Class XII Chemistry Question Bank PDF

केन्द्रीय विद्यालय संगठन KENDRIYA VIDYALAYA SANGATHAN आंचवलक विक्षा एिं प्रविक्षण संस्थान , मैसरू ZONAL INSTITUTE OF EDUCATION AND TRAINING, MYSORE CLASS XII CHEMISTRY QUESTION BANK KENDRIYA VIDYALAYA SANGATHAN ZONAL INSTITUTE OF EDUCATION AND TRAINING MYSURU SUPPORT MATERIAL CLASS -XII SUBJECT- CHEMISTRY (043) SESSION- 2024-25 PREPARED BY PARTICIPANTS OF PGT(CHEM.)-LDCE INDUCTION COURSE HELD AT ZIET MYSURU COURSE DIRECTOR Ms. MENAXI JAIN DEPUTY COMMMISSIONER, KVS & DIRECTOR ZIET MYSURU ASSOCIATE COURSE DIRECTOR SH. JYOTHI MOHAN N V PRINCIPAL PM SHRI KENDRIYA VIDYALAYA RUBBER BOARD KOTTAYAM ERNAKULUM REGION COORDINATOR Sh. DINESH KUMAR TA (PHYSICS) ZIET- MYSURU RESOURCE PERSONS SMT. SHAYLA P. PGT CHEMISTRY KV 3 Port Trust COCHI ERNAKULUM REGION SH. SIBU JOHN PGT CHEMISTRY KV KOLLAM (ERNAKULUM REGION) CLASS XII CHAPTER 1: SOLUTIONS Solution: A homogeneous mixture of two or more chemically non- reacting substances whose composition can be varied within certain limit. Solute (component present in lesser amount) Solution Solvent (component present in larger amount) Types of Solution (depending upon physical state of solute & solvent) Expressing Concentration of Solutions Concentration: Amount of solute present in a specified amount of solvent or solution. i. Mass percentage (w/w) = Mass of the component in the solution X 100 Total mass of the solution ii. Volume percentage (V/V) =Volume of the component X 100 Total volume of the solution iii. Mass by volume percentage (w/V) = Mass of the component in the solution X 100 Volume of solution iv. Parts per million (ppm) = _____no. of parts of the component____ X 106 Total no. of parts of all the components v. Mole fraction (×) = _____no. of moles of the component____ Total no. of moles of all the components vi. Molarity (M) = _____Moles of solute_____ Volume of solution in litre vii. Molality (m) = ____Moles of solute___ Mass of solvent in kg Solubility The solubility of a substance depends upon i. Nature of solute and solvent ii. Temperature iii. Pressure (for solution of gas dissolved in liquid) Nature of solvent: A polar solute will be soluble in polar solvent and a non-polar solute will be soluble in non-polar solvent Temperature: If dissolution is exothermic, solubility decreases with increase in temperature If dissolution is endothermic, solubility increases with increase in temperature. Pressure: Pressure has no effect on solubility of a solid or liquid in liquid. Henry’s law The amount of gas that is dissolved in a liquid is directly proportional to the partial pressure of that gas above the liquid when the temperature is kept constant. [Henry’s law] APPLICATIONS: Soft drinks are sealed under high pressure to increase solubility of CO2. To avoid BENDS, the tanks used by scuba divers are filled with air diluted with helium (less soluble than N2) Anoxia problem at higher altitudes due to low pressure low concentration of O2 in Blood cells. Raoult’s Law The vapour pressure of any volatile component in the solution is directly proportional to its mole fraction. Psolution = Χsolvent P0solvent QUESTIONS 1. KH value for Ar(g), CO2(g), HCHO (g) and CH4(g) are 40.39, 1.67, 1.83 × 10-5 and 0.413 respectively. Arrange these gases in the order of their increasing solubility. a) HCHO < CH4 < CO2 < Ar b) HCHO < CO2 < CH4 < Ar c) Ar < CO2 < CH4 < HCHO d) Ar < CH4 < CO2 < HCHO 2. When a non-volatile solid is added to pure water it will: a) boil above 100oC and freeze above 0oC b) boil below 100oC and freeze above 0oC c) boil above 100oC and freeze below 0oC d) boil below 100oC and freeze below 0oC 3. Molarity of a solution at 60oC is-------- than molarity at 30oC a) More b) less c) same d) no effect of temperature ANSWERS 1 (a), 2 (a), 3 (b), 4. Aquatic species are more comfortable in cold waters rather than in warm waters. Give reason. Answer Solubility of gases increases with decrease in temperature as it is an exothermic process. 5. Aarav Sharma is very fond of a special drink made by his grandmother using different fruits available in their hometown. It has an outstanding taste and also provides great health benefits of natural fruits. He thought of utilizing his grandmother recipe to create a new product in the beverage market that provide health benefits and also contain fizziness of various soft drinks available in the market. Based on your understanding of solutions chapter, help Aarav Sharma to accomplish his idea by answering following: (a) How he can add fizz to the special drink made by his grandmother? 1 (b) What is the law stated in the chapter that can help Aarav to make his drink fizzy? 1 (c) What precautions he should take while bottling so that his product does not lose fizz during storage and handling across long distances? 2 Answer a) Carbon dioxide is a gas which provides fizz and tangy flavour. He can dissolve Carbon dioxide gas in the drink. b) Henry’s law which states that solubility of a gas in liquid is directly proportional to partial pressure of the gas. (c) Bottles should be sealed under high pressure of CO and capping should be done perfectly to avoid leakage of CO2 as any loss of partial pressure will result into decrease in solubility. CHAPTER 2: ELECTROCHEMISTRY PART-A: MULTIPLE CHOICE QUESTIONS Q1. Galvanised iron sheets are coated with (a) Carbon (b) Copper (c) Zinc (d) Nickel Q2. How many coulombs are required for the oxidation of 1 mole of H2O to O2? (a) 1.93 × 105 C (b) 9.65 × 104 C (c) 3.86 × 105 C (d) 4.825 × 105 C Q3. Rust is a mixture of (a) FeO and Fe (OH)3 (b) FeO and Fe (OH)2 (c) Fe2O3 and Fe (OH)3 (d) Fe3O4 and Fe (OH)3 Q4. The Standard electrode potentials for the half cell reactions are as follows Zn ? Zn2+ + 2e– [E° = 0.41 V] Fe ? Fe2+ + 2e– [E° = 0.76 V] (a) -0.35 V (b) 0.35 V (c) + 1.17 V (d) -1.17 V Q5. The standard reduction potentials of Cu2+/Cu and Cu2+/Cu+ are 0.337 and 0.153 respectively. The standard electrode potential of Cu+/Cu half cell is (a) 0.184 V (b) 0.827 V (c) 0.521V (d) 0.490 V Q6. The standard reduction potentials of X, Y, Z metals are 0.52, -3.03, -1.18 respectively. The order of reducing power of the corresponding metals is: (a) Y > Z > X (b) X > Y > Z (c) Z > Y > X (d) Z > X > Y Q7. Which of the following is not a good conductor? (a) Cu (b) NaCl (aq) (c) NaCl (molten) (d) NaCl(s) Q8. The e.m.f. of the cell Zn/Zn2+ (0.01 M) || Fe2+ (0.001 M) Fe at 298 K is 0.2905 volt. Then the value of equilibrium constant for the cell reaction is: (a) e0.32/0.0295 (b) 100.32/0.0295 (c) 100.26/0.0295 (d) 100.32/0.0591 Q9. The volume of H2 gas at NTP obtained by passing 4 amperes through acidified H+O for 30 minutes is: (a) 0.0836 L (b) 0.0432 L (c) 0.1672 L (d) 0.836 L Q10. 4.5 g of aluminium (at. mass 27 amu) is deposited at cathode from Al3+ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from H+ ions in solution by the same quantity of electric charge will be: (a) 44.8 L (b) 11.2 L (c) 22.4 L (d) 5.6 L Q11. The amount of electricity required to deposit 1 mol of aluminium from a solution of AlCl3 will be (a) 0.33 F (b) 1 F (c) 3 F (d) 1 ampere Q12. A current of 2.0 A passed for 5 hours through a molten metal salt deposits 22.2 g of metal (at. wt. = 177). The oxidation state of the metal in the metal salt is (a) +1 (b) +2 (c) +3 (d) +4 Q13. Ionic mobility of Ag+ ions? [Ag+ = 5 × 10-4 ohm-1 cm² eq-1] is (a) 5.2 × 10-9 (b) 2.4 × 10-9 (c) 1.52 × 10-9 (d) 8.25 × 10-9 Q14. For a cell reaction involving two electron change the standard emf of the cell is found to be 0.295 V at 25°C. The equilibrium constant for the reaction at 25°C will be: (a) 2.95 × 10² (b) 10 (c) 1 × 1010 (d) 1 × 10-10 Q15. If the equivalent conductance of 1 M benzoic acid is 12.8 ohm-1 cm² and if the conductance of benzoate ion and H+ ion are 42 and 288.42 ohm-1 cm² respectively, its degree of dissociation is: (a) 39% (b) 3.9% (c) 0.35% (d) 0.039% ANSWERS FOR MCQS: 1. ZINC 2. 1.93 × 105 C 3. Fe2O3 and Fe (OH)3 4. 0.35 V 5. 0.521V 6. Y > Z > X 7. NaCl(s) 8. 100.32/0.0295 9. 0.0836 L 10. 44.8 L 11. 3 F 12. +3 13. 5.2 × 10-9 14. 1 × 1010 15. 3.9% PART-B: 2 MARKS QUESTIONS Q1. Why should we use mercury cells? Q2. What do you understand about normal hydrogen electrodes? Why is it important? Q3. Write the equation that shows the effect of concentration on the electrode potential. Q4. A cell reaction is considered spontaneous if the overall emf of the cell is positive. Comment. Q5. What is corrosion? Q6. Describe the oxidation and reduction potential. Q7. What is a cell constant? How do you determine it? Q8. Electrolysis of KBr solution gives Br2 ions at the anode, but the solution of KF does not produce any F2 ions. Give a reason. Q9. Corrosion of motor cars is a common problem in winter when salt is spread on roads to melt the snow. Give a reason. Q10. What is the reason that in concentrated solutions, strong electrolytes deviate from the Debye-Huckel- Onsager equation? ANSWER FOR 2 MARKS QUESTIONS: Ans1. In a mercury cell, the overall cell reaction does not include any ions from the solution. So, the concentration of the solution remains constant. As a result, mercury cells give a constant voltage. Therefore, mercury cells are more advantageous. Ans2. Standard Hydrogen Electrode or better known as the normal hydrogen electrode is used as a reference for half-cell potential reactions. By using a hydrogen electrode, it is possible to calculate cell potentials using different electrodes. This is also a standard measurement of electrode potential for the thermodynamic scale of redox potential. Ans3. The equation that shows the effect of concentration on electrode potential is known as the Nernst Equation. It can be written as follows: Ecell = E⁰cell – ( RT/nF) ln Q Where, Ecell = the reduction potential at current potential Ecell = the standard reduction potential relative to the reduction potential of hydrogen at 25⁰C R = universal gas constant T =temperature in K n = the moles of electrons transferred between the positive and negative terminals of an electrochemical system. F =Faraday’s Constant Q = reaction quotient Ans4. When the emf of a cell is positive, then the Gibbs free energy of the overall reaction is less than zero. ∆G = -nFEcell Hence, the cell reaction is considered spontaneous. Ans 5. Corrosion can be defined as the process of deterioration of metals because of their reaction with air and water. In this process, sulphides, oxides, carbonates, hydroxides, etc. are produced that slowly damage the metals. Rusting of iron is an example of corrosion. Rust is nothing but an oxide of iron (Fe2O3, x H2O). Ans 6. The oxidation potential of a chemical element refers to its propensity to be oxidised by losing one or more electrons at an electrode. Reduction potential, on the other hand, refers to the tendency of a chemical element to be reduced at the electrode by gaining one or more electrons.\ Ans 7. Cell constant can be defined as the ratio of the distance between the electrodes of a conductivity cell to their surface area. It can be determined by calculating the resistance of a cell of known conductivity. Ans 8. We get the negative ions at the anode because it is the electrode where oxidation takes place. The oxidation process becomes easy if the oxidation potential of the solution is high. Br- has the highest oxidation potential, followed by H2O, and finally F–. So, in the aqueous solution of KBr, Br− ions are oxidised to Br2 in preference to H2O. On the other hand, in the aqueous solution of KF, H2O is oxidised in preference to F−. Hence, in this case, oxidation of H2O at the anode gives O2 and thus no F2 is produced at the anode. Ans 9. When two metals are brought together under the surface of an electrolyte, a short-circuited cell is formed. Motor cars contain metals like lead, chromium and salt (NaCl) sprinkled over ice that acts as an electrolyte. Thus, a short-circuited cell is formed when cars move on the salt spread on the ice. So, corrosion on cars becomes a great problem in winter. Ans 10. Strong electrolytes in concentrated solutions deviate from the Debye-Huckel-Onsager equation due to the large interionic forces of attraction. PART-C: 3 MARKS QUESTIONS Q1. Write three differences between potential difference and emf. Q2. What is meant by “electrolytic conductance”? Name the factors that determine electrolytic conduction. What is the effect of temperature on it? Q3. How is electrolytic conductance measured? Q4. Iron does not rust even if a zinc coating is broken in a galvanised iron pipe, but rusting occurs much faster if the tin coating over iron is broken. Why? Q5. What do you mean by Kohlrausch’s law: from the following molar conductivities at infinite dilution λ m Ba (OH)2 = 457.6 Ω-1cm2mol-1 λ m BaCl2 = 240.6 Ω-1cm2mol-1 λ m NH4Cl = 129.8 Ω-1cm2mol-1 Calculate λ m for NH4OH. Q6. State Faraday’s Laws of electrolysis? Q7. How many g of chlorine can be produced by the electrolysis of molten NaCl with a current of 1 ampere for 15 min? Q8. What do you mean by primary and secondary battery? Q9. a) Why does the conductivity of a solution decrease with dilution? b) Can you store copper sulphate solutions in a zinc pot? Q10.At 298 K , the molar conductivities at infinite dilution of NH4OH, NaOH and NaCl are 129.8 , 217.4 and respectively.It molar conductivity of 0.01M NH4OH solution is ,calculate the degree of dissociation of at this NH4OH dilution? ANSWER FOR 3 MARKS QUESTIONS Ans 1. Potential difference is the difference of potential of electrodes when there is a flow of electricity in the circuit; whereas emf refers to the electrode potential of two electrodes when there is no flow of electricity in the circuit. The value of the potential difference is less than the maximum voltage obtained from the cell. Emf, on the other hand, is the maximum voltage obtained from a cell. Potential difference is not responsible for a steady flow of current. Emf is responsible for the steady flow of the current in a circuit. Ans 2. The ability of the electrolytic solutions to let the electric current pass through them is called electrolytic conductance. The factors on which electrolytic conductance depends are as follows: 1. Temperature 2. The concentration of ions in the solution 3. Nature of the electrolyte The ability of electrolytes to get dissolved in a solution changes with the change in temperature. When temperature increases, the solubility also increases, and hence, the electrolytic conductance also increases. Ans 3. The resistance between two nodes helps determine the conductance of electrolytes. When electricity passes through the solution, it produces positive ions and negative ions in the solution. The conductance of electrolytes can be measured by using galvanic cells or the method of electrolysis. Ans 4. Iron is less electropositive than zinc. So, the coating of zinc on the surface of the iron pipe acts as an anode and the iron pipe itself acts as a cathode. As a result, the iron pipe does not get damaged easily. Thus, rusting on galvanised iron is prevented by the coating of zinc. On the other hand, zinc is more electropositive than tin. If the coating of tin is broken anywhere, or any pores or breaks are observed on the pipe, the parts where the iron is exposed are rusted easily. Ans 5. According to Kohlrausch’s Law, “An infinite dilution of each ion migrates independently of its co-ion and makes its own contribution to the total molar conductivity of an electrolyte irrespective of nature.” λmNH4OH = λ m ++ λ m- = 2NH4+ + 2Cl– + Ba2+ +2 OH– – Ba2+ + 2 Cl– = 2 λm(NH4Cl) + λmBa(OH)2 – λmBaCl2 =2 λm(NH4OH) = 238.3 Ω-1cm2mol-1 Ans 6. Faraday’s Laws of electrolysis First Law: The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte. Second Law: The amount of different substances liberated by the same quantity of electricity passing through the electrolytic solution is proportional to their chemical equivalent weights. Ans 7. Q = It The reaction is 2mol 1mol 2mol 1mol of 900 C will produce. Ans 8. In the primary batteries, the reaction occurs only once and after the use over a period of time battery becomes dead and cannot be reused again. A secondary battery , after used, can be recharged by passing current through it in the opposite direction so that it can be used again. Ans 9. a) The conductivity of a solution is the conductance of ions present in a unit volume of the solution. The number of ions (responsible for carrying current) decreases when the solution is diluted. As a result, the conductivity of a solution decreases with dilution. b) Zinc is more reactive than copper. Therefore, zinc can displace copper from its salt solution. If copper sulphate solution is stored in a zinc pot, then zinc will displace copper from the copper sulphate solution. Ans 10. = 129.8 +217.4 – 108.9 Degree of dissociation, = = 0.039 or 3.9 %. CASE BASED QUESTIONS CASE BASED QUESTION 1. Boiling point or freezing point of liquid solution would be affected by the dissolved solids in the liquid phase. A soluble solid in solution has the effect of raising its boiling point and depressing its freezing point. The addition of non-volatile substances to a solvent decreases the vapor pressure and the added solute particles affect the formation of pure solvent crystals. According to many researches the decrease in freezing point directly correlated to the concentration of solutes dissolved in the solvent. This phenomenon is expressed as freezing point depression and it is useful for several applications such as freeze concentration of liquid food and to find the molar mass of an unknown solute in the solution. Freeze concentration is a high quality liquid food concentration method where water is removed by forming ice crystals. This is done by cooling the liquid food below the freezing point of the solution. The freezing point depression is referred as a colligative property and it is proportional to the molar concentration of the solution (m), along with vapor pressure lowering, boiling point elevation, and osmotic pressure. These are physical characteristics of solutions that depend only on the identity of the solvent and the concentration of the solute. The characters are not depending on the solute’s identity. (Jayawardena, J. A. E. C., Vanniarachchi, M. P. G., & Wansapala, M. A. J. (2017). Freezing point depression of different Sucrose solutions and coconut water.) 1. When anon volatile solid is added to pure water it will: a. boil above 100oC and freeze above 0oC b. boil below 100oC and freeze above 0oC c. boil above 100oC and freeze below 0oC d. boil below 100oC and freeze below 0oC 2. Colligative properties are: a. dependent only on the concentration of the solute and independent of the solvent’s and solute’s identity. b. dependent only on the identity of the solute and the concentration of the solute and independent of the solvent’s identity. c. dependent on the identity of the solvent and solute and thus on the concentration of the solute. d. dependent only on the identity of the solvent and the concentration of the solute and independent of the solute’s identity. 3. Assume three samples of juices A, B and C have glucose as the only sugar present in them. The concentration of sample A, B and C are 0.1M,.5M and 0.2 M respectively. Freezing point will be highest for the fruit juice: a. A b. B c. C d. All have same freezing point 4. Identify which of the following is a colligative property : a. freezing point b. boiling point c. osmotic pressure d. all of the above Ans. 1 (b) 2 (d) 3 (a) 4(c) CASE BASED QUESTION 2. All chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules are present in a few gram of any chemical compound varying with their atomic/molecular masses. To handle such large number conveniently, the mole concept was introduced. All electrochemical cell reactions are also based on mole concept. For example, a 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrode. The amount of products formed can be calculated by using mole concept. (i)The total number of moles of chlorine gas evolved is (a) 0.5 (b) 1.0 (c) 1.5 (d) 1.9 (ii) If cathode is a Hg electrode, then the maximum weight of amalgam formed from this solution is (a) 300g (b) 446 g (c) 396 g (d) 256 g or The total charge (coulomb) required for complete electrolysis is (a) 186000 (b) 24125 (c) 48296 (d) 193000 (iii) In the electrolytes, the number of moles of electrons involved are (a) 2 (b) 1 (c) 3 (d) 4 (iv) In electrolysis of aqueous NaCl solution when Pt electrode is taken, then which gas is liberated at cathode? (a) H2 gas (b) Cl2 gas (c) O2 gas (d) None of these ANSWERS: (i) b(ii) bord (iii) a(iv) a CASE BASED QUESTION 3. The potential of each electrode is known as electrode potential. Standard electrode potential is the potential when concentration of each species taking part in electrode reaction is unity and the reaction is taking place at 298 K. By convention, the standard ectrode potential of hydrogen (SHE) is 0.0 V. The electrode potential value for eacfi electrode process is a measure of relative tendency of the active species in the process to remain in the oxidisedlreduced form. The negative electrode potential means that the redox couple is stronger reducing agent than H+/H2 couple. A positive electrode potential means that the redox couple is a weaker reducing agent than the H+/H2 couple. Metals which have higher positive value of standard reduction potential form the oxides of greater thermal stability. In these questions (i-iv), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (i) Assertion : An electrochemical cell can be set-up only if the redox reaction is spontaneous. Reason : A reaction is spontaneous if the free energy change is negative. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. (ii) Assertion : The standard electrode potential of hydrogen is 0.0 V. Reason : It is by convention. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. (iii) Assertion : The negative value of standard reduction potential means that reduction takes place on this electrode with reference to hydrogen electrode. Reason : The standard electrode potential of a half cell has a fixed value. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. (iv) Assertion : The absolute value of electrode potential cannot be determined experimentally. Reason : The electrode potential values are generally determined with respect to SHE. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. ANSWER: (i) (b) (ii) (a) (iii) (d) : A negative value of standard reduction potential means that oxidation takes' place on the electrode with reference to SHE. (iv) (a) CASE BASED QUESTION 4. The study of the conductivity of electrolyte solutions is important for the development of electrochemical devices, for the characterization of the dissociation equilibrium of weak electrolytes, and for the fundamental understanding of charge transport by ions. The conductivity of the electrolyte is measured for electrolyte solution with concentrations in the range of 10–3 to 10–1 mol L–1, as a solution in this range of concentrations can be easily prepared. The molar conductivity (Λm) of strong electrolyte solutions can be nicely fit by the Kohlrausch equation. Λm = Λ°m – K √C …(i) Where Λ°m is the molar conductivity at infinite dilution and C is the concentration of the solution. K is an empirical proportionality constant to be obtained from the experiment. The molar conductivity of weak electrolytes, on the other hand, is dependent on the degree of dissociation of the electrolyte. At the limit of a very dilute solution, the Ostwald dilution law is expected to be followed, where CA is the analytical concentration of the electrolyte and Kd is the dissociation constant. The molar conductivity at infinite dilution can be decomposed into the contributions of each ion. Where, λ+ and λ– are the ionic conductivities of positive and negative ions, respectively and v+ and v– are their stoichiometric coefficients in the salt molecular formula. Which statement about the term infinite dilution is correct? (a) Infinite dilution refers to a hypothetical situation when the ions are infinitely far apart. (b) The molar conductivity at infinite dilution of NaCl can be measured directly in solution. (c) Infinite dilution is applicable only to strong electrolytes. (d) Infinite dilution refers to a real situation when the ions are infinitely far apart. Ans: (a) Infinite dilution refers to a hypothetical situation when the ions are infinitely far apart. Which of the following is a strong electrolyte in aqueous solution? (a) HNO2 (b) HCN(c) NH3 (d) HCl Answer: (d) HCl OR Which of the following is a weak electrolyte in aqueous solution? (a) K2SO4 (b) Na3PO4(c) NaOH (d) H2SO3 Answer: (d) H2SO3 If the molar conductivities at infinite dilution for NaI, CH3COONa and (CH3COO)2Mg are 12.69, 9.10 and 18.78 S cm2 mol–1 respectively at 25°C, then the molar conductivity of MgI2 at infinite dilution is (a) 25.96 S cm2 mol–1 (b) 390.5 S cm2 mol–1 (c) 189.0 S cm2 mol–1 (d) 3.89 × 10–2 S cm2 mol–1 Answer: (a) 25.96 S cm2 mol–1 Which of the following is the correct order of molar ionic conductivities of the following ions in aqueous solutions? (a) Li+ < Na+ < K+ < Rb+ (b) Li+ > Na+ > K+ > Rb+ (c) Rb+ < Na+ < Li+ < K+ (d) Li+ < Rb+< Na+ < K+ Answer: (a) Li+ > Na+ > K+ > Rb+ 5 MARLS QUESTIONS QUS 1. Two half cell reactions of an electrochemical cell are given below : MnO–4(aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (I), E° = + 1.51 V Sn2+ (aq) → 4 Sn4+ (aq) + 2e–, E° = + 0.15 V Construct the redox equation from the two half cell reactions and predict if this reaction favours formation of reactants or product shown in the equation. (All India 2009) Answer: The reactions can be represented at anode and at cathode in the following ways : At anode (oxidation) : Sn2+ → = Sn4+ (aq) + 2e– ] × 5 E° = + 0.15 V At cathode (reduction) : MnO–4(aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (I)] × 2 E° = + 1.51 V The Net R × M = 2MnO–4(aq) + 16H+ + 5Sn2+ → 2Mn2+ + 5Sn4+ + 8H2O Now E°cell = E°cathode – E°anode = 1.51 – 0.15 = + 1.36 V ∴ Positive value of E°cell favours formation of product. QUS 2. A zinc rod is dipped in 0.1 M solution of ZnSO4. The salt is 95% dissociated at this dilution at 298 K. Calculate the electrode potential. [ E°Zn2+ /Zn = – 0.76 V] (Comptt. Delhi 2012) Answer: The electode reaction is given as Zn+2 + 2e → Zn Using Nernest Equation QUS 3. a) Write the reactions taking place at cathode and anode in lead storage battery when the battery is in use. What happens on charging the battery ? b) The standard electrode potential (E°) for Daniel cell is +1.1 V. Calculate the ΔG° for the reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) (1 F = 96500 C mol-1). Answer: a) At Anode: Pb + SO4-2 → PbSO4 + 2e at Cathode : PbO2 + SO4-2 + 4H+ + 2e → PbSO4 + 2H2O On charging the battery, the reaction is reversed and PbSO4 on anode and cathode is converted into Pb and PbO2 respectively. b) We know, ΔG° = -nFE°cell Given :E°cell = 1.1 volt ∴ΔG° = -2 × 96500 C mol-1 × 1.1 volt = -212300 CV mol-1 = -212300 J mol-1 = -212.3 KJ mol-1 QUS 4. The conductivity of 0.001 M acetic acid is 4 × 10-5 S/cm. Calculate the dissociation constant of acetic acid, if molar conductivity at infinite dilution for acetic acid is 390 S cm2/mol. Answer: Given : K = 4 × 10-5 S/cm, M = 0.001 M Λ°m = 390 S cm2/mol, k = ? Using the formula CHAPTER 3: CHEMICAL KINETICS SECTION A MULTIPLE CHOICE QUESTIONS 1. What will be the fraction of molecules having energy equal to or greater than activation energy, Ea? (a) K (b) A (c) Ae-Ea/Rt (d) e-Ea/Rt 2. What type of reaction is this? (a) Second order (b) Unimolecular (c) Pseudo-unimolecular (d) Third order 3. Which among the following is a false statement? (a) Rate of zero order reaction is independent of initial concentration of reactant. (b) Half life of a third order reaction is inversely proportional to square of initial concentration of the reactant. (c) Molecularity of a reaction may be zero or fraction. (d) For a first order reaction, t1/2=0.693K 4. Which of the following statements about the catalyst is true? (a) A catalyst accelerates the rate of reaction by bringing down the activation energy. (b) A catalyst does not participate in reaction mechanism. (c) A catalyst makes the reaction feasible by making ∆G more negative. (d) A catalyst makes equilibrium constant more favourable for forward reaction. 5. An endothermic reaction with high activation energy for the forward reaction is given by the diagram. 6. For the reaction N2 + 3H2 → 2NH3 if Δ[NH3]/Δt = 2 × 10-4 mol L-1s-1, the value of −Δ[H2]/Δt would be (a) 1 × 10-4 mol L-1s-1 (b) 3 × 10-4 mol L-1s-1 (c) 4 × 10-4 mol L-1s-1 (d) 6 × 10-4 mol L-1s-1 7. The rate of a certain hypothetical reaction A + B + C → products is given by r = −d[A]/dtK[A]1/2[B]1/3[C]1/4. The order of the reaction is (a) 13/11 (b) 13/14 (c) 12/13 (d) 13/12 8. In the formation of S02 by contact process; 2SO2 + O2 → 2SO3, the rate of reaction was measured as −d[O2]/dt=2.5 × 10-4 mol L-1s-1. at The rate of formation of of S03 will be (a) -5.0 × 10-4 mol L-1s-1 (b) -1.25 × 10-4 mol L-1s-1 (c) 3.75 × 10-4 mol L-1s-1 (d) 5.00 × 10-4 mol L-1s-1 9. For a chemical reaction A→B, it is found that the rate of reaction doubles when the concentration of A is increased four times. The order of reaction is (a) Two (b) One (c) Half (d) Zero 10. The half life of the first order reaction having rate constant K = 1.7 x 10-5s-1 is (a) 12.1 h (b) 9.7 h (c) 11.3 h (d) 1.8 h SHORT ANSWER QUESTIONS(2M) 11) Define ‘rate of a reaction’ 12) Define ‘order of a reaction’ 13) Define ‘activation energy’ of a reaction 14) If the rate constant of a reaction is k = 3 × 10-4 s-1, then identify the order of the reaction. 15) Write the unit of rate constant for a zero order reaction. 16) Define rate constant (K) 17) For a reaction R → P, half-life (t1/2) is observed to be independent of the initial concentration of reactants. What is the order of reaction? 18) Define Elementary step in a reaction 19)Define Activation energy of a reaction 20)What is meant by rate constant SHORT QUESTIONS(3M) 21) Write two differences between ‘order of reaction’ and ‘molecularity of reaction’. 22)Define the following terms : (a) Pseudo first order reaction. (b) Half life period of reaction (t1/2). 23) Explain the following terms : (i) Rate constant (k) (ii) Half life period of a reaction (t1/2) 24) For a chemical reaction R → P, the variation in the f concentration (R) vs. time (f) plot is given as (i) Predict the order of the reaction. (ii) What is the slope of the curve? (All India 2014) 25) ) For a reaction, A + B → Product, the rate law is given by, Rate = k[A]1[B]2. What is the order of the reaction? (b) Write the unit of rate constant ‘k’ for the first order reaction 26) For a reaction A + B → P, the rate is given by Rate = k[A] [B]2 (i) How is the rate of reaction affected if the concentration of B is doubled? 27) What is the physical significance of energy of activation ? Explain with diagram 28) Following data are obtained for the reaction: N2O5 → 2NO2 + 1/2O2 t/s 0 300 600 [N205]/mol L-1 1.6 × 10-2 0.8 × 10-2 0.4 × 10-2 (a) Show that it follows first order reaction. (b) Calculate the half-life. (Given log 2 = 0.3010 log 4 = 0.6021) 29) The rate constant for a first order reaction is 60 s-1. How much time will it take to reduce the initial concentration of the reactant to its l/10th value? 30) The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.2 – 1.0×104TK Calculate Ea for this reaction and rate constant k if its half-life period be 200 minutes. (Given: R = 8.314 J K-1 mol-1) SECTION D(LONG ANSWER QUESTIONS)-5M 31) For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained: t/s 0 10 20 [CH3COOCH3]/mol L-1 0.10 0.05 0.025 (i) Show that it follows pseudo first order reaction, as the concentration of water remains constant. (ii) Calculate the average rate of reaction between the time interval 10 to 20 seconds. (Given : log 2 = 0.3010, log 4 = 0.6021) (All India 2015) 32) a) For a reaction A + B → P, the rate is given by Rate = k[A] [B]2 (i) How is the rate of reaction affected if the concentration of B is doubled? (ii) What is the overall order of reaction if A is present in large excess? (b) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction. 33) (a) For a reaction A + B → P, the rate is given by Rate = k[A] [B]2 (i) How is the rate of reaction affected if the concentration of B is doubled? (ii) What is the overall order of reaction if A is present in large excess? (b) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction. (log 2 = 0.3010) 34) The decomposition of A into products has a value of K as 4.5 × 103 s-1 at 10°C and energy of activation 60 kj mol-1. At what temperature would K be 1.5 × 104 s-1? (b) (i) If half life period of a first order reaction is x and 3/4,th life period of the same reaction is y, how are x and y related to each other? (ii) In some cases it is found that a large number of colliding molecules have energy more than threshold energy, yet the reaction is slow. Why? 35) A first order reaction takes 100 minutes for completion of 60% of the reaction. Find the time when 90% of the reaction will be completed. (b) With the help of diagram explain the role of activated complex in a reaction SECTION E (CASE BASED QUESTIONS)-4M 36) The integrated rate law for the second-order reaction A → products is 1/[A]_t = kt + 1/[A]_0. Because this equation has the form y = mx + b, a plot of the inverse of [A] as a function of time yields a straight line. The rate constant for the reaction can be determined from the slope of the line, which is equal to k. A reaction is second order in A and first order in B. (36.i) Write the differential rate equation(1m), (36.ii) How is the rate affected on increasing the concentration of A three times?(1m) (36.iii) How is the rate affected when the concentrations of both A and B are doubled?(2m) 37) Molecularity is the number of reacting species involved in simultaneous collisions in an elementary or simplest reaction. Order is an experimentally determined quantity. It may be equal to zero, positive, negative, whole number or fractional number. Molecularity is a theoretical concept, it is always in whole numbers. Explain the following terms : (37.i) Order of a reaction(1m) (37.ii) Molecularity of a reaction(1m) (37.iii) The rate of a reaction increases four times when the temperature changes from 300 K to 320 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature. (R = 8.314 J K-1 mol-1)(2m) 38) For the reaction 2NO(g) + Cl2(g) → 2NOCl(g) the following data were collected. All the measurements were taken at 263 K: Initial Experiment No. Initial [NO](M) Initial rate of tlisa/ipcarance of Cl2 (M/min) [Cl2](M) 1 0.15 0.15 0.60 2 0.15 0.30 1.20 3 0.30 0.15 2.40 4 0.25 0.25 ? (a) Write the expression for rate law. (b) Calculate the value of rate constant and specify its units. (c) What is the initial rate of disappearance of Cl2 in exp. 4? 39) For a chemical reaction R → P, variation in ln[R] vs time (f) plot is given below: For this reaction: (39.i) Predict the order of reaction (39.ii) What is the unit of rate constant (k)? Answer: (39.iii) Write the rate equation for the reaction A2 + 3B2 → 2C, if the overall order of the reaction is zero. ANSWERS TO CHEMICAL KINETICS PART-1 SECTION A (MCQ) 1.d 2.c 3. c 4.a 5.c 6.b 7.d 8.d 9.c 10.c Short answers(2m) 11) Rate of a reaction: Either, The change in the concentration of any one of the reactants or products per unit time is called rate of a reaction. Or, The rate of a chemical reaction is the change in the molar concentration of the species taking part in a reaction per unit time. 12) The sum of powers of the concentration of the reactants in the rate law expression is called the order of reaction 13) The minimum extra amount of energy absorbed by the reactant molecules to form the activated complex is called activation energy. The activation energy of the reaction decreases by the use of catalyst. 14) S-1 is the unit for rate constant of first order reaction. 15) Mol L-1 S-1 is unit of rate constant for a zero order reaction 16) Rate constant. It is defined as the rate of reaction when the concentration of reaction is taken as unity. 17) The t1/2 of a first order reaction is independent of initial concentration of reactants. 18) Those reactions which take place in one step are called elementary reactions. Example : Reaction between H2, and I2 to form 2HI H2 + I2 → 2HI 19) The minimum extra amount of energy absorbed by the reactant molecules to form the activated complex is called activation energy. SECTION B(Short Answer questions (3m)) 21) Order of reaction Molecularity of reaction It is the number of atoms, ions or molecules that must (i) It is the sum of tire concentration terms on collide with one another simultaneously so as to result which die rate of reaction actually depends. into a chemical reaction. (ii) It can be fractional as well as zero. it is always a whole number. 22) (a) Those reactions which are not truly of the first order but under certain conditions become first order reactions are called pseudo first order reaction. (b) The time taken for half of the reaction to complete is called half life period. 23) i) Rate constant (k): It is a proportionality constant and is equal to the rate of reaction when the molar concentration of each of the reactants is unity. (ii) Half life period of a reaction (t1/2): The time taken for half of the reaction to complete is called half life penod.(R)t 24) ) It is zero order reaction. (ii) Slope of the curve = -K 25) a) For a reaction, A + B Rate = k [A]1 [B]2 This is the third order of reaction. (b) Unit of rate constant for first order reaction is S-1 26) For the reaction A + B → P rate is given by Rate = k[A]1[B]2 (i) r1 = k[A]1 [B]2 r2 = k[ A]1[2B]2 = r2 = k[A]1 [2B]2=4k[A]1 [B]2 r1 = 4r2, rate will increase four times of actual rate. 27) The minimum extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to threshold value is called activation energy. Less is the activation energy, faster is the reaction or greater is the activation energy, slower is the reaction 28) (a) Show that it follows first order reaction. (b) Calculate the half-life. (Given log 2 = 0.3010 log 4 = 0.6021) (Delhi 2016) Answer: (a) For first order reaction: 29) Given : k = 60 s-1, t = ? If initial concentration is [A0] 30) Given: t1/2 = 200 min Ea = ?, T = ? Using Arrhenius equation SECTION C (LONG ANSWERS-5M) 31) As k1 and k2 are equal, hence pseudo rate constant is same. It follows the pseudo first order reaction. (ii) Average rate of reaction between 10 to 20 seconds = −Δ[R]Δt=−(0.025−0.05)(20−10)=0.02510 = 0.0025 mol lit-1 sec-1 32) a) For the reaction A + B → P rate is given by Rate = k[A]1[B]2 (i) r1 = k[A]1 [B]2 r2= k[A]1 [2B]2 = 4k[A]1 [B]2 r1 = 4r2 (rate of reaction becomes 4 times) (ii) When A is present in large amounts, order w.r.t. A is zero. Hence overall order = 0 + 2 = 2 33) a) For the reaction A + B → P rate is given by Rate = k[A]1[B]2 (i) r1 = k[A]1 [B]2 r2 = k[ A]1[2B]2 = r2 = k[A]1 [2B]2=4k[A]1 [B]2 r1 = 4r2, rate will increase four times of actual rate. (ii) When A is present in large amount, order w.r.t. A is zero. Hence overall order = 0 + 2 = 2, second order reaction. 34) (a) Given : K1 = 4.5 × 103 s-1, T1 = 10K + 273K = 283K K2 = 1.5 × 104 s-1, T2 = ? Ea = 60 KJ mol-1 Using formula : ∴ Temperature, T2 will be = 297° – 273° = 24° C (b) (i) t1/2 = 0.693K (For first order reaction) t3/4 = K ⇒ t3/4 = 1.3864K According to condition (The value 1.3864 is double of 0.693) From the above equation it is clear that t3/4 = 2t1/2 ∴ y = 2X (ii) It is due to improper orientation of the colliding molecules at the time of collision. 35) ) For the first order reaction, (b) In order that the reactants may change into products, they have to cross an energy barrier as shown in the diagram This diagram is obained by plotting potential energy vs. reaction coordinate. It is believed that when the reactant molecules absorb energy, their bonds are loosened and new bonds are formed between them. The intermediate complex thus formed is called activated complex. It is unstable and immediately dissociates to form the stable products. 36.i) Differential rate equation : dxdt = K [A]2[B] (36.ii) When concentration of A is increased to three times, the rate of reaction becomes 9 times r = K[3A]2B ∴ r = 9KA2B i.e. = 9 times (36.iii) r = K[2A]2[2B] ∴ r = 8KA2B i.e. = 8 times 37.i) Order of a reaction: It is the sum of powers of the molar concentrations of reacting species in the rate equation of the reaction. (37.ii) Molecularity of a reaction : It is the total number of reacting species (molecules, atoms or ions) which bring the chemical change. It is always a whole number. It is a theoretical concept. It is meaningful only for simple reactions or individual steps of a complex reaction. It is meaningless for overall complex reaction. (37.iii) Given : T1 = 300 K T2 = 320 K K1 = K (Consider) K2 = 4 K R = 8.314 Ea = ? Substituting these values in the formulae, ∴ Energy of activation, Ea = 55327.46 = 55.3 KJ mol-1 38.i) Rate law = K[NO]2 [Cl2] (38.ii) 0.60 M min-1 = K[0.15]2 [0.15] M3 ∴ K = 177.7 M-2 min-1 (38.iii) Initial rate of disappearance of Cl2 in exp. 4 Formula : Rate = K[NO]2 [Cl2] ∴ Initial rate = 177.7M-2 min-1 × (0.25)2 × (0.25)M3 = 2.8 M min-1 39.i) It is zero order reaction. (39.ii) The unit of rate constant (k) is mol L-1 S-1. 39.iii) (ii) A2 + 3B2 → 2C Rate = (dxdt) = K[A]0 [B]0 = K (rate constant) S NO. CHAPTER 4: THE D & F BLOCK ELEMENTS MARKS Q.1 Which of the following is a strong oxidising agent? (At. No. Mn =25 Zn=30, Cr=24 1 Sc=21) (a) Mn3+ (b) Zn2+ (c) Cr3+ (d) Sc 3+ Q.2 In which of the following pairs both the ions are coloured in aqueous solution? 1 (a) Sc3+ Ti3+ (b) Sc3+Co2+ (c) Ni2+ Cu+ (d) Ni2+Ti3+ Q.3 Which of the following has the same ionic size? 1 (a)Zr4+, Hf4+ (b)Zn2+, Hf4+ (c) Fe2+, Ni2+ (d)Zr4+ , Ti4+ Q.4 Magnetic moment of 2.83 B.M. is given by which of the following ion? 1 (a)Ti3+ (b) Ni2+ (c) Cr3+ (d)Mn2+ Q.5 The correct order of decreasing second ionising enthalpy of Ti(22) V(23) Cr(24) Mn(25) 1 (a) V > Mn < Cr>Ti (b) Mn Mn (d) Cr> Mn > V >Ti Q.6 Which property of transition metals enables them to behave as a catalyst ? 1 (a) high melting point (b) high ionisation energy (c) alloy formation (d) variable oxidation state Q.7 The incorrect statement about interstitial compounds is: 1 (a) they are chemically reactive (b) they are very hard (c) they retain metallic conductivity (d) they have high melting point Q.8 Out of the following transition elements, the maximum number of oxidation states is 1 shown by which element: (a) Sc [Z=21] (b) Mn [Z=25] (c) Cr [Z=24] (d) Fe [Z=26] Q.9 KMnO4 is not acidified by HCl instead of H2SO4 because 1 (a) H2SO4 is stronger acid than HCl (b) HCl is oxidised to Cl2 by KMnO4 (c) H2SO4 is dibasic (d) rate is faster in presence of H2SO4 Q.10 The oxidation state of Cr in final product formed by reaction of KI and acidified 1 dichromate solution is- (a) +4 (b) +6 (c) +2 (d) +3 Q.11 The difference between the oxidation number of Cr in chromate and dichromate 1 ion is (a) 1 (b) 2 (c) 3 (d) 0 Q.12 Which of the following are d-block elements but not regarded as transition 1 elements? (a) Cu, Ag, Au (b) Zn, Cd, Hg (c) Fe, Co, Ni (d) Ru, Rh, Pd Q.13 Transition elements form alloys easily because they have 1 (a) Same atomic number (b) Same electronic configuration (c) Nearly same atomic size (d) None of the above Q.14 The properties which is not characteristic of transition elements- 1 (a) Variable oxidation state (b) Tendency to form complexes (c) Formation of colour compounds (d) natural radioactivity. Q.15 Which of the following has magnetic moment value of 5.9? 1 (a) Fe2+ (b) Fe3+ (c) Ni2+ (d) Cu2+ Q.16 Why transition elements exhibit variable oxidation states? 2 Q.17 Calculate the spin only moment of Co+2 by writing the electronic configuration 2 of Co ( Z=27) and Co+2. Q.18 Give reason and select one atom or ion which exhibit asked property 2 (a) Sc+3 or Cr+3 ( diamagnetic behaviour) (b) Cr or Cu (high melting and boiling point) Q.19 2 Explain the followings: (a) Why transition elements act as a catalyst? (b) Why transition elements form alloys? Q.20 Calculate the magnetic moment of a divalent ion in aqueous solution if its at.no is 25. 2 Zn2+ salts are white while Cu2+ salts are coloured. Why? Q.21 2 2+ 2+ Which is a stronger reducing agent Cr or Fe and why? Q.22 In the following ions: Mn3+, V3+, Cr3+, Ti4+ (Atomic no.: Mn=25, V=23, Cr=24, Ti=22) 2 A. Which ion is most stable in an aqueous solution? B. Which ion is strongest oxidizing agent? C. Which ion is colorless? D. Which ion has the highest number of unpaired electrons? Q.23 The elements of 3d transition series are given as: Sc, Ti, V, Cr, Mn, Fe, Co, Ni, 2 Cu, Zn Answer the following: A. Copper has exceptionally positive Eo M+²/M value. Why? B. Which element is a strong reducing agent in +2 oxidation state and why? Q.24 How would you account for the following:- 2 The oxidising power of the following three oxo ions in the series follows the order VO2+ 2Mn2+ + 8H2O + 10CO2 (c) Hydrogen sulphide is oxidised, sulphur being precipitated: H2S —> 2H+ + S2− 5S2− + 2MnO4 − + 16H+ ——> 2Mn2+ + 8H2O + 5S 2.. Name the metal of the 1 st row transition series that i) has highest value for magnetic moment ii) has zero spin only magnetic moment in its +2-oxidation state. iii) exhibit maximum number of oxidation states. Transition metals form a large number of complex compounds. Give reason. An i) Chromium ii) Zinc iii) Manganese s ii) Transition metals for complex compounds due to, i) small sizes of metal cations ii) their ionic charges and iii) availability of d orbitals for bond formation. 3. Write two comparisons of variability in oxidation states of transition metals and non transition elements (p- block elements) ? What happens when (a) A lanthonoid reacts with dilute acids ? (b) A lanthonoid reacts with water? An 1. In transition elements, variable oxidation state differ from each other by s unity, whereas in case of non transition elements, oxidation state differ by units of two.(For example Fe exhibits o.s of +2 and +3. similarly copper exhibits two o.s of +1 and +2. on the other hand, Sn, Pb exhibit o.s of +2 and +4.) 2. 2. In transition elements, higher o.s are more favoured in elements of higher atomic mass, whereas in p-block elements lower o.s are favoured by heavier members ( due to inert pair effect, For example Mo(VI) and W(VI) are more stable than Cr(VI). On the other hand Pb(II) is more stable than Sn(II)) 3..(a) When lanthonoid reacts with dilute acids , it liberates hydrogen gas. (b)When lanthonoid reacts with water , it forms lanthanoid hydroxide and liberate hydrogen gas. 4. (a). Calculate the ‘spin only’ magnetic moment of M2+ (aq) ion (Z = 27). (b) The second ionisation enthalpy is high for Cr and Cu , why? © Why first ionisation enthalpy of Cr is lower than that of Zn ? (d) Give two characteristics of transition metal alloys. An (a) M (z= 27 , 3d7 4s2 ) → M+2 (3d7 4s0 ) hence it has 3 unpaired electrons n= s 3 = √3(3+2) = 3.87 BM (b) The second ionisation enthalpy is unusually high values for Cr and Cu because when M+ ion ionize to M+2 ion , the d5 and d10 configurations of the M+ ions (i.e Cr+ or Cu+ ) are disrupted, with considerable loss of exchange energy. (c) IE1 of Cr is lower, because removal of an electron from Cr does not change the d (3d5 4s1 to 3d5 4s0 ) configuration. Cr (z= 24 , 3d5 4s1 ) → Cr+ (3d5 4s0 ) ------ IE1 IE1 value for Zn is higher, because removal of electron from 4s level needs more energy. Zn (z= 30 , 3d10 4s2 ) → Zn+ (3d10 4s1 ) ------ IE1 ∴ IE1 (Zn) > IE1 (Cr) (d) The alloys are hard and have high melting points. CHAPTER 5: COORDINATION COMPOUDS Coordination compounds have a central atom (or cation) that is coordinated to a number of anions or neutral molecules, and they usually retain their identity in both solution and solid state. These can be positively, negatively, or neutrally charged species, such as [Co(NH3)6]3+, [NiCl4]2-, [Ni(CO)4], and so on. In 1893, Werner proposed a theory to explain coordination compound structure and bonding. According to this theory metals have two types of valencies in coordination compounds: primary valency and secondary valency. Primary valencies are ionisable. Secondary valencies cannot be ionized. On the basis of Werner’s theory the structure of [Co(NH3)5Cl] Cl2 is: DIFFERENCE BETWEEN DOUBLE SALTS AND COORDINATION COMPOUNDS. COORDINATION COMPOUNDS DOUBLE SALT A coordination compound contains a central When two salts in stoichiometric ratio are metal atom or ion surrounded by number of crystallised together from their saturated oppositely charged ions or neutral solution they are called double salts molecules. These ions or molecules are bonded to the metal atom or ion by a coordinate bond. Example: K4 [Fe(CN)6 ] Example: FeSO4. (NH4)2SO4.6H2O (Mohr’s salt) They do not dissociate into simple ions They dissociate into simple ions when when dissolved in water. dissolved in water. Ligands: Definition: Ligands are molecules or ions that donate electrons to a metal ion and form coordinate covalent bonds. Ligands can be classified into various types based on their structure and properties. Classification based on denticity: Ligands can be classified as monodentate, bidentate, tridentate, etc., based on the number of donor atoms they can use to bind to a metal ion. Chelating ligands: Chelating ligands are ligands that can bind to a metal ion through more than one donor atom. Chelating ligands form stable complexes and play an important role in many biological processes. Polydentate ligands: Polydentate ligands are ligands that have multiple donor atoms and can bind to more than one metal ion. Polydentate ligands play an important role in the formation of coordination polymers and metal complexes with high stability. Coordination sphere & Counter Ions. The central atom/ion and the ligands attached to it are enclosed in square bracket and is collectively termed as coordination sphere and the ionizable groups written outside the coordination sphere is called counter ions etc. In K4[Fe(CN6)] [Fe(CN)6]4− is coordination sphere K+ is counter ion Coordination number: The coordination number in coordination compounds is defined as the number of ligand (donor) atoms/ions surrounding the central metal atom in a complex ion. For example, the coordination number of cobalt in [Co(NH3)6]3+ is six. Similarly the coordination number of Ag+ in [Ag(NH3)2]+ is 2, that of Cu2+ in [Cu(NH3)4]2+ is 4, and that of Fe3+ in [Fe(CN)6]3– is 6. Coordination polyhedra: the spatial arrangement of the ligand atoms which are directly attached to the central atom / ion. Oxidation number: Another important property of coordination compound is the oxidation number of the central metal atom. The net charge on a complex ion is the sum of the charges on the central atom and its surrounding ligands. In the [PtCl6]2- ion for example, each chloride ion has an oxidation number of–1, so the oxidation number of Pt must be +4. If the ligands do not bear net charges the oxidation number of the metal is equal to the charge of the complex ion. Thus in [Cu(NH3)4]2+ each NH3 is neutral, so the oxidation number of copper is +2. Magnetic properties of coordination compounds: A coordination compound is paramagnetic in nature, if it has unpaired electrons and diamagnetic if all the electrons in the coordination compound are paired. Magnetic moment = [n (n + 2)]1/2 where n is number of unpaired electrons. Rules for naming mononuclear coordination compounds 1. Order of naming ions The positive ion (cation) whether simple or complex, is named first followed by the negative ion (anion). The name is started with a small letter and the complex part is written as one word. 2. Naming ligands 3. i) Negative ligands (organic or inorganic) end in −O, eg, CN− (cyano), Cl− (chlorido), Br− (bromido), F- (fluorido), NO2− (nitro), OH- (hydroxo), O2− (oxo), H- (hydrido). If the name of the anionic ligands ends in −ide, −ite or −ate, the last ‘e’ is replaced by ‘O’ giving −ido, −ito and −ato eg, SO42− (sulphato), C2O42− (oxalato), NH2− (amido), NH2− (imido), ONO− (nitrito). ii) Neutral ligands have no special ending, NH3 (ammine), H2O (aqua), CO (carbonyl) iii) Positive ligands (which are very few) end in -ium, NO+ (nitrosoium), NO2+ (nitronium) 3. Numerical prefixes to indicate number of ligands If there are several ligands of the same type, the prefixes like di, tri, tetra, penta and hexa are used to indicate the number of ligands of that type. When the name of polydentate ligand includes a number e.g., ethylendediamine, then bis, tris, tetrakis are used as prefixes. 4. Order of naming of ligands All ligands whether negative, neutral or positive are named first in the alphabetical order followed by the name of the metal atom/ion. 5. Naming of the complex ion and ending of the central atom Ligands are named first followed by the metal atom. a) If the complex ion is a cation or the coordination compund is non-ionic, the name of the central metal ion is written as such followed by its oxidation state indicated by Roman numeral (such as II, III, IV) in the parentheses at the end of the name of the metal without any space between the two. b) If the complex ion is anion, the name of the central metal atom is made to end in-ate followed by the oxidation number in brackets without any space between them. Rules for Writing Formula from the Name of the Mononuclear Complex Formula of the cation (whether simple or complex) is written first followed by that of the anion. The formula of the complex ion (coordination entity) (whether charged or not) is written in square brackets called coordination sphere. Within the coordination sphere, the symbol of the metal atom is written first followed by the symbols/formulas of the ligands arranged alphabetically according to their names irrespective of the charge present on them. While listing the ligands alphabetically, the following rules should be followed: i) Polydentate ligands are also listed in the alphabetical order. ii) The position of abbreviated ligands in the alphabetical order is determined from the first letter of the abbreviation. iii) The position of ligands with special names (such as aqua for water) in the alphabetical order is determined from the first letter of the special name. iv) Abbreviations used for the ligands and the formulas of the polyatomic ligands are enclosed in parentheses separately. v) The metal atom as well as all the ligands are listed without any space between them. If the formula of the complex ion is to be written without writing the counter ion, the charge on the complex ion is indicated outside the square bracket as a right superscript with the number before the sign. For example, [Fe(CN)6]3-, [Cu(NH3)4]2+, etc. The number of cations or anions to be written in the formula is calculated on the basis that total positive charge must be equal to the total negative charge, as the complex as a whole is electrically neutral. Example : Tetraammineaquachloridocobalt(III) chloride has complex ion = [Co(NH3)4(H2O)Cl] and simple ion = Cl- To balance the charge, the formula will be [Co(NH3)4(H2O)Cl]Cl2 SECTION A: MCQ (1 MARKS): 1. The correct name of the compound [Cu(NH3)4](NO3)2 according to IUPAC system is (a) Cuprammonium nitrate (b)Tetrammonium copper (II) dinitrate (c) tetraamminecopper (Il) nitrate (d) tetrammine copper (II) dinitrite 2. Coordination number and oxidation number of Cr in K2[Cr(C2O4)3] are, respectively? (a) 4and +2 (b) б апd +3 ©З апd +3 (d) З апd 0 3. Which of the following is a negatively charged bidentate ligand? (a) Dimethylglyximato (b) Cyano (c) Ethylenediamine (d) Acetato 4. The oxidation number of Cobalt in K[Co (CO)4] is (a) +1 (b) +3 (c) -1 (d) -3 5. Amongst the following ions, which one is highly paramagnetic? (a) [Cr (H2O)6] 3+ (b) [Fe(H2O)6] 2+ (c) [Cu(H2O)6] 2+ (d) [Zn(H2O)6] 2+ 6. The spin only magnetic moment of [Ni(Cl4)]2- is (a) 1.82 BM (b) 5.46 BM (c) 2.82BM (d) 1.41 BM 7. The IUPAC name of complex ion ,[Fe(CN)6] 3- is (a) Hexacyanidoiron(III)ion (b) Hexacyanatoferrate (III)ion (c) Hexacyanidoferrate (III)ion (d) Tricyanoiron(III)ion 8. The geometry and magnetic behaviour of the complex [Ni(CO)4] are (a) Square planar and paramagnetic (b) Tetrahedral and diamagnetic (c) Square planar and diamagnetic (d) Tetrahedral and paramagnetic 9. The hybridisation involved in the complex [Ni(CN)4] 2- is (a) d 2 sp2 (b) d 2 sp3 (c) dsp2 (d) sp3 10. The geometry of [Ni(CN)4] 2- and [NiCl4] 2- are (a) Both square planar (b) Both tetrahedral (c) Tetrahedral and square planar respectively (d) Square planar and tetrahedral respectively 11. The diamagnetic species is (a) [Ni(CN)4] 2- (b) [NiCl4] 2- (c) [CuCl4] 2- (d) [CoF6] 3- 12. Correct increasing order for the wavelengths of absorption in the visible region in the complexes of Co3+ is (a) [Co(H2O)6] 3+ , [Co (en)3] 3+ , [Co(NH3)6] 3+ (b) [Co(H2O)6] 3+, [Co(NH3)6] 3+ , [Co (en)3] 3+ (c) [Co(NH3)6] 3+, [Co (en)3] 3+, [Co(H2O)6] 3+ (d) [Co (en)3] 3+,[Co(NH3)6] 3+, [Co(H2O)6] 3+ ASSERTION – REASON TYPE QUESTIONS In the following questions, a statement of Assertion followed by a statement of Reason is given. Choose the correct option out of the following choices. (a) Assertion and Reason both are true, Reason is the correct explanation of Assertion. (b) Assertion and Reason both are true but Reason is not the correct explanation of Assertion (c) Assertion is true, Reason is false. (d) Assertion is false, Reason is true. 13. Assertion: The crystal field theory is successful in explaining the formation, structure, colour and magnetic properties of coordination compounds. Reason: crystal field theory considers the metal-ligand bond to be ionic. Ans: (b) Assertion and Reason both are true but Reason is not the correct explanation of Assertion 14. Assertion: Linkage isomerism arises in coordination compounds containing ambidentate ligands Reason: Ambidentate ligand has two different donor atoms Ans: (a) Assertion and Reason both are true, Reason is the correct explanation of Assertion. 15. Assertion: CO is stronger ligand than NH3 for many metals Reason: NH3 can form pi bonds by back bonding Ans: (c) Assertion is true, Reason is false. SECTION B: (2MARKS) 16. Give reasons (i)CuSO4 5H2O is blue in colour while CuSO4 is colourless. Ans. In CuSO4 5H2O , water acts as ligand and causes crystal field splitting. Hence d-d transitionis possible and shows colour. (ii)Low spin tetrahedral complexes not formed Ans:In a tetrahedral complex, the d-orbital is split too small as compared to octahedral complex. For the same metal and same ligand Δt = 4/9Δo.Hence, the orbital energies are not enough to force pairing. As a result, low spin configurations are rarely observed in tetrahedral complexes. 17. Arrange the following complexes in the increasing order of conductivity of their solution: [Co(NH3)3Cl3], [Co(NH3)4Cl2]CI, [Co(NH3)6]Cl3, [Cr(NH3)5Cl]Cl2 Ans: Co(NH3)3Cl3] < [Co(NH3)4Cl2]Cl < [Cr(NH3)5Cl]Cl2 < [Co(NH3)6]Cl3 Here, the number of ions increases and conductivity increases. 18. A coordination compound CrCl3. 4H2O precipitates silver chloride when treated with silver nitrate. The molar conductance of its solution corresponds to a total of two ions. Write the structural formula of the compound and name it. Ans: Silver chloride precipitates when treated with silver nitrate because there is chloride ion outside the complex. There is only one chloride ion outside the complex since only two ions are generated. As a result, the compound’s structural formula and nomenclature are: [Co(H2O)4Cl2]Cl – Tetraaquadichloridocobalt(III) chloride 19. Explain why [Fe(H2O)6] 3+ has a magnetic moment value of 5.92 BM whereas [Fe(CN)6] 3- has a value of only 1.74 BM. Ans:[Fe(CN)6] 3– involves d2 sp3 hybridisation with one unpaired electron and [Fe(H2O)6] 3+ involves sp3 d 2 hybridisation with five unpaired electrons. This difference is due to the presence of strong ligand CN– and weak ligand H2O in these complexes. 20. [Ni(CN)4] 2- is colourless where as [Ni(H2O)6] 2+ is green. Why? Ans:In [Ni(CN)4] 2- , Ni is in +2 oxidation state with electronic configuration 3d8. In the presence of strong CNligand the two unpaired electron in 3d orbital pair up. As there is no unpaired electron, it is colourless. In [Ni(H2O)6] 2+ Ni is +2 oxidation state and electronic configuration 3d8. The two unpaired electrons do not pair up in the presence of weak ligand H2O. The d-d transition absorbs red lightand complementary green light is emitted. 21.Give the formula of each of the following coordination entities: (i)Co 3+ ion bound to one Cl- , one NH3 molecule and two ethylene diamine molecules. (ii)Ni 2+ ion is bound to two water molecules and two oxalate ions. Ans. (i) [Co (NH3)Cl (en)2] 2+ (ii)Ni(H2O)2(ox)2] 2- 22. Explain the following terms giving a suitable example in each case : (i) Ambident ligand (ii) Denticity of a ligand Ans: Denticity : The number of coordinating groups present in a ligand is called the denticity of ligand. For example, bidentate ligand ethane-1, 2-diamine has two donor nitrogen atoms which can link to central metal atom. Ambidentate ligand : A unidentate ligand which can coordinate to central metal atom through two dierent atoms is called ambidentate ligand. For example NO2 – ion can coordinate either through nitrogen or through oxygen to the central metal atom/ion. 23. Using IUPAC norms write the formulae for the following coordination compounds : (i) Hexaamminecobalt(III)chloride (ii) Potassiumtetrachloridonickelate(II) Ans: (i) [Co(NH3)6]Cl3 (ii) K2[NiCl4] 24. Write the name and draw the structures of each of the following complex compounds: (i) [Co(NH3)4(H2O)2]Cl3 (ii) [Pt(NH3)4][NiCl4] 25. For the complex [Fe(H2O)6] 3+, write the hybridization magnetic character and spin of the complex. (At number Fe = 26) Ans: The complex ion has outer orbital octahedral geometry (high spin) and is paramagnetic due to the presence of five unpaired electrons. SECTION C: (3MARKS) 26..[Ni(CO)4] possesses tetrahedral geometry while [Ni(CN)4] 2- is square planar. Why? Ans. In [Ni(CO)4] Ni is in 0 oxidation state and its electronic configuration is 3d8 4s2. CO is strong ligand. Hybridization is sp3 and it is tetrahedral. 27. For the complex [Fe(en)2Cl2], Cl, (en = ethylene diamine), identify (i) the oxidation number of iron, (ii) the hybrid orbitals and the shape of the complex, (iii) the magnetic behaviour of the complex, Ans: (i) [Fe(en)2Cl2]Cl x + 0 × 2 + (–1) × 2 + (–1) × 1 = 0 \ x = 3 Oxidation number of iron = 3 (ii) d2 sp3 hybridisation and octahedral shape. (iii) Paramagnetic due to presence of one unpaired electron. 28. Compare the following complexes with respect to their shape, magnetic behaviour and the hybrid orbitals involved : (i) [CoF4] 2- (ii) [Cr(H2O)2(C2O4)2]– (iii) [Ni(CO)4] (Atomic number : Co = 27, Cr = 24, Ni = 28) 29. Write the IUPAC name, deduce the geometry and magnetic behaviour of the complex K4[Mn(CN)6]. [Atomic no. of Mn = 25] 30. Write the state of hybridization, the shape and the magnetic behaviour of the following complex entities : (i) [Cr(NH3)4Cl2]Cl (ii) [Co(en)3]Cl3 (iii) K2[Ni(CN)4] 31. How is the stability of a co-ordination compound in solution decided? How is the dissociation constant of a complex defined? Ans: The stability of a complex in solution refers to the degree of association between the two species involved in the state of equilibrium. -e magnitude of the equilibrium constant for the association, quantitatively expresses the stability. -e instability constant or dissociation constant of coordination compound is dened as the reciprocal of the formation constant. 32. Explain the following : (i) Low spin octahedral complexes of nickel are not known. (ii) -e p-complexes are known for transition elements only. (iii) CO is a stronger ligand than NH3 for many metals. Ans: (i) Nickel forms octahedral complexes mainly in +2 oxidation state which has 3d8 configuration. In presence of strong field ligand also it has two unpaired electrons in eg orbital. Hence, it does not form low spin octahedral complexes. (ii) the transition metals/ions have empty d orbitals into which the electron pairs can be donated by ligands containing p electrons. For example: CH2= CH2 and C6H6, C5H5 –. (iii) Co is stronger ligand than NH3 because CO has vacant molecular orbitals with which it can form p-bond with metal through back donation. 33. (i) Write down the IUPAC name of the following complex. [Cr (en)3]Cl3 (ii) Write the formula for the following complex. Potassium trioxalato chromate (III) Ans: Tris(ethylenediammine)chromium(III) chloride (ii) K3[Cr(ox)3] 34. Write the hybridization and shape of the following complexes: (i) [CoF6] 3– (ii) [Ni(CN)4] 2 35. For the complex [NiCl4] 2–, write (i) the IUPAC name (ii) the hybridization type (iii) the shape of the complex. (Atomic no. of Ni = 28) (i) Tetrachloridonickelate(II) ion SECTION D: (4MARKS) CASE BASED QUESTIONS: 36. Read the given passage and answer the questions that follow: Complex compounds play an important role in our daily life. Werner’s theory of complex compounds says every metal atom or ion has primary valency (oxidation state) which is satisfied by –vely charged ions, ionisable where secondary valency (coordination number) is nonionisable, satisfied by ligands (+ve, –ve, neutral) but having lone pair. Primary valency is nondirectional, secondary valency is directional. Complex compounds are name according to IUPAC system. Valence bond theory helps in determining shapes of complexes based on hybridisation, magnetic properties, outer or inner orbital complex. Complex show ionisation, linkage, solvate and coordination isomerism also called structural isomerism. Some of them also show stereoisomerism i.e. geometrical and optical isomerism. Ambidentate ligand are essential to show linkage isomerism. Polydentate ligands form more stable complexes then unidentate ligands. There are called chelating agents. EDTA is used to treat lead poisoning, cis-platin as anticancer agents. Vitamin B12 is complex of cobalt. Haemoglobin, oxygen carrier is complex of Fe2+ and chlorophyll essential for photosynthesis is complex of Mg2+. (a) What is the oxidation state of Ni in [Ni(CO)4]? Ans. Zero (b) One mole of CrCl3. 6H2O reacts with excess of AgNO3 to yield 2 mole of AgCl. Write formula of complex. Write IUPAC name also. Ans. [Cr(H2O)5Cl]Cl2. H2O, pentaaquachloridochromium (III) chloride. (c) Out Cis – [Pt(en)2 Cl2] 2+ and trans [Pt(en)2Cl2] 2+ which one shows optical isomerism? Ans. Cis – [Pt(en)2 Cl2] 2+ shows optical isomerism. (d) Name the hexadentate ligand used for treatment of lead poisoning. Ans. EDTA4– (ethylenediamine tetraacetate) (e) What is hybridisation of [CoF6] 3–? [Co = 27] Give its shape and magnetic properties. Ans. Sp3 d 2 , octahedral, paramagnetic. (f) What type of isomerism is shown by [Cr(H2O)6] Cl3 and [Cr(H2O)5 Cl] Cl2.H2O? Ans. Solvate isomerism. 37. Observe the diagram of splitting of d-orbitals in octahedral field and answer the questions based on the diagrams and related studied concepts. (a) What is crystal field splitting anergy? Ans. The energy difference between the two sets of d-orbitals is called crystal field splitting energy denoted by ∆0. (b) Arrange the following complex ions in the increasing order of Crystal field splitting energy [CrCl6] 3- , [CrCN6] 3-, [Cr(NH3)6 ] 3+ (c) Ans. [CrCl6] 3- , [Cr(NH3)6 ] 3+ , [CrCl6] 3- (d) What is relationship between ∆0 (CFSE) and strength of ligand? Ans. Greater the ∆0(CFSE), more will be strength of ligand (e) What is electronic configuration of d5 ion if ∆ 0< P? Ans. t2g 3 eg 2 38. Read the following paragraph and answer the questions. In 1823, Werner put forth this theory to describe the structure and formation of complex compounds or coordination compounds. It is because of this theory that he got the Nobel prize and is known as the father of coordination chemistry. According to his theory The central metals of coordination compounds exhibit two types of valencies, primary valency and secondary valency. The primary valencies are ionizable. These are written outside the coordination sphere. These are nondirectional and do not give any geometry to complex compound. The secondary valency of metals is either by negative ions or neutral molecules or both. In modern terminology it represents the coordination number of the metal. Secondary valencies are written inside the coordination sphere. These are directional in nature and give definite geometry to the complex. These are non-ionisable. (a)A coordination compound CrCl3.4H2O precipitates silver chloride when treated with silver nitrate. The molar conductance of its solution corresponds to a total of 2 ions. Write the structural formula of the complex. Ans. [Co(H2O)4Cl2]Cl (b)The increasing order of conductivity of the following complexes in their solutions [Co(NH3)3Cl3] , [Co(NH3)6]Cl3 ,[Co (NH3)4Cl2]Cl Ans. [Co(NH3)3Cl3], ,[Co (NH3)4Cl2]Cl, [Co(NH3)6Cl3 (c)Write the correct formula for the following coordination compound CrCl3.6H2O with 3 chloride ions precipitated as AgCl. Ans.[Cr(H2O)6]Cl3 (f) How many ions are produced from the complex [Co(NH3)5Cl]Cl2 in solution ? Ans.3 39. Read the following paragraph and answer the questions. Valence bond theory considers the bonding between metal and the ligands as purely covalent. On the other hand the crystal field theory considers the bond to be ionic purely arising from electrostatic interactions between the metal ion and the ligands. In coordination compounds the interaction between the metal ion and the ligands causes the 5 d orbitals to split up. This is called crystal field splitting and the energy difference between the two sets of orbitals is called crystal field splitting energy. CFSE depends upon the nature of the ligand. (a)Write the electronic configuration of [CoF6] 3- on the basis of crystal field theory Ans. Co3+ has ( d 6 ) ; t2g 4 eg 2 configuration (b)Arrange the following complex ions in the increasing order of crystal field splitting energy [CrCl6] 3- , [Cr(CN)6] 3- ,[Cr(NH3)6] 3+ Ans. [CrCl6] 3- , [Cr(NH3)6] 3+,[Cr(CN)6] 3 (dCAmong the ligands ,NH3 ,en and CO which ligand is having the highest field strength ? Ans.CO (d)What will be the increasing order for the wave lengths of absorption in the visible region of the following : [Ni (NO2)6] 4- , [Ni (NH3)6] 2+,[Ni (H2O)6] 2+ Ans. [Ni (H2O)6] 2+, [Ni (NH3)6] 2+, [Ni (NO2)6] 4- SECTION E (5 MARKS) 40. (a) Give two examples of coordination compounds used in industries. (b) Using valence bond theory, explain the geometry and magnetic behavior of [Co(NH3)6]3+ (At. no. of Co = 27) Ans: (a) Examples: (i) Pure Ni can be obtained from Ni(CO)4 (ii) Gold and Ag are extracted by the use of complex formation like Na[Ag(CN)2]. 41. (a) How is a double salt different from a complex? (b) Write IUPAC names of the following: (i) K3[Fe(C2O4)3] (ii) [Pt(NH3)6]Cl4. (c) Draw the structure of cis isomer of [CO(NH3)4Cl2]+ Ans: (a) Double salt dissociates completely into its constituent ions in their aqueous solution. Example : KCl.MgCl2.6H2O dissociates into K+, Cl–, Mg2+ and H2O Complex does not dissociate into its constituent ions. Example : K4[Fe(CN)6] → 4K+ + Fe(CN)6]4- (b) (i) K3[Fe(C2O4)3] IUPAC name : Potassium trioxalatoferrate (III) (ii) [Pt(NH3)6]Cl4 IUPAC name : Hexaammine Platinum (IV) chloride (c) Structure of cis isomer of [CO(NH3)4Cl2]+ 42. When a coordination compound CrCl3.6H2O is mixed with AgNO3 solution, 3 moles of AgCl are precipitated per mole of the compound. Write : (i) Structural formula of the complex (ii) IUPAC name of the complex (iii) Magnetic and spin behaviour of the complex Inner orbital complex so it is low spin complex. Since 3 unpaired electrons are present, it is paramagnetic in nature. 43. (i) What type of isomerism is shown by the complex [Co(NH3)6][Cr(CN)6]? (ii) Why a solution of [Ni(H2O)6]2+ is green while a solution of [Ni(CN)4]2- is colourless? (At. no. of Ni = 28) (iii) Write the IUPAC name of the following complex: [CO(NH3)5(CO3)]Cl. Ans: (i) Coordination isomerism (ii) [Ni(H2O)6]2+ is an outer orbital complex due to weak field ligand H2O and the presence of unpaired electrons undergoes d—d transition by absorbing red light and shows green colour while [Ni(CN)4]2- is an inner orbital complex and has no unpaired electrons hence colourless. (iii) Pentaamminecarbonatocobalt (III) Chloride*********************************************************************** COORDINATION COMPOUNDS (2ND HALF) Section A Section A - 1 Mark's MCQ (15) Section B - 2 Mark's questions (10) Section C - 3 Mark's questions (10) Section D - 4 Mark's case study based questions (4 ) Section E - 5 Mark's questions (4) MCQ carrying 1 mark each Q.1. The crystal field splitting energy for octahedral (∆o) and tetrahedral (∆t) complexes is related as - a) ∆t = 2/9∆o b) ∆ t = 5\9∆o c) ∆t = 4/9∆o d) ∆ t = 2∆o Q.2. What are the number of unpaired electrons in the square planer [Pt(CN)4]2- ion? a) 0 b) 1 c) 4 d)6 Q.3. Magnetic nature and spin of [Co(NH3)6]3+ is - a) Paramagnetic, High spin complex b) Diamagnetic, low spin complex c) Paramagnetic, low spin complex d) Diamagnetic, high spin complex Q.4. What was the term proposed by Werner for the number of groups bound directly to the metal ion in a coordination complex? a) Primary valence b) Secondary valence c) Oxidation number d) Polyhedra Q.5. Werner postulated that octahedral, tetrahedral and square planar geometrical shapes are more common in coordination compounds of ________ a) alkali metals b) lanthanides c) actinides d) transition metals Q.6. What is the sum of the oxidation number of cobalt in [Co(H2O)(CN)(en)2]2+and [CoBr2(en)2]+? a) +3 b) +4 c) +5 d) +6 Q.7. Hybridisation of [Ni(CN)4]2- is (a) dsp2 (b) d2sp2 (c) sp3 (d) d2sp3 Q.8. Which of the following is true for the formation of stable bonds according to valence bond theory? (a) Greater overlapping between atomic orbitals (b) Close proximity between two atoms (c) Pairing of electrons having opposite spins (d) All of the above Q.9. A coordination complex’s core atom/ion is also known as ________ a) Bronsted-Lowry acid b) Lewis base c) Lewis acid d) Bronsted-Lowry base Q.10. Which of the following d orbitals take part in the octahedral complex with d2sp3 hybridisation? (a) dxy, dyz (b) dxz, dx2−y2 (c) dx2−y2, dz2 (d) dz2, dxz Q.11. The correct increasing order of splitting power of ligands according to spectrochemical series is (a) Cl– < OH– < CN– (b) Cl– < CN– < OH– (c) OH– < Cl– < CN– (d) OH– < CN– < Cl– Q.12. Which of the following is paramagnetic? (a) [CoBr]42- (b) Mo(Co)6 (c) [Pt(en)Cl2] (d) [Co(NH3)6]3+ Q.13. Hardness of water is estimated by simple titration with which compound? a) Na2(EDTA) b) Fe(EDTA) c) Mg(EDTA) d) Co(EDTA) Q.14. Haemoglobin is a complex compound of which metal ion? a) Fe3+ b) Fe2+ c) Co2+ d) Co3+ Q.15. The complex ion [Ag(S2O3)2]3- is associated with which field? a) Electroplating b) Medicine c) Water treatment d) Photography Section - B Q.16. Write the hybridization and shape of the following complexes - i) [CoF6]3- ii) [Ni(CN)4]2- Q.17. Why crystal field splitting energy more in case of octahedral complexes as compared to tetrahedral complexes? Q.18. Explain crystal field splitting in an octahedral complex. Q.19. Explain the term spectrochemical series. Q.20. Using valence bond approach, explain the magnatic character of [Co(NH3)6]3+ ion. (At. No. Of Co = 27) Q.21. Write the hybridization and shape of the complex [NiCl4 ]2-. Q.22. Differentiate between weak field and strong field ligand. Q.23. Why are low spin tetrahedral complexes not formed? Q.24. Why are different colours observed in Octahedral and tetrahedral complexes for the same metal and same ligands? Q.25. CuSO4.5H2O is blue in colour while CuSO4 is colourless. Why? Section - C Q.26. Using valence bond theory, explain the following in relation to the complexes given below: [Mn(CN)6]3-, [Co(NH3)6]3+, [Cr(H2O)6]3+ (i) Type of hybridisation. (ii) Inner or outer orbital complex. (iii) Magnetic behaviour. Q.27. NH3 acts as complexing agent but NH4+ does not, explain. Q.28. Fe(H2O)6]3− is strongly paramagnetic whereas [Ni(NH3)6]2+ weakly paramagnetic. Explain. Q.29. Name the central metal atom/ion present in -Chlorophyll, Haemoglobin, Vitamin B-12. Q.30.Write the limitations of Valence Bond Theory. Q.31. (i) Low spin octahedral complexes of nickel are not known. (ii) The π-complexes are known for transition elements only. Q.32. What is meant by crystal field splitting energy? On the basis of crystal field theory, write the electronic configuration of d4 in terms of t2g and eg in an octahedral field when (i) Δ0 > P (ii) Δ0 < P Q.33. Discuss the nature of bonding in metal carbonyl. Q.34. State and describe the factors which govern stability of complexes. Q.35. Discuss briefly giving an example in each case the role of coordination compound in (1) biological systems (2) medicinal chemistry (3) analytical chemistry Section - D Q.36. Read the passage given below and answer the following questions: To explain bonding in coordination compounds various theories were proposed. One of the important theory was valence bond theory. According to that, the central metal ion in the complex makes available a number of empty orbitals for the formation ofcoordination bonds with suitable ligands. The appropriate atomic orbitals of the metal hybridise to give a set of equivalent orbitals of definite geometry. The d-orbitals involved in the hybridisation may be either inner d-orbitals i.e., (n - 1) d or outer d-orbitals i.e., nd. For example, CO3+ forms both inner orbital and outer orbital complexes, with ammonia it forms [Co(NH3)6]3+ and with fluorine it forms [CoF6]3- complex ion. The following questions are multiple choice questions. Choose the most appropriate answer: (i) Which of the following is not true for [CoF6]3- ? (a) It is paramagnetic. (b) It has coordination number of 6. (c) It is outer orbital complex. (d) It involves d2sp3 hybridisation. (ii) Which of the following is true for [Co(NH3)6]3+ ? (a) It is an octahedral, dimagnetic and outer orbital complex. (b) It is an octahedral, paramagnetic and outer orbital complex. (c) It is an octahedral, paramagnetic and inner orbital complex. (d) It is an octahedral, dimagnetic and inner orbital complex. (iii) The paramagnetism of [CoF6]3- is due to (a) 3 electrons (b) 4 electrons (c) 2 electrons (d) 2 electrons (iv) Which of the following is an inner orbital or low spin complex? (a) [Ni(H2O)6]3+ (b)[FeF6]3− (c) [Co(CN)6]3− (d) [NiCl4]2− Q.37. Valence bond theory considers the bonding between the metal ion and the ligands as purely covalent. On the other hand, crystal field theory considers the metal-ligand bond to be ionic arising from electrostatic interaction between the metal ion and the ligands. In coordination compounds, the interaction between the ligand and the metal ion causes the five d-orbitals to split-up. This is called crystal field splitting and the energy difference between the two sets of energy level is called crystal field splitting energy. The crystal field splitting energy (Δo) depends upon the nature of the ligand. The actual configuration of complexes is divided by the relative values of Δo and P (pairing energy) If Δo < P, then complex will be high spin. If Δo > P, then complex will be low spin. The following questions are multiple choice questions. Choose the most appropriate answer : (i) Which of the following ligand has lowest Δo value? (a) CN- (b) CO (c) F- (d) NH3 (ii) The crystal field splitting energy for octahedral (Δo) and tetrahedral (Δt) complex is related as (a) Δt=1/2Δo (b) Δt=4/9Δo (c) Δt=3/5Δo (d) Δt=2/5Δo (iii) On the basis of crystal field theory, the electronic configuration of d4 in two situations: (i) Δo > P and (ii) Δo< P are (iv) Using crystal field theory, calculate magnetic moment of central metal ion of [FeF6]4-. (a) 1.79 B.M. (b) 2.83 B.M. (c) 3.85 B.M. (d) 4.9 B.M. Q.38. balance bond theory describe the bonding in complexes in terms of co-ordinate covalent bonds resulting from overlap filled ligand orbitals with vacant metal hybrid orbitals. The theory explain magnetic behaviour and geometrical shape of coordination compounds. Magnetic moment of a complex can be determined experimentally and theoretically by using spin only formula. Magnetic momentum √n(n+2)BM (n is no. Of unpaired electrons) (i) write the state of hybridization, shape and the magnetic behaviour of the [Cr(H2O)2(C2O4)2 ]-. (ii) Why s orbital does not show preference to any direction. (III) why is [CoF6]3- is paramagnetic but [Co(NH3)6]3+ is diamagnetic in nature? (IV) Describe the type of hybridization, shape and magnetic property of [Co(NH3)4Cl2]Cl. Q.39. coordination compound plays many important role in animals and plants they are essential in the storage and transport of oxygen, as electron transfer agents as catalysts and in photosynthesis. White range of application in daily life takes place through formation of complexes photo photographic fixing qualitative and quantitative analysis purification of water, metallurgical extraction are some specific worth mentioning. (I) Exalic acid is commonly used (A)To make medicines (B) To remove rust and stains (C)as a thinner or in polishing the wooden material (D)all of the above (Ii) Ziegler- NATA catalyst is used to convert (A) ethylene (b) ethyne (c) ethanal (d)ethanol (III) calcium dihydrogen salt of EDTA is used as an antidote for lead poisoning because (A) excess EDTA remove Ca2+ ions from body (b) calcium ions co-ordinates with Lead in the body (c) excess EDTA will not remove Ca2+ ions from body (d) b and C (IV) cis - platin, stands for (A) cis - diamminechloroplatinum (II) (b) cis - dichloridodiammineplatinum(I) (c) cis - dimethyldichloridoplantinum (II) (d)bcis - dichloridodiammineplatinum (II) Section - E Q.40. Using valence bond theory, explain the following in relation to the complexes given below: [Mn(CN)6]3-, [Co(NH3)6]3+, [FeCl6]4- (i) Type of hybridisation. (ii) Inner or outer orbital complex. (iii) Magnetic behaviour. (iv) Spin only magnetic moment value. Q.41. Explain Werner's theory of coordinate compounds with suitable examples? Q.42. What is crystal field splitting? Explain crystal field splitting in tetrahedral complexes. Q.43. a) Explain why [Fe(HO)6]3+ has a magnetic moment value of 5.92 BM whereas [Fe(CN)6]3- has a value of only 1.74 BM. b) Why do compounds having similar geometry have different magnetic moments? MCQ answer key Q.1. c) Q.2. a) Q.3. b) Q.4. b) Q.5. d) Q.6. d) Q.7. a) Q.8. d) Q.9. c) Q.10. c) Q.11. a) Q.12. a) Q.13. a) Q.14. b) Q.15. d) Section - B Answer key Q.16. i) In [CoF6]3- cobalt is in+3 O.S.. F- is a weak field ligand. It does not cause pairing. Therefore cobalt undergoes sp3d2 hybridization and have octahedral geometry. ii) In [Ni(CN)4]2- Ni is in+2 O. S. and has the electronic configuration 3d84s0. As CN- is strong field ligand it causes pairing of electrons resulting one 3d orbital empty, with one 4s and two 4p it gives dsp2 hybridization. And shape is square planer. Q.17. Crystal field splitting energy in Octahedral complexes is more than in tetrahedral complexes, because - i) in Octahedral complexes there are 6 ligand which face repulsion from valence electrons of central metal atom. But only 4 ligand in tetrahedral complexes. ii) Also because in Octahedral complexes 2 ligands approach Central metal atom along the direction of 2 d orbitals (dx2-y2, dz2). But in tetrahedral complexes all ligands approach the central metal atom between the axes not along any axes with d orbitals. Q.18. Crystal field splitting in Octahedral field -In free transition metal ion all d orbitals are equal in energy called degenerate orbitals. But as soon as ligand approaches the central metal ion, repulsion between electron's of central metal ion and electron's of ligand causes splitting of 5 degenerate d orbitals into two sets of orbitals. eg orbitals with 2 orbitals are high in energy and t2g with 3 orbitals low in energy as follows Q.19. Spectrochemical series - The arrangement of ligands in the order of increasing field strength or increasing crystal field splitting energy values is known as spectrochemical series. Br-, Cl-,SCN-,F-, OH-, H2O, NH3, en, CN- , CO Q.20. In [Co(NH3)6]3+ ion O.S. of Co is +3 hybridization in presence of NH3 is strong field ligand Q.22. Weak field ligand - the ligand which have small value of crystal field splitting energy (∆o). For these ligands ∆oP. These ligands produce low spin complexes. Q.23. For tetrahedral complexes, the crystal field splitting energy is smaller than pairing energy, so pairing doesn't take place and high spin complexes are formed instead of low spin. Q.24. ∆t=(4/9)∆o So, a higher wavelength of light is absorbed in octahedral complexes than tetrahedral complexes for same metal and ligands. Thus, different colours are observed. Q.25. Water acts as a ligand in CuSO4.5H2O, causing crystal field splitting. As a result, in CuSO4.5H2O, a d—d transition is possible and coloured. Crystal field splitting is impossible in anhydrous CuSO4 due to the lack of water (ligand), hence there is no colour. Section - C Q.26. [Mn(CN)6]3- (i) Type of hybridisation = d2sp3 hybridisation (ii) Inner or outer orbital complex = Inner orbital complex because (n-1)d – orbitals are used. (iii) Magnetic behaviour = Paramagnetic, as two unpaired electrons are present. [Co(NH3)6]3+ (i) Type of hybridisation = d2sp3 hybridisation (ii) Inner or outer orbital complex = Inner orbital complex (as(n-1)d-orbitals take part.) (iii) Magnetic behaviour = Diamagnetic (as three paired electrons are present.) [Cr(H2O)6]3+ (i) Type of hybridisation = d2sp3 hybridisation (ii) Inner or outer orbital complex = Inner orbital complex as(n-1)d-orbitals take part.) (iii) Magnetic behaviour = Paramagnetic (as three unpaired electrons are present.) Q.27. Because ammonia only has one lone pair of electrons, it can accept electrons from coordination complexes. Ammonia is transformed into an ion called ammonium, which lacks a lone pair to donate and form complexes. As a result, whereas ammonia can be an excellent ligand, the ammonium ion does not form complexes. Q.28. The ligand NH3 is neither a strong field ligand nor a weak field ligand in the [Ni(NH3)6]2+ complex ion. It is, in fact, a weak strong field ligand. However, because the crystal field stabilisation energy is smaller than the pairing energy, the ligand NH3 acts as a weak field ligand in the [Ni(NH3)6]2+ ion. As a result, the electronic configuration under the effect of an octahedral crystal field is t2 g6eg2. The complex has two unpaired electrons, according to the aforementioned electrical structure. As a result, the [Ni(NH3)6]2+ complex is weakly paramagnetic. In the complex [Fe(H2O)6]3+ the oxidation state of Fe is +3, having the structure 3 d5. Water (H2O) is a weak ligand, and 3 d electrons do not couple up in its presence. The sp3d2 hybridization results in an outer orbital octahedral complex with 5 unpaired electrons. As a result, it is very paramagnetic. Q.29. The central metal ion in the structure of chlorophyll is magnesium. The central metal ion in the structure of haemoglobin is iron. The central metal ion in the structure of vitamin B-12 is cobalt. Q.30. The limitations of valence bond theory are: (i) VBT fails in explaining the tetravalency of carbon. (ii) The geometries of ammonia, water, methane, etc cannot be properly explained by this theory. (iii) Bond angles of molecules such as carbon dioxide, water, ammonia, etc could not properly be given by VBT. (iv) VBT does not explain the magnetic properties of complex Q.31. (i) The electronic configuration of Ni is [Ar] 3d8 4s2 which shows that it can only form two types of complexes i.e. square planar (dsp2) in presence of strong ligand and tetrahedral (sp3) in presence of weak ligand. There are four empty orbitals in Ni while octahedral complexes require six empty orbitals. (ii) Due to presence of empty d-orbitals in transition metals, they can accept electron pairs from ligands containing π electrons and hence can form ic-bonding complexes. Q.32. d-orbital is degenerate which split into two levels eg and t2g in the presence of ligands. This splitting is due to the presence of ligands. This is called the crystal-field splitting and the energy difference between the two levels (eg and t2g) is called the crystal-field splitting energy, Δ0 Electrons are singly filled in the t2g energy levels first and the remaining electrons are filled based on the crystal field splitting energy and pairing energy. Filling up of the remainig electrons takes place in two ways. (i) If the CFSE (Δ0) is greater than that of the pairing energy (p), electrons will be filled in the 'eg' orbitals. (ii) If the CFSE (Δ0) is lesser than that of the pairing energy (p), electrons will be paired up in the t2g energy levels. Q.33. The metal carbon bond in metal carbonyl have both sigma and pi character. The metal carbon sigma bond is formed by the donation of lone pair of electrons of the carbonyl carbon to a vacant orbital of the metal. The metal carbon pi bond is formed by the donation of a pair of electrons from a filled d orbital of metal into the vacant antibonding pi molecular orbitals of carbon monoxide. The metal to ligand bonding creates a synergic effect which strengthen the bond between CO and the metal. Q.34. I) Charge on the central metal ion - greater the charge, greater is the stability of complex. II) Nature of the metal ion- group 3-6 form more stable complexes. III) Basic nature of the ligands. IV) presence of chelate rings increases the stability of complex. V) Multidentate cyclic ligands without any steric effect increases the stability of complex further. Q.35. Biological systems - (A) hemoglobin the oxygen carrier in blood is a complex of Fe2+ with porphyrin. (B) the pigment chlorophyll in plants responsible for photosynthesis is a complex of Mg2+ with porphyrin. (C) vitamin B12 is a complex of Cobalt. Medicinal chemistry (A) the platinum complex cis platin is used in the treatment of cancer. (B) the excess of copper and

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