrO42– , MnO4– –2– –– (ii) ClO, CrO,MnO, CrO 34 42 –––2– (iii) CrO, ClO , MnO, CrO 23 44 –– (iv) CrO2–, MnO– , CrO, ClO 4 423 9. The largest oxidation number exhibited by an element depends on its outer electronic configuration. With which of the following outer electronic configurations the element will exhibit largest oxidation number? (i) 3d14s 2 (ii) 3d34s 2 3d54s 1 (iii) (iv) 3d54s 2 10. Identify disproportionation reaction (i) CH + 2O⎯→ CO + 2HO 42 22 (ii) CH4 + 4Cl2 ⎯→ CCl4 + 4HCl (iii) 2F + 2OH– ⎯→ 2F– + OF + HO 2 22 (iv) 2NO2 + 2OH – ⎯→ NO2– + NO3– + H2O 11. Which of the following elements does not show disproportionation tendency? (i) Cl (ii) Br (iii) F (iv) I II. Multiple Choice Questions (Type-II) In the following questions two or more options may be correct. 12. Which of the following statement(s) is/are not true about the following decomposition reaction. 2KClO3 → 2KCl + 3O2 (i) Potassium is undergoing oxidation (ii) Chlorine is undergoing oxidation (iii) Oxygen is reduced (iv) None of the species are undergoing oxidation or reduction 13. Identify the correct statement (s) in relation to the following reaction: Zn + 2HCl → ZnCl2 + H2 (i) Zinc is acting as an oxidant (ii) Chlorine is acting as a reductant (iii) Hydrogen ion is acting as an oxidant (iv) Zinc is acting as a reductant 14. The exhibition of various oxidation states by an element is also related to the outer orbitalelectronic configuration ofits atom. Atom(s) having which of the following outermostelectronic configurations will exhibit more than one oxidation state in its compounds. (i) 3s 1 (ii) 3d14s 2 (iii) 3d24s 2 (iv) 3s 23p 3 15. Identify the correct statements with reference to the given reaction P4 + 3OH– + 3H2O → PH3 + 3H2PO2– (i) Phosphorus is undergoing reduction only. (ii) Phosphorus is undergoing oxidation only. (iii) Phosphorus is undergoing oxidation as well as reduction. (iv) Hydrogen is undergoing neither oxidation nor reduction. 16. Which of the following electrodes will act as anodes, when connected to Standard Hydrogen Electrode? EV (i) Al/Al3+ = –1.66 (ii) Fe/Fe2+ EV = – 0.44 EV (iii) Cu/Cu2+ = + 0.34 EV (iv) F (g)/2F– (aq) = + 2.87 2 III. Short Answer Type 17. The reaction Cl2 (g) + 2OH– (aq) ⎯→ ClO– (aq) + Cl– (aq) + H2O (l ) represents the process of bleaching. Identify and name the species that bleaches the substances due to its oxidising action. 2– – 18. MnO4 undergoes disproportionation reaction in acidic medium but MnO4 does not. Give reason. 19. PbOand PbOreact with HCl according to following chemical equations : 2 2PbO + 4HCl ⎯→ 2PbCl2 + 2H2O PbO2 + 4HCl ⎯→ PbCl2 + Cl2 + 2H2O Why do these compounds differ in their reactivity? 20. Nitric acid is an oxidising agent and reacts with PbO but it does not react with PbO. Explain why? 2 21. Write balanced chemical equation for the following reactions: (i) Permanganate ion (MnO– ) reacts with sulphur dioxide gas in acidic 4 medium to produce Mn2+ and hydrogensulphate ion. (Balance by ion electron method) (ii) Reaction of liquid hydrazine (N2H4) with chlorate ion (ClO3– ) in basic medium produces nitric oxide gas and chloride ion in gaseous state. (Balance by oxidation number method) (iii) Dichlorine heptaoxide (Cl2O7) in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion (ClO2– ) and oxygen gas. (Balance by ion electron method) 22. Calculate the oxidation number of phosphorus in the following species. 2– 3– (a) HPO and (b) PO 3 4 23. Calculate the oxidation number of each sulphur atom in the following compounds: (a) NaSO(b) NaSO(c) NaSO (d) NaSO 22 3 246 2324 24. Balance the following equations by the oxidation number method. (i) Fe2+ + H+ + Cr2O2– ⎯→ Cr3+ + Fe3+ + H2O (ii) I + NO– ⎯→ NO + IO– 7 2323 2– 2– (iii) I2 + S2O ⎯→ I– + S4O 36 (iv) MnO2 + C2 O2– ⎯→ Mn2+ + CO2 4 25. Identify the redox reactions out of the following reactions and identify the oxidising and reducing agents in them. (i) 3HCl(aq) + HNO (aq) ⎯→ Cl (g) + NOCl (g) + 2HO (l ) 32 2 (ii) HgCl2 (aq) + 2KI (aq) ⎯→ HgI2 (s) + 2KCl (aq) (iii) FeO (s) + 3CO (g) Δ→ 2Fe (s) + 3CO (g) 23⎯⎯⎯ 2 (iv) PCl (l) + 3HO (l) ⎯→ 3HCl (aq) + H PO (aq) 32 33 (v) 4NH3 + 3O2 (g) ⎯→ 2N2 (g) + 6H2O (g) 26. Balance the following ionic equations (i) CrO2– + H+ + I– ⎯→ Cr3+ + I + HO 27 22 (ii) Cr2O2– + Fe2+ + H+ ⎯→ Cr3+ + Fe3+ + H2O 7 – 2– 2– (iii) Mn O + SO + H+ ⎯→ Mn2+ + SO + HO 4342–– (iv) Mn O + H+ + Br ⎯→ Mn2+ + Br2 + H2O 4 IV. Matching Type 27. Match Column I with Column II for the oxidation states of the central atoms. Column I Column II (i) CrO2– (a) + 3 (ii) MnO4– (b) + 4 27 (iii) VO– (c) + 5 3 (iv) FeF3– (d) + 6 6 (e) + 7 28. Match the items in Column I with relevant items in Column II. Column I Column II (i) Ions having positive charge (a) +7 (ii) The sum of oxidation number (b) –1 of all atoms in a neutral molecule (c) +1 (iii) Oxidation number of hydrogen ion (H+) (d) 0 (iv) Oxidation number of fluorine in NaF (e) Cation (v) Ions having negative charge (f) Anion V. Assertion and Reason Type In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question. 29. Assertion (A) : Among halogens fluorine is the best oxidant. Reason (R) : Fluorine is the most electronegative atom. (i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true but R is not the correct explanation of A. (iii) A is true but R is false. (iv) Both A and R are false. 30. Assertion (A): In the reaction between potassium permanganate and potassium iodide, permanganate ions act as oxidising agent. Reason (R) : Oxidation state of manganese changes from +2 to +7 during the reaction. (i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true but R is not the correct explanation of A. (iii) A is true but R is false. (iv) Both A and R are false. 31. Assertion (A) : The decomposition of hydrogen peroxide to form water and oxygen is an example of disproportionation reaction. Reason (R) : The oxygen of peroxide is in –1 oxidation state and it is converted to zero oxidation state in O and –2 oxidation state 2 in H2O. (i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true but R is not the correct explanation of A. (iii) A is true but R is false. (iv) Both A and R are false. 32. Assertion (A) : Redox couple is the combination of oxidised and reduced form of a substance involved in an oxidation or reduction half cell. EV EV Reason (R) : In the representation 3+ 2+ and 2+ , Fe3+/ Fe2+ Fe /Fe Cu /Cu and Cu2+ / Cu are redox couples. (i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true but R is not the correct explanation of A. (iii) A is true but R is false. (iv) Both A and R are false. VI. Long Answer Type 33. Explain redox reactions on the basis of electron transfer. Give suitable examples. 34. On the basis of standard electrode potential values, suggest which of the following reactions would take place? (Consult the book for EV value). (i) Cu + Zn2+ ⎯→ Cu2+ + Zn (ii) Mg + Fe2+ ⎯→ Mg2+ + Fe (iii) Br + 2Cl– ⎯→ Cl + 2Br– 22 (iv) Fe + Cd2+ ⎯→ Cd + Fe2+ 35. Why does fluorine not show disporportionation reaction? 36. Write redox couples involved in the reactions (i) to (iv) given in question 34. 37. Find out the oxidation number of chlorine in the following compounds and arrange them in increasing order of oxidation number of chlorine. NaClO4, NaClO3, NaClO, KClO2, Cl2O7, ClO3, Cl2O, NaCl, Cl2, ClO2. Which oxidation state is not present in any of the above compounds? 38. Which method can be used to find out strength of reductant/oxidant in a solution? Explain with an example. ANSWERS I. Multiple Choice Questions (Type-I) 1. (iv) 2. (iv) 3. (iv) 4. (iv) 5. (i) 6. (i) 7.(ii) 8.(i) 9. (iv) 10. (iv) 11. (iii) II. Multiple Choice Questions (Type-II) 12. (i),(iv) 13. (iii), (iv) 14. (iii),(iv) 15. (iii), (iv) 16. (i), (ii) III. Short Answer Type 17. Hypochlorite ion 18. In MnO –4 , Mn is in the highest oxidation state i.e. +7. Therefore, it does 2– not undergo disproportionation. MnO4 undergoes disproportionation as follows : – 3MnO2– + 4H+ ⎯→ 2MnO + MnO + 2HO 4 422 19. 2PbO + 4HCl ⎯→ 2PbCl2 + 2H2O (Acid base reaction) PbO + 4HCl ⎯→ PbCl + Cl + 2HO (Redox reaction) 2 222 (Hint : Note the oxidation number of lead in the oxides) 20. PbO is a basic oxide and simple acid base reaction takes place between PbO and HNO. On the other hand in PbO lead is in + 4 oxidation state 3 2 and cannot be oxidised further. Therefore no reaction takes place. Thus, PbO is passive, only PbO reacts with HNO. 2 3 2PbO + 4HNO3 ⎯→ 2Pb (NO3)2 + 2H2O (Acid base reaction) 22. (a) +3, (b) +5 23. (a) +2 (b) +5, 0, 0, +5 (c) +4 (d) +6 Justification : Write Lewis structure of each ion then assign electron pair shared between atoms of different electronegativity to more electronegative atom and distribute the electron pair shared between atoms of same element equally. Now count the number of electrons possessed by each atom. Find out the difference in number of electrons possessed by neutral atom and that possessed by atom in the compound. This difference is the oxidation number. If atom present in the compound possesses more electrons than the neutral atom, the oxidation number is negative. If it possesses less electrons then oxidation number is positive. (i) Lewis structure of S2O42– can be written as follows : Electron pair shared between sulphur and oxygen is assigned to oxygen atoms because of more electronegativity of oxygen. Thus each sulphur atom is deficient of two electrons with respect to neutral sulphur atom hence, each sulphur atom is in +2 oxidation state. Each oxygen atom gets two excess electrons hence, it is in –2 oxidation state. Lewis structure of SO2– 46 can be written as follows : To find out oxidation state of each atom we distribute electrons of electron pair shared between two sulphur atoms equally (i.e. one electron is assigned to each sulphur atom). Both the electrons of electron pair shared between sulphur and oxygen atom are assigned to oxygen as oxygen is more electronegative. Thus we find that each of the central sulphur atoms obtain six electrons. This number is same as that in the outer shell of neutral sulphur atom hence oxidation state of central sulphur atoms is zero. Each of the sulphur atoms attached to oxygen atoms obtain only one electron as its share. This number is less by five electrons in comparison to the neutral sulphur atom. So, outer sulphur atoms are in +5 oxidation state. Therefore average oxidation state of sulphur atoms is : +++05 10 50 ==. 25 44 By using the formula we obtain average oxidation state of the particular type of atoms. Real oxidation state can be obtained only by writing the complete structural formula. Similarly we can see that each oxygen atom is in – 2 oxidation state. In the same way one can find out the oxidation state of each atom in SO32– and SO2– ions. Oxidation state of metal atoms will be +1 as these will lose 4 one electron in each case. IV. Matching Type 27. (i)→ (d) (ii)→ (e) (iii) → (c) (iv)→(a) 28. (i)→ (e) (ii) → (d) (iii) → (c) (iv) → (b) (v) → (f) V. Assertion and Reason Type 29. (ii) 30. (iii) 31. (i) 32. (ii)

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