+2 Chemistry Notes 2024-2025 PDF

Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® +2 CHEMISTRY NOTES (For the Academic year 2024 - 25) PREPARED BY: ANIL KUMAR K L HSST...

Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® +2 CHEMISTRY NOTES (For the Academic year 2024 - 25) PREPARED BY: ANIL KUMAR K L HSST CHEMISTRY PHSS VANDIPERIYAR IDUKKI. Ph: 9496688551  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® CONTENTS No. Units Page No. 1. Solutions 2 – 11 2. Electrochemistry 12 - 21 3. Chemical Kinetics 22 - 31 4. d and f Block Elements 32 - 40 5. Co-ordination Compounds 41 - 52 6. Haloalkanes and Haloarenes 53 - 64 7. Alcohols, Phenols and Ethers 65 – 77 8. Aldehydes, Ketones and Carboxylic acids 78 – 91 9. Amines 92 – 99 10. Biomolecules 100 - 107 +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 1  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® 1. SOLUTIONS Solutions are homogeneous mixtures containing two or more components. Generally, the component that is present in larger quantity is called solvent. Solvent determines the physical state of the solution. One or more components present in the solution other than solvent are called solutes. [Or, the substance which is dissolved is called solute and the substance in which solute is dissolved is called solvent]. Solutions containing only two components are called binary solutions. Here each component may be solid, liquid or in gaseous state. Based on this, solutions are of the following types: Types of Solution Solute Solvent Examples Gas Gas Mixture of O2 and CO2 Gaseous solutions Liquid Gas Chloroform mixed with nitrogen gas, water-vapour in air Solid Gas Camphor in nitrogen gas, naphthalene in air Gas Liquid Oxygen dissolved in water, soda water Liquid solutions Liquid Liquid Alcohol dissolved in water, dilute acids and alkalies Solid Liquid Salt in water, glucose in water Gas Solid Hydrogen in Pd, Pt, Ni etc Solid solutions Liquid Solid Amalgam of mercury with sodium Solid Solid Gold ornaments, alloys of metals Concentration of Solutions Composition of a solution can be expressed in terms of concentration. Concentration is defined as the number of moles of solute present per litre of the solution. The concentration of a solution can be expressed by the following ways: (i) Mass percentage (w/w): It is defined as the mass of the component present in 100g of the solution. Mass of the component in the solution × 100 i.e. Mass % of a component = Total mass of the solution For e.g. 10% aqueous solution of glucose by mass means that 10 g of glucose is dissolved in 90 g of water resulting in a 100 g solution. Concentration described by mass percentage is commonly used in industrial chemical applications. (ii) Volume percentage (v/v): It is defined as the volume of a component present in 100 mL of the solution. Volume of the component ×100 i.e. Volume % of a component = Total volume of the solution For example, 10% ethanol solution in water means that 10 mL of ethanol is dissolved in 90 mL of water such that the total volume of the solution is 100 mL. Concentration of solutions containing liquids is commonly expressed in this unit. (iii) Mass by volume percentage (w/v): It is the mass of solute dissolved in 100 mL of the solution. It is commonly used in medicine and pharmacy. 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑥 100 Mass/volume % of a component = 𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (iv) Parts per million (ppm): When a solute is present in trace quantities (i.e. very small amounts), its concentration is expressed in parts per million (ppm). It is defined as the number of parts of a particular component in million parts of the solution. Number of parts of the component × 106 i.e. Parts per million (ppm) = Total number of parts of all the components of the solution +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 2  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® Concentration in parts per million can be expressed as mass to mass, volume to volume and mass to volume. The concentration of pollutants in water or atmosphere is expressed in terms of μg mL –1 or ppm. [μg is microgram] (v) Mole fraction (χ): It is defined as the ratio of the number of moles of a particular component to the total number of moles of the solution. Number of moles of the component Mole fraction of a component = Total number of moles of solution For example, in a binary solution, if the number of moles of A and B are nA and nB respectively, nA then, the mole fraction of A (χA) = nA + nB nB and that of the component B (χB ) = nA + nB nA nB χA + χB = n + nB + nA + nB =1 A i.e. in a given solution sum of the mole fractions of all the components is unity. If there are i components, then, χ1 + χ2 + χ3 +.................. + χ𝑖 = 1 Mole fraction is useful in describing the calculations involving gas mixtures. (vi) Molarity (M): It is defined as the number of moles of solute dissolved per litre of solution. Number of moles of solute (n) i.e. Molarity (M) = Volume of solution in litre (V) For example, 1 M (molar) NaOH solution means that 1 mol of NaOH (40g) is dissolved in one litre of solution. (vii) Molality (m): It is defined as the number of moles of the solute present per kilogram (kg) of the solvent. Number of moles of solute i.e. Molality (m) = Mass of solvent in kg For example, 1 molal (m) solution of KCl means that 1 mol (74.5 g) of KCl is dissolved in 1 kg of water. Among the different methods for expressing the concentration of solution, mass percentage, ppm (in terms of mass), mole fraction and molality are temperature independent; while molarity, mass by volume percentage and volume percentage are temperature dependent. This is because volume of solution depends on temperature and the mass does not. SOLUBILITY Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a particular temperature. It depends upon the nature of solute, nature of the solvent, temperature and pressure. Solubility of a Solid in a Liquid It is observed that polar solutes dissolve in polar solvents and non-polar solutes in non-polar solvents. In general, a +solute dissolve in a solvent if the intermolecular interactions are similar in the two or the general principle related to solubility is that “like dissolves like”. Saturated and Unsaturated solutions When a solid solute is added to the solvent, some solute dissolves and its concentration increases in solution. This process is known as dissolution. Some solute particles in solution collide with the solid solute particles and get separated out of solution. This process is known as crystallisation. After sometime, the rate of dissolution and crystallization becomes equal and a dynamic equilibrium is reached. At this stage the concentration of solute in the solution remain constant and such a solution is called saturated solution. A solution in which no more solute can be dissolved at the same temperature and pressure is called a saturated solution. For such a solution, the concentration of the solution is equal to its solubility. A solution in which more solute can be dissolved at the same temperature is called an unsaturated solution. +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 3  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® Effect of temperature on the solubility of a solid in a liquid The solubility of a solid in a liquid mainly depends on temperature. Since the dissolution of a solid in a liquid is an equilibrium process, it should follow Le Chatelier’s Principle. In general, in a nearly saturated solution, if the dissolution process is endothermic (Δsol H > 0), the solubility should increase with rise in temperature and if it is exothermic (Δsol H > 0) the solubility should decrease with temperature. Effect of pressure on the solubility of a solid in a liquid Since solids and liquids are highly incompressible, pressure does not have any significant effect on solubility of solids in liquids. Solubility of a Gas in a Liquid Solubility of gases in liquids is greatly affected by pressure and temperature. The solubility of a gas increases with increase of pressure. A quantitative relation between pressure and solubility of a gas in a liquid was first given by Henry, which is known as Henry’s law. “The law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas”. Or, “the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (χ) in the solution” and is expressed as: p = KH χ Here KH is the Henry’s law constant. The value of KH depends on the nature of the gas and temperature. As the value of KH increases, the solubility of the gas in the liquid decreases. A graph of partial pressure (p) of the gas against mole fraction (χ) of the gas in solution is a straight line as follows. The slope of the graph gives the value of KH. Vapour pressure (p) Mole fraction (χ) As the temperature increases solubility of a gas in a liquid decreases. It is due to this reason that aquatic species are more comfortable in cold water rather than in warm water. Applications of Henry’s law 1. To increase the solubility of CO2 in soft drinks and soda water, the bottle is sealed under high pressure. 2. A medical condition known as bends in scuba divers. To avoid bends, the cylinders used by scuba divers are filled with air diluted with helium (The composition of the air in the cylinders used by scuba divers is 32.1% oxygen, 56.2% nitrogen and 11.7% helium). 3. A medical condition known as anoxia in people living at high altitudes and in mountaineers or climbers. Effect of Temperature on the solubility of a gas in a liquid Solubility of gases in liquids decreases with rise in temperature. When dissolved, the gas molecules are present in liquid phase. So the process of dissolution can be considered similar to condensation, which is exothermic. Hence solubility decreases with increase of temperature. +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 4  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® Vapour Pressure of a liquid In a liquid, the molecules with higher energy are escaped to vapour phase. This process is called evaporation. As the density of the vapour increases, the molecules collide with each other and so their energy decreases and returns to the liquid state. This process is called condensation. After some time, the rate of evaporation becomes equal to rate of condensation and the two processes attain equilibrium. At this condition, the pressure exerted by the vapour is called vapour pressure. So vapour pressure is defined as the pressure exerted by the vapour in equilibrium with its own liquid. It depends on the nature of the liquid and the temperature. As the temperature increases, the vapour pressure also increases. Vapour Pressure of Liquid Solutions In liquid solutions, the solvent is always a liquid. The solute can be a gas, a liquid or a solid. Generally, the liquid solvent is volatile. The solute may or may not be volatile. Based on the volatility of solute, the vapour pressure of the solution is greater or less than that of the solvent. Vapour Pressure of Liquid-Liquid Solutions – Raoult’s Law A quantitative relationship between the vapour pressure and mole fraction of solute in a solution was first given by a French chemist Francois Marte Raoult (F M Raoult) and it is known as Raoult’s Law. It states that for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction. Consider a binary solution of two volatile liquids 1 and 2. Let p 1 and p2 be the partial vapour pressures of the two components 1 and 2 respectively and ptotal be the total vapour pressure. Let χ1 and χ2 be the mole fractions of the two components 1 and 2 respectively. Then according to Raoult’s law, for component 1, p1 ∝ χ1 or, p1 = 𝑝10. χ1 and for component 2, p2 ∝ χ2 or, p2 = 𝑝20. χ2 Where p10 and p20are the vapour pressures of the pure components 1 & 2 respectively. According to Dalton’s law of partial pressures, the total pressure (ptotal ) will be the sum of the partial pressures of the components of the solution. So, ptotal = p1 + p2 Substituting the values of p1 and p2, we get ptotal = χ1. 𝑝10 + χ2. 𝑝20 = (1 – χ2 )𝑝10 + χ2. 𝑝20 Or, 𝐩𝐭𝐨𝐭𝐚𝐥 = 𝒑𝟎𝟏 + (𝒑𝟎𝟐 – 𝒑𝟎𝟏 ) 𝛘𝟐 Plots of p1 or p2 against the mole fractions χ1 and χ2 give straight lines (I and II). Similarly the plot of ptotal versus χ2 (line III) is also linear. +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 5  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® The composition of vapour phase in equilibrium with the solution is determined from the partial pressures of the components. If y1 and y2 are the mole fractions of the components 1 and 2 respectively in the vapour phase then, using Dalton’s law of partial pressures: p1 = y1 ptotal and p2 = y2 ptotal In general, pi = yi ptotal Raoult’s Law as a special case of Henry’s Law According to Raoult’s law, the vapour pressure of a volatile component in a solution is given by P1 = χ1. p10. According to Henry’s law, solubility of a gas in a liquid is given by p = KH χ. If we compare the equations for Raoult’s law and Henry’s law, we can see that the partial pressure of the volatile component or gas is directly proportional to its mole fraction in solution. Only the proportionality constant KH differs from p10. Thus, Raoult’s law becomes a special case of Henry’s law in which KH becomes equal to p10. Ideal and non-ideal solutions Liquid – liquid solutions can be classified into ideal and non-ideal solutions on the basis of Raoult’s law. 1. Ideal solutions: These are solutions which obey Raoult’s law over the entire range of concentration. For such solutions, the vapour pressure of a component in the solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution. i.e. p1 = 𝑝10. χ1 and p2 = 𝑝20. χ2 For such solutions, no heat change occurs during the mixing of the pure components [i.e. ∆mixH = 0]. Also, the volume of the solution is equal to the sum of the volumes of the components. Or, there is no change in the volume of the solution during mixing the components [i.e. ∆mixV = 0] Thus for an ideal solution, p1 = 𝒑𝟎𝟏. 𝝌𝟏 , p2 = 𝒑𝟎𝟐. 𝝌𝟐 , ∆mixH = 0 and ∆mixV = 0 Ideal behaviour can be explained by considering two components A and B. In pure components, the inter molecular attractive interactions will be of types A-A and B-B. In solution, in addition to these two interactions, A-B type of interaction will also be present. If the A-A and B-B interactions are nearly equal to the A-B interaction, the solution behaves ideally. i.e. solute-solute interactions and solvent-solvent interactions are nearly equal to solute-solvent interaction. A perfectly ideal solution is rare. But some solutions are nearly ideal in behaviour. E.g. solutions of n-hexane and n-heptane, bromoethane and chloroethane, benzene and toluene etc. 2. Non-ideal solutions: These are solutions which do not obey Raoult’s law over the entire range of concentration. The vapour pressure of such a solution is either higher or lower than that predicted by Raoult’s law. If it is higher, the solution shows positive deviation from Raoult’s law and if it is lower, it shows negative deviation from Raoult’s law. (i) Solutions which show positive deviation from Raoult’s law: For such solutions, p1 > 𝒑𝟎𝟏. 𝛘𝟏 , p2 > 𝒑𝟎𝟐. 𝛘𝟐 , ∆mixH > 0 and ∆mixV > 0 Here A-B interactions are weaker than A-A and B-B interactions. i.e., in this case solute-solvent interactions are weaker than solute-solute and solvent-solvent interactions. So more molecules are escaped to vapour phase and hence the vapour pressure of the solution increases. E.g. solutions of ethanol and acetone, acetone and CS2, acetone and CCl4 etc. (ii) Solutions which show negative deviation from Raoult’s law: For such solutions, p1 < 𝒑𝟎𝟏 𝛘𝟏 , p2 < 𝒑𝟎𝟐 𝛘𝟐 , ∆mixH < 0 and ∆mixV < 0 Here A-B interactions are stronger than A-A and B-B interactions. i.e. solute-solvent interactions are stronger than solute-solute interaction and solvent-solvent interaction. So number of molecules escaped to vapour phase decreases and hence the vapour pressure of the solution decreases. E.g. solution of phenol and aniline, chloroform and acetone etc. +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 6  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® The plots of vapour pressure curves for ideal solution, solutions which show positive deviation and negative deviation from Raoult’s law are as follows: Solutions showing Positive Solutions showing Negative Ideal Solutions deviation from Raoult’s law deviation from Raoult’s law Azeotropes [Constant Boiling Mixtures] These are binary mixtures having the same composition in liquid and vapour phase and boil at a constant temperature. For such solutions, it is not possible to separate the components by fractional distillation. There are two types of azeotropes: minimum boiling azeotrope and maximum boiling azeotrope. The solutions which show a large positive deviation from Raoult’s law form minimum boiling azeotrope at a particular composition. E.g. 95% aqueous ethanol solution by volume at 78.20C [351.35 K]. The solutions which show large negative deviation from Raoult’s law form maximum boiling azeotrope at a particular composition. E.g. a mixture of 68% Nitric acid and 32% water by mass forms a maximum boiling azeotrope at 393.5 K. Vapour Pressure of Solutions of Solids in Liquids The vapour pressure of a liquid is the pressure exerted by the vapour in equilibrium with its own liquid. If a non-volatile solute is added to a pure solvent, the vapour pressure of the resulting solution is always lower than that of the pure solvent. This is because in a pure solvent, there are only solvent molecules, which can vapourise. But when a non-volatile solute is added to the solvent, a fraction of the surface is occupied by solute molecules. So the number of solute molecules passing to the vapour phase decreases and hence the vapour pressure also decreases. The decrease in the vapour pressure of solvent depends on the quantity of non-volatile solute present in the solution and not on its nature. For such a solution the Raoult’s law can be stated as, for any solution, the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction in solution. Consider a binary solution containing a solvent 1 and solute 2. Since the solute is non-volatile, only the solvent molecules are present in vapour phase and contribute to vapour pressure. Let p1 be the vapour pressure of the solvent, χ1 be its mole fraction, p10 be its vapour pressure in the pure state. Then according to Raoult’s law, p1 ∝ χ1 or, p1 = 𝑝10 χ1 A graph between the vapour pressure and the mole fraction of the solvent is linear as follows: +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 7  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® COLLIGATIVE PROPERTIES AND DETERMINATION OF MOLARMASS The properties which depend only on the relative number of solute particles and not on their nature are called Colligative properties. The important colligative properties are: Relative lowering of Vapour pressure, Elevation of Boiling point, Depression of Freezing point and Osmotic Pressure. 1. Relative lowering of Vapour Pressure When a non-volatile solute is added to a pure solvent, the vapour pressure (V.P) of the resulting solution is lower than that of the pure solvent. The difference between the vapour pressure of pure solvent and that of the solution is called lowering of vapour pressure (∆p). Consider a binary solution containing a non-volatile solute 2 dissolved in a solvent 1. Let 𝑝10 be the vapour pressure of pure solvent 1 and p1 be the vapour pressure of solution [V.P of solution = V.P of the solvent in the solution, since the solute is non-volatile]. Then according to Raoult’s law, p1 = 𝑝10. χ1 The lowering of vapour pressure of the solvent (∆p) = 𝑝10 – p1 = 𝑝10 – 𝑝10 χ1 Or, ∆p = 𝑝10 (1 - χ1) But χ1 + χ2 = 1. Therefore 1-χ1 = χ2 So ∆p = 𝑝10. χ2 ∆p Or, = χ2 , the mole fraction of the solute. 𝑝10 ∆p Where is called relative lowering of vapour pressure. It is defined as the ratio of the lowering of vapour 𝑝10 pressure to the vapour pressure pure solvent. n2 But χ2 = n1 + n2 Where n1 and n2 are the number of moles of solvent and solute respectively. For dilute solutions, n2 1. Thus for NaCl, i =2, for K2SO4, i = 3, for CaCl2, i = 3 and for acetic acid in benzene, i = ½ Inclusion of van’t Hoff factor modifies the equations for colligative properties as follows: ∆𝐩 𝐰 𝐌 1. Relative lowering of vapour pressure, 𝒑𝟎 = 𝑖. 𝐰𝟐 x 𝐌𝟏 𝟏 𝟏 𝟐 2. Elevation of Boiling point (∆Tb) = i.Kb.m 3. Depression of freezing point (∆Tf) = i.Kf.m 4. Osmotic Pressure (π) = i.CRT +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 11  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® 2. ELECTROCHEMISTRY It is a branch of chemistry that deals with the relationship between chemical energy and electrical energy and their inter conversions. ELECTROCHEMICAL CELLS These are devices that convert chemical energy of some redox reactions to electrical energy. They are also called Galvanic cells or Voltaic cells. Example for Galvanic cell is Daniel cell. It is constructed by dipping a Zn rod in ZnSO4 solution and a Cu rod in CuSO4 solution. The two solutions are connected externally by a metallic wire through a voltmeter and a switch and internally by a salt bridge. A salt bridge is a U-tube containing an inert electrolyte like NaNO3 or KNO3 in a gelly like substance. The functions of a salt bridge are: 1. To complete the electrical circute. 2. To maintain the electrical neutrality in the two half cells. The reaction taking place in a Daniel cell is Zn(s) + Cu2+(aq) → Zn2+ (aq) + Cu(s) This reaction is a combination of two half reactions: (i) Cu2+ + 2 e- → Cu(s) (reduction half reaction) (ii) Zn(s) → Zn2+ + 2 e- (oxidation half reaction) These reactions occur in two different portions of the Daniel cell. The reduction half reaction occurs on the copper electrode while the oxidation half reaction occurs on the zinc electrode. These two portions of the cell are also called half-cells or redox couples. The copper electrode may be called the reduction half-cell and the zinc electrode, the oxidation half-cell. Electrode Potential The tendency of a metal to lose or gain electron when it is in contact with its own solution is called electrode potential. When the concentrations of all the species involved in a half-cell is unity then the electrode potential is known as standard electrode potential. According to IUPAC convention, standard reduction potential is taken as the standard electrode potential. In a galvanic cell, the half-cell in which oxidation takes place is called anode and it has a negative potential. The other half-cell in which reduction takes place is called cathode and it has a positive potential. In a cell, the electrons flow from negative electrode to positive electrode and the current flows in the opposite direction. The potential difference between the two electrodes of a galvanic cell is called the cell potential and is measured in volts. The cell potential is the difference between the electrode potentials (reduction potentials) of the cathode and anode. The cell electromotive force (emf) of the cell is the potential difference between the two electrodes, when no current is flow through the cell. It is denoted by Ecell. i.e. Ecell = Ecathode – Eanode By convention, while representing a galvanic cell, the anode is written on the left side and the cathode on the right side. Metal and electrolyte solution are separated by putting a vertical line and a salt bridge is denoted by putting a double vertical line. [The concentration of the electrolyte is shown in simple bracket]. For Daniel cell, the cell representation is: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) Under this convention the emf of the cell is positive and is given by the potential of the half-cell on the right hand side (cathode) minus the potential of the half-cell on the left hand side (anode). i.e. Ecell = Eright – Eleft Or, Ecell = ER – EL +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 12  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® Measurement of Electrode Potential The potential of an individual half-cell cannot be measured. We can measure only the difference between the two half-cell potentials that gives the emf of the cell. For this purpose a half-cell called Standard Hydrogen Electrode (SHE) or Normal Hydrogen Electrode (NHE) is used. It consists of a platinum foil coated with platinum black (finely divided Pt). The electrode is dipped in an acidic solution of one molar concentration and pure hydrogen gas at 1 bar pressure and 298K is bubbled through it. It is represented as Pt(s)|H2(g)|H+(aq). By convention, the electrode potential of SHE is taken as zero. To determine the electrode potential of an electrode, it is connected In series with the standard hydrogen electrode through a volt meter. Here SHE is always connected on LHS and the emf of the resulting cell is determined by the equation, Ecell = ER – EL Since the electrode potential of SHE is zero, the value of Ecell is equal to the electrode potential of the given electrode. If the standard electrode potential of an electrode is greater than zero (i.e. +ve), then its reduced form is more stable compared to hydrogen gas. Similarly, if the standard electrode potential is negative, then hydrogen gas is more stable than the reduced form of the species. Electrochemical series It is a series in which various electrodes are arranged in the decreasing order of their reduction potential. In this table, fluorine is at the top indicating that fluorine gas (F2) has the maximum tendency to get reduced to fluoride ions (F–). Therefore fluorine gas is the strongest oxidising agent and fluoride ion is the weakest reducing agent. Lithium has the lowest electrode potential indicating that lithium ion is the weakest oxidising agent while lithium metal is the most powerful reducing agent in an aqueous solution. Nernst Equation It is an equation relating electrode potential or emf of a cell with electrolytic concentration. (i) Nernst equation relating Electrode potential and Electrolytic concentration: Consider the electrode reaction: Mn+(aq) + ne- –→ M(s) Nernst equation for the electrode potential can be given by: RT [Mn+ ] E(Mn+|M) = E0(Mn+|M) + ln nF [M] Since the concentration of any solid is taken as unity, the above equation becomes: RT E(Mn+| M) = E0(Mn+|M) + nF ln[M n+ ] 2.303 RT Or, E(Mn+| M) = E0(Mn+|M) + log [Mn+] nF 0 2.303 RT or, Eel. = Eel + log [Mn+] nF 0 2.303 RT 1 or, Eel. = Eel – log [Mn+] nF 0 Where Eel is the standard electrode potential of the electrode, R is the universal gas constant (R = 8.314 JK mol ), F is Faraday constant (96500 C/mol), T is temperature in Kelvin and [M n+] is the –1 –1 concentration of the ion Mn+. 0 0.0591 On substituting the values of R and F at 298K, the equation becomes: Eel = Eel + n log [Mn+] +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 13  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® (ii) Nernst equation relating emf of a cell and electrolytic concentration: A Daniel cell can be represented as: Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s) The electrode reactions are: Cu2+ + 2 e- → Cu(s) (cathode reaction) Zn(s) → Zn + 2 e- (anode reaction) 2+ The electrode potentials are: For cathode: RT E(Cu2+|Cu) = E0(Cu2+|Cu) + 2F ln [Cu2+] RT For anode: E(Zn2+|Zn) = E0(Zn2+|Zn) + 2F ln [Zn2+] The cell potential, Ecell = E(Cu2+|Cu) – E(Zn2+|Zn) RT RT = {E0(Cu2+|Cu) + 2F ln [Cu2+]} – {E0(Zn2+|Zn) + 2F ln [Zn2+]} RT [Cu2+ ] = [E0(Cu2+|Cu) – E0(Zn2+|Zn) ] + 2F ln [Zn2+ ] RT [𝑍𝑛2+ ] Or, Ecell = [E0(Cu2+|Cu) – E0(Zn2+|Zn) ] – 2F ln [Cu2+] 0 RT [𝑍𝑛2+ ] 0 Or, Ecell = Ecell – ln [Cu2+] (Since E0(Cu2+|Cu) – E0(Zn2+|Zn) = Ecell ) 2F On changing the base of logarithm, we get 0 2.303 RT [𝑍𝑛2+ ] Ecell = Ecell – log [Cu2+] 2F On substituting the values of R (8.314 JK–1 mol–1), F (96500 C mol–1) at 298K, the above equation becomes, 0 0.0591 [𝑍𝑛2+ ] Ecell = Ecell – log [Cu2+ ] 2 For a general electrochemical reaction of the type: 𝑛𝑒 − a + bB → cC + dD Nernst equation can be written as: 0 0.0591 [𝐶]𝑐 [𝐷]𝑑 Ecell = Ecell – log [𝐴]𝑎[𝐵]𝑏 n (where n is the no. of electrons involved in the cell reaction) Equilibrium Constant from Nernst Equation For a Daniel cell, the emf of the cell at 298K is given by: 0 0.0591 [𝑍𝑛2+ ] Ecell = Ecell – log [Cu2+ ] 2 When the cell reaction attains equilibrium, Ecell = 0 0 0.0591 [𝑍𝑛2+ ] So, 0 = Ecell – log [Cu2+] 2 0 0.0591 [𝑍𝑛2+ ] Or, Ecell = log [Cu2+ ] 2 [𝑍𝑛2+ ] But at equilibrium, [Cu2+ ] = Kc, the equilibrium constant. 0 0.0591 So the above equation becomes, Ecell = log Kc 2 0 2.303 RT In General, Ecell = log Kc nF On substituting the values of R (8.314 JK–1 mol–1), F (96500 C mol–1) at 298K, the above equation becomes, 0 0.0591 Ecell = log Kc 𝑛 +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 14  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® Electrochemical Cell and Gibbs Energy of the Reaction Electrical work done in one second is equal to electrical potential multiplied by total charge passed. Also the reversible work done by a galvanic cell is equal to decrease in its Gibbs energy. Therefore, if the emf of the cell is Ecell and nF is the amount of charge passed, then the Gibbs energy of the reaction, ΔG = – nFEcell 0 If the concentration of all the reacting species is unity, then Ecell = Ecell. 0 0 So, ΔG = – nFEcell 0 Thus, from the measurement of Ecell , we can calculate the standard Gibbs energy of the reaction. By knowing ΔG , we can calculate equilibrium constant by the equation: ΔG0 = – RT ln Kc. 0 Or, ΔG0 = – 2.303RT log Kc Conductance of Electrolytic Solutions Resistance (R): The electrical resistance is the hindrance to the flow of electrons. Its unit is ohm (Ω). The resistance of a conductor is directly proportional to the length of the conductor (Ɩ) and inversely proportional to the area of cross-section (A) of the conductor. i.e. R α Ɩ /A or, R = a constant x Ɩ/A or, R = ρ x Ɩ/A, where ρ (rho) is a constant called resistivity. It is defined as the resistance offered by a conductor having unit length and unit area of cross-section. Its unit is ohm-metre (Ω m) or ohm-centimetre (Ω cm). 1 Ω m = 100 Ω cm, 1 Ω cm = 10-2 Ω m Conductance (G): It is the reciprocal of resistance. 1 i.e. Conductance, G = Resistance (R) Its unit is ohm-1 or mho or siemens (S) 1 A Or, G = 𝜌 x Ɩ A 1 Or, G = ƙ x [where ƙ (kappa) = , the conductivity. Ɩ 𝜌 It is defined as the conductance of a conductor having unit length and unit area of cross-section. Its unit is ohm-1 m-1 or mho m-1 or S m-1. 1 S cm-1 = 100 S m-1 1 S m-1 = 10-2 S cm-1 There are two types of conductance - electronic or metallic conductance and electrolytic or ionic conductance. Electrical conductance through metals is called metallic or electronic conductance and it is due to the movement of electrons. It depends on the nature and structure of the metal, the no. of valence electrons per atoms and temperature. For electronic conductance, when temperature increases, conduction decreases. The conductance of electricity by ions present in solutions is called electrolytic or ionic conductance. It depends on i) the nature of electrolyte ii) size of the ion produced and their solvation iii) the nature of the solvent and its viscosity iv) concentration of the electrolyte and v) temperature (As temperature increases electrolytic conduction also increases). Note: Substances which allow the passage of electricity in molten state or in solution state are called electrolytes. On the passage of electricity, they undergo chemical decomposition. Measurement of the conductivity of ionic solutions A We know that, conductance, G = ƙ x Ɩ Ɩ So conductivity, ƙ = G x A The quantity Ɩ/A is called cell constant (G*). It depends on the distance between the electrodes and their area of cross-section. Its unit is m-1. i.e. conductivity = conductance x cell constant +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 15  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® So, in order to determine the conductivity of an electrolytic solution, first determine the resistance by using a Wheatstone bridge. It consists of two resistances R3 and R4, a variable resistance R1 and the conductivity cell having the unknown resistance R2. It is connected to an AC source (an oscillator, O) and a suitable detector (a headphone or other electronic device, P). Direct current (DC) cannot be used since it causes the decomposition of the solution. The bridge is balanced, when no current passes through the detector. R R Under this condition, R2 = R4 1 3 R1 R4 Therefore, the unknown resistance, R2 = R3 By knowing the resistance, we get the value of conductance and conductivity. Conductivity cell It consists of two platinum electrodes coated with platinum black (finely powdered platinum). The electrodes are separated by a distance Ɩ and their area of cross-section is A. Two types of conductivity cells The solution confined between the electrodes can be considered as a column of length l and area of cross section A. The cell constant of a conductivity cell is usually determined by measuring the resistance of the cell containing a solution whose conductivity is already known (e.g. KCl solution). Molar conductivity (Λm – lambda m): It is the conductivity of a solution containing 1 mole of an electrolyte, placed in a conductivity cell with unit distance between the electrodes. OR, Molar conductivity of a solution at a given concentration is the conductance of ‘V’ volume of a solution containing one mole of electrolyte kept between two electrodes with area of cross section A and distance of unit length. It is related to conductivity of the solution by the equation, +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 16  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® ƙ Λm = (where C is the concentration of the solution) C 1000 ƙ Or, Λm = (where M is the molarity of the solution) M The unit of molar conductivity is Ω-1cm2 mol-1 or S cm2 mol-1. 1S m2 mol-1 = 104 S cm2 mol-1 1S cm2 mol-1 = 10– 4 S m2 mol-1 Variation of conductivity and Molar conductivity with concentration (dilution) Both conductivity and molar conductivity change with the concentration of the electrolyte. We know that when a solution is diluted, its concentration decreases. For both strong and weak electrolytes, conductivity always decreases with dilution. This is because conductivity is the conductance of unit volume of electrolytic solution. As dilution increases, the number of ions per unit volume decreases and hence the conductivity decreases. For both strong and weak electrolytes, the molar conductivity increases with dilution (or decreases with increase in concentration), but due to different reasons. For strong electrolytes, as dilution increases, the force of attraction between the ions decreases and hence the ionic mobility increases. So, molar conductivity increases. When dilution reaches maximum or concentration approaches zero, the molar conductivity becomes maximum and it is called the limiting molar conductivity (Λ0𝑚 ). For strong electrolytes, the relation between Λm and concentration can be given as: Λm = 𝚲0𝑚 - A√𝒄 Where ‘c’ is the concentration and A is a constant depending on temperature, the nature of the electrolyte and the nature of the solvent. All electrolytes of a particular type have the same value for ‘A’. For weak electrolytes, as dilution increases, the degree of dissociation increases. So the number of ions and hence the molar conductivity increases. The variation of Λm for strong and weak electrolytes is shown in the following graphs: Weak Electrolyte Λm Λ0m Strong Electrolyte √𝑐 For strong electrolytes, the value of Λ0𝑚 can be determined by the extending the graph to y-axis. But for weak electrolytes, it is not possible, since the graph is not a straight line. So their Λ0𝑚 values are calculated by applying Kohlrausch’s law of independent migration of ions. Kohlrausch’s law of independent migration of ions The law states that the limiting molar conductivity of an electrolyte is equal to the sum of the individual contributions of the anion and the cation of the electrolyte. Thus if an electrolyte on dissociation gives n(+) cations and n(-) anions, its limiting molar conductivity is given as: Λ0m = n(+)λ0(+) + n(-)λ0(-) For an electrolyte like AxBy the dissociation can be denoted as: AxBy → xAy+ + yBx- Λ0m(AxBy) = x.λ0(Ay+) + y.λ0(Bx-) +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 17  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® For NaCl, Λ0m (NaCl) = λ0(Na+) + λ0(Cl-) For CaCl2, Λ0m (CaCl2) = λ0(Ca2+) + 2 x λ0(Cl-) Applications of Kohlrausch’s law 1) Determination of λ0m of weak electrolytes By knowing the Λ0𝑚 values of strong electrolytes, we can calculate Λ0𝑚 of weak electrolytes. For e.g. we can determine the Λ0𝑚 of acetic acid (CH3COOH) by knowing the Λ0𝑚 of CH3COONa, NaCl and HCl as follows: Λ0𝑚 (CH3COONa) = λ0CH3COO– + λ0Na+ …………. (1) Λ0𝑚 (HCl) = λ0H+ + λ0 Cl– …………….. (2) Λ0𝑚 (NaCl) = λ0Na+ + λ0Cl– ………….. (3) (1) + (2) – (3) gives: Λ0𝑚 (CH3COONa) + Λ0𝑚 (HCl) – Λ0𝑚 (NaCl) = λ0CH3COO–+ λ0Na+ + λ0H+ + λ0Cl– – λ0Na+ – λ0Cl– = Λ0m(CH3COOH) 2) Determination of degree of dissociation of weak electrolytes By knowing the molar conductivity at a particular concentration (Λ𝑐𝑚 ) and limiting molar conductivity Λ𝑐𝑚 (Λ0𝑚 ), we can calculate the degree of dissociation (α) as, α = Λ0𝑚 cα2 By using α, we can calculate the dissociation constant of weak acid (Ka) as: Ka = 1−α Electrolytic Cells and Electrolysis In an electrolytic cell, the electrical energy is converted to chemical energy. The dissociation of an electrolyte by the passage of electricity is called electrolysis. For e.g. when CuSO4 solution is electrolysed by Cu electrodes, Cu is deposited at the cathode and Cu2+ ions are liberated from the anode. Quantitative Aspects of electrolysis – Faraday’s laws 1) Faraday’s first law It states that the amount of substance deposited or liberated at the electrodes (m) is directly proportional to the quantity of electricity (Q) flowing through the electrolyte. Mathematically, mαQ Or, m = zQ Equivalent weight Where z is a constant called electrochemical equivalent (ECE). Z = 96500 Atomic weight Equivalent weight of an ion = valency But quantity of electricity is the product of current in ampere (I) and time in second (t). i.e. Q = It Therefore, m= zIt 1 Faraday is the charge of 1 mole of electron or it is the amount of electricity required to deposit one gram equivalent of any substance. Its value is 96487 C mol-1 or, 96500 C mol-1. For the deposition of 1 mole of Na, the amount of charge required = 1 F (Since Na + + e- → Na) For Ca, Q = 2F (since Ca2+ + 2e- → Ca) 2) Faraday’s second law It states that when the same quantity of electricity is passed through solutions of different substances, the amount of substance deposited or liberated is directly proportional to their chemical equivalence. +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 18  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® For e.g. when same quantity of electricity is passed through solutions of two electrolytes A and B, then Mass of A deposited Equivalent mass of A = Mass of B deposited Equivalent mass of B Products of electrolysis The products of electrolysis depend on the following factors: i) The nature of the electrolyte: The electrolyte may be in molten state or in aqueous solution state. For e.g. if molten NaCl is electrolysed, Na is deposited at the cathode and chlorine is liberated at the anode. NaCl → Na+ + Cl- At cathode: Na+ + e- → Na At anode: Cl- → ½ Cl2 + e- If NaCl solution is electrolysed, we get H2 gas at the cathode and Cl2 gas at the anode. NaCl solution contains 4 ions – Na+, Cl-, H+ and OH- Cathode reaction: H+ + e- → ½ H2 Anode reaction: Cl- → ½ Cl2 + e- NaOH is formed in the solution. ii) The type of electrodes used: If the electrode is inert (e.g. Pt, gold, graphite etc.), it does not participate in the electrode reaction. While if the electrode is reactive, it also participates in the electrode reaction. iii) The different oxidising and reducing species present in the electrolytic cell and their standard electrode potentials. Some of the electrochemical processes are very slow and they do not take place at lower voltages. So some extra potential (called overpotential) has to be applied, which makes such process more difficult to occur. For e.g. during the electrolysis of NaCl solution, the possible reactions at anode are: Cl– (aq) → ½ Cl2 (g) + e–; E0cell = 1.36 V 2H2O (l) → O2 (g) + 4H+(aq) + 4e–; E0cell = 1.23 V At anode, the reaction with lower value of E0cell is preferred and so water should get oxidised in preference to Cl– (aq). However, on account of overpotential of oxygen, the first reaction is preferred and hence Cl2 is formed at anode. When dilute H2SO4 is electrolysed, O2 gas is liberated from the anode. 2H2O(l) → O2 (g) + 4H+ (aq) + 4e– while if we use conc. H2SO4, peroxodisulphate ion is formed at the anode. 2SO42– (aq) → S2O82– (aq) + 2e– In both cases H2 gas is liberated from the cathode. Batteries A battery is basically a galvanic cell in which the chemical energy of a redox reaction is converted to electrical energy. They are of mainly 2 types – primary batteries and secondary batteries. a) Primary cells: These are cells which cannot be recharged or reused. Here the reaction occurs only once and after use over a period of time, they become dead. E.g. Dry cell, mercury button cell etc. 1. Dry Cell It is a compact form of Leclanche cell. It consists of a zinc container as anode and a carbon (graphite) rod surrounded by powdered manganese dioxide (MnO2) and carbon as cathode. The space between the +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 19  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® electrodes is filled by a moist paste of ammonium chloride (NH4Cl) and zinc chloride (ZnCl2). The electrode reactions are: Anode: Zn(s) → Zn2+ + 2e– Cathode: MnO2+ NH4++ e– → MnO(OH) + NH3 Ammonia produced in this reaction forms a complex with Zn 2+ and thus corrodes the cell. The cell has a potential of nearly 1.5 V. 2. Mercury cell Here the anode is zinc – mercury amalgam and cathode is a paste of HgO and carbon. The electrolyte is a paste of KOH and ZnO. The electrode reactions are: Anode reaction: Zn(Hg) + 2OH– → ZnO(s) + H2O + 2e– Cathode reaction: HgO + H2O + 2e– → Hg(l ) + 2OH– The overall reaction is : Zn(Hg) + HgO(s) → ZnO(s) + Hg(l ) The cell has a constant potential of 1.35 V, since the overall reaction does not involve any ion in solution. b) Secondary cells A secondary cell can be recharged and reused again and again. Here the cell reaction can be reversed by passing current through it in the opposite direction. E.g.: Lead storage cell, Ni – Cd cell (Nicad cell). Lead storage cell: It is used in automobiles and invertors. It consists of lead as anode and a grid of lead packed with lead dioxide (PbO2) as the cathode. The electrolyte is 38% H2SO4 solution. The cell reactions are: Anode: Pb(s) + SO42- (aq) → PbSO4(s) + 2e- Cathode: PbO2(s) + SO42-(aq) + 4H+(aq) + 2e- → PbSO4 (s) + 2H2O (l ) The overall cell reaction is: Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l) On charging the battery, the reaction is reversed and PbSO4(s) on anode and cathode is converted into Pb and PbO2, respectively. Another example for a secondary cell is nickel – cadmium cell. Here the overall cell reaction is: Cd (s)+2Ni(OH)3 (s) → CdO (s) +2Ni(OH)2 (s) +H2O(l ) Differences between Primary cell and Secondary cell Primary cell Secondary cell Cannot be recharged or reused. Can be recharged and reused. The cell reaction cannot be reversed. The cell reaction can be reversed. E.g. Dry cell, Mercury cell E.g.: Lead storage cell, Ni-Cd cell Fuel cells These are galvanic cells which convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy. One example for fuel cell is Hydrogen – Oxygen fuel cell, which is used in the Apollo space program. Here hydrogen and oxygen are bubbled through porous carbon electrodes into concentrated aqueous sodium hydroxide solution. To increase the rate of electrode reactions, catalysts like finely divided platinum or palladium metal are filled into the electrodes. The electrode reactions are: Cathode: O2(g) + 2H2O(l) + 4e–→ 4OH–(aq) Anode: 2H2(g) + 4OH–(aq) → 4H2O(l) + 4e– Overall reaction is: 2H2(g) + O2(g) → 2 H2O(l) +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 20  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® Advantages of Fuel cells 1. The cell works continuously as long as the reactants are supplied. 2. It has higher efficiency as compared to other conventional cells. 3. It is eco-friendly (i.e. pollution free) since water is the only product formed. 4. Water obtained from H2 – O2 fuel cell can be used for drinking. Corrosion: It is the process of formation of oxide or other compounds of a metal on its surface by the action of air, water-vapour, CO2 etc. Some common examples are: The rusting of iron, tarnishing of silver, formation of green coating on copper and bronze (verdigris) etc. Most familiar example for corrosion is rusting of iron. Chemistry of rusting of iron: Rusting of iron is a redox reaction, which occurs in presence of air and water. At a particular spot of the metal iron, oxidation takes place and that spot behaves as anode. Here Fe is oxidized to Fe 2+. 2 Fe (s) → 2 Fe2+ + 4 e– Electrons released at anodic spot move through the metal and go to another spot on the metal and reduce oxygen in presence of H+ [H+ is formed by dissolving atmospheric CO2 in moisture]. This spot behaves as cathode. The reaction taking place at the cathodic spot is: O2(g) + 4 H+(aq) + 4 e– → 2 H2O (l ) The overall reaction is: 2Fe(s)+O2(g) + 4H+(aq) → 2Fe2 +(aq)+ 2 H2O (l ) The ferrous ions (Fe2+) are further oxidised to ferric ions (Fe3+) and finally to hydrated ferric oxide (Fe2O3. x H2O), which is called rust. Methods used to prevent corrosion 1. By coating the metal surface with paint, varnish etc. 2. By coating the metal surface with another electropositive metal like zinc, magnesium etc. The coating of metal with zinc is called galvanisation and the resulting iron is called galvanized iron. 3. By coating with anti-rust solution. 4. An electrochemical method used is connecting the iron object with a sacrificial electrode of another metal (like Mg, Zn, etc.), which corrodes itself but saves the object (sacrificial protection). ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 21  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® 3. CHEMICAL KINETICS The branch of Chemistry that deals with the rate of chemical reaction, factors affecting the rate and the mechanism of a reaction is called Chemical Kinetics. Rate of a chemical reaction The rate of a chemical reaction is the change in concentration of any one of the reactants or products in unit time. Or, it is the rate of decrease in concentration of any one of the reactants or the rate of increase in concentration of any one of the products. For a chemical reaction, if ∆x is the change in concentration of reactant or product in time interval ∆t, ∆x the rate of reaction (r) = ∆t. For a hypothetical reaction R → P, the rate of reaction can be expressed by rate of decrease in concentration of reactant or by rate of increase in concentration of product. i.e. Rate of reaction (r) = Rate of disappearance of R Decrease in concentration of R Or, r = Time taken ∆[R] Or, r = – ∆t [Since, concentration of reactant decreases with time, ∆[R] is negative. But rate of reaction cannot be negative. So in order to make it positive, it is multiplied with –1]. Also, rate of reaction = Rate of appearance (formation) of P Increase in concentration of P i.e. r = Time taken ∆[P] Or, r = ∆t The above rate expressions are also called Average rate of reaction. For a gaseous reaction, at constant temperature, concentration is directly proportional to the partial pressure. Hence the rate of reaction can also be expressed as the rate of change in partial pressure of the reactant or product. Units of rate of reaction If concentration is expressed in mol L-1 and time is in second, the unit of rate of reaction is mol L-1 s-1. In general the dimension of rate of reaction is concentration time-1. For gaseous reaction, the concentration is replaced by partial pressure and so the unit of rate of reaction is atm s-1. Instantaneous rate of a reaction The rate of a reaction at a particular instant of time is called Instantaneous rate of a reaction. Or, it is the rate of a reaction when the time interval approaches zero (i.e. for the smallest time interval). ∆x dx i.e., Instantaneous rate of a reaction = lim = dt ∆t→0 ∆t d[R] d[P] For the reaction, R → P, the instantaneous rate is given by, rinst = – = dt dt 1 d[𝑁2 𝑂5 ] 1 d[N𝑂2 ] d[𝑂2 ] For the reaction, 2N2O5 → 4NO2 + O2, rinst = – 2 = = dt 4 dt dt For the reaction, 5 Br– (aq) + BrO3 – (aq) + 6 H+ (aq) → 3 Br2 (aq) + 3 H2O (l) 1 d[Br− ] d[BrO− 3] 1 d[[H]+ ] 1 d[Br2 ] 1 d[H2 O] rinst = – 5 = − = −6 = = dt dt dt 3 dt 3 dt For a general reaction aA + bB → cC + dD, the instantaneous rate is given by: 1 d[A] 1 d[B] 1 d[C] 1 d[D] rinst = − a = −b = = dt dt c dt d dt +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 22  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® Determination of Instantaneous rate of a reaction Instantaneous rate of a reaction can be determined graphically. For this first conduct the chemical reaction and find out the concentration of any one of the reactants or products at a regular interval of time. Then plot a graph between concentration along y-axis and time along x-axis. In order to determine the instantaneous rate at a particular time, mark the point on the graph at that time and draw a tangent at the point. The slope of this tangent gives the instantaneous rate at that time. Concentration Factors affecting rate of a reaction The important factors which affect the rate of a chemical reaction are: 1. Nature of the reactants 2. Concentration of the reactants 3. Temperature 4. Pressure (for gaseous reaction only) 5. Catalyst 6. Radiation or light Dependence of Rate of reaction on concentration It is found experimentally that rate of reaction is directly proportional to the concentration of reactants. Thus for a general reaction, aA + bB → cC + dD, Rate α [A]x[B]y Or, r = k [A]x[B]y (where x and y may or may not be equal to the stoichiometric coefficients a and b) This expression is known as rate law or rate equation. Thus rate law is a “mathematical equation relating the rate of a chemical reaction and concentration of reactants, in which each concentration term is raised to a power which may or may not be equal to the stoichiometric coefficients in the balanced chemical equation.” In the above equation ‘k’ is a constant called rate constant. It is defined as the rate of the reaction when the concentration of each of the reactants is unity. Order of reaction (n) Order of a reaction is the sum of the powers of the concentration terms of the reactants in the rate law. It is an experimental quantity. It can have the values 0,1,2,3,…… or a fraction. It is applicable to both elementary and complex reactions. For a general reaction, aA + bB → cC + dD; r = k[A]x[B]y Here ‘x’ is the order with respect to the reactant A, ‘y’ is the order with respect to B and (x + y) is the overall order of the reaction. If the order of a reaction is zero, it is called zero order reaction. E.g.: Decomposition of ammonia at the surface of platinum at high pressure. +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 23  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® If the order of a reaction is one, it is called first order reaction. E.g. Decomposition of hydrogen peroxide. If the order of a reaction is two, it is called second order reaction and so on. E.g. for fractional order reactions are: Decomposition of acetaldehyde: CH3-CHO → CH4 + CO; r = k[CH3-CHO]3/2 Chlorination of chloroform: CHCl3 + Cl2 → CCl4 + HCl; r = k[CHCl3][Cl2]1/2 Molecularity of a reaction It is the total number of reacting species collides simultaneously in a chemical reaction. It is a theoretical quantity. It can have values 1,2,3 etc. It cannot be zero or fractional. It is applicable only to elementary reactions. If the molecularity of a reaction is 1, it is called unimolecular reaction. e.g. Decomposition of ammonium nitrite NH4NO2 → N2 + 2H2O If the molecularity of a reaction is 2, it is called bimolecular reaction. e.g. Decomposition of Hydrogen Iodide 2HI → H2 +I2 Differences between Order and Molecularity of a reaction Order Molecularity 1. It is the sum of the powers of the It is the total number of reactant concentration terms in the rate law species collide simultaneously in a expression chemical reaction 2. It is an experimental quantity It is a theoretical quantity 3. It can be zero or fractional It cannot be zero or fractional 4. It is applicable to both elementary and It is applicable only to elementary complex reactions reactions Elementary and complex reactions A reaction that takes place in a single step is called elementary reaction. While a reaction that occurs in more than one steps is called a complex reaction. In a complex reaction one of the steps is slower than the other steps. The overall rate of the reaction is controlled by this step and this step is called the rate determining step (rds). The sequence of steps by which a chemical reaction occurs is called the mechanism of the reaction. Consider the decomposition of hydrogen peroxide which is catalysed by iodide ion in alkaline medium. 2H2O2 I– 2H2O + O2 Alkaline Medium d[H2 O2 ] The rate equation for this reaction is found to be r = – = k[H2O2][I –] dt This reaction is first order with respect to both H2O2 and I–. The reaction takes place in two steps and its mechanism can be written as: 𝑆𝑙𝑜𝑤/𝑟𝑑𝑠 (1) H2O2 + I– → H2O + IO– 𝑓𝑎𝑠𝑡 (2) H2O2 + IO– → H2O + I– + O2 Here the first step is slower than the second. So it is the rate determining step (rds). +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 24  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® Unit of Rate constant Different ordered reactions have different units for k. For an nth order reaction n A → products, rate = k[A]n rate Therefore, k = [A]n mol L−1 s−1 Unit of k = = (mol L-1)1-n s-1 = mol1-n Ln-1 s-1 (mol L−1 )n The units of rate constant for different ordered reactions are: Reaction Order (n) Unit of rate constant Zero order reaction 0 mol L-1s-1 First order reaction 1 s-1 Second order reaction 2 mol-1 L s-1 Integrated Rate Equations These are equations relating the rate of a reaction and concentration of reactants. Different ordered reactions have different integrated rate law equations. 1. Integrated Rate Equation for a Zero Order Reaction Zero order reactions are reactions in which the rate of reaction is independent of concentration of the reactants. Consider a zero order R→ P d[R] The rate expression for the above reaction is r = – dt ………………. (1) Rate law for the above reaction is r= k[R]0 = k ………………… (2) [Since [R]0 = 1] d[R] From equations (1) & (2), we can write – =k dt This equation is known as differential rate equation for a zero order reaction. Or, d[R] = – k.dt On integrating the above equation we get, [R] = –kt +C …………………. (3) Where C is the constant of integration. To evaluate C, consider the initial conditions. i.e., when t=0, [R] = [R] 0 Substitute these values in eqn. (3) [R]0 = – k x 0 + C Or, C = [R]0 Substituting ‘C’ in equation (3), we get [R] = – kt + [R]0 …………. (4) Or, kt = [R]0 – [R] [𝑅] − [𝑅] Or, k= 0 𝑡 ⬚ ………..… (5) Equation (4) is of the form y = mx + c, equation for a straight line, So if we plot [R] against t, we get a straight line with slope = –k and intercept equal to [R]0. +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 25  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® Example for zero order reaction is the decomposition of gaseous ammonia on a hot platinum surface at high pressure. 𝑃𝑡/ℎ𝑖𝑔ℎ 𝑃 2 NH3(g) → N2(g) + 3 H2(g) ; r = k[NH3]0 In this reaction, platinum metal acts as a catalyst. At high pressure, the metal surface gets saturated with gas molecules. So, a further change in pressure does not change the rate of the reaction. Hence it becomes a zero order reaction. Another e.g. is the thermal decomposition of HI on gold surface. 𝐴𝑢/ℎ𝑖𝑔ℎ 𝑃 2HI(g) → H2(g) + I2(g); r= k[HI]0 2. Integrated Rate Equation for a First order reaction For a first order reaction, the rate of the reaction is proportional to the first power of the concentration of the reactant. Consider a first order reaction, R → P d[R] The rate expression for the above reaction is r = – dt ………………. (1) Rate law for the above reaction is r = k[R] ………………… (2) d[R] From equations (1) & (2), we can write – dt = k[R] This equation is known as differential rate equation for a 1st order reaction. d[R] Or, – [R] = k.dt On integrating the above equation we get, – ln[R] = kt + C …………………. (3) Where ‘C’ is the constant of integration. To evaluate C, consider the initial conditions. i.e., when t = 0, [R] = [R]0 Substitute these values in eqn. (3) – ln[R]0 = k x 0 +C Or, C = – ln[R]0 Substituting the value of ‘C’ in equation (3), we get – ln[R] = kt – ln[R]0 …………. (4) Or, ln[R]0 – ln[R] = kt [R] Or, ln [R]0 = kt …………….. (5) 1 [R]0 Or, k = ln ……………. (6) 𝑡 [R] 𝟐.𝟑𝟎𝟑 [𝐑]𝟎 On changing the base of logarithm, we get: k = log ………………… (7) 𝐭 [𝐑] This equation is known as integrated rate equation for a first order reaction. From eqn. (4), ln[R] = – kt + ln[R]0 This equation is of the form y = mx + c, equation for a straight line. So if we plot a graph between ln[R] against time, t we get straight line graph as follows: +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 26  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® [R]0 From the eqn. (5) ln = kt [R] [R]0 Or, 2.303 log = kt [R] [R]0 kt Or, log = [R] 2.303 [R]0 So, if we plot a graph between log against ‘t’, we get a straight line graph as shown below: [R] Examples for 1st order reactions are: 1. Hydrogenation of ethene: C2H4(g) + H2(g) → C2H6(g); r = k[C2H4] 2. All natural and artificial radioactive decay. 3. Decomposition of N2O5 and N2O N2O5 → 2NO2 + ½ O2; r = k [N2O5] Half life of a reaction (t1/2) The half-life of a reaction is the time taken for the concentration of a reactant is reduced to one half of its initial concentration. It is represented as t1/2. 1. Half-life of a zero order reaction [R]0 − [R] For a zero order reaction, the integrated rate law is: k = t When t = t1/2, [R] = ½ [R]0 On substituting these values in the above equation, [R]0 − [R]0 ⁄2 k= t1⁄ 2 [𝑹]𝟎 or, t1/2 = 𝟐𝒌 i.e. half-life of a zero order reaction is directly proportional to the initial concentration of the reactants and inversely proportional to the rate constant. 2. Half-life of a first order reaction 𝟐.𝟑𝟎𝟑 [𝐑] For a first order reaction, k = 𝐭 log [𝐑]𝟎 When t = t1/2, [R] = [R]0/2 Substitute these values in the above equation: 𝟐.𝟑𝟎𝟑 [𝐑] k= log [𝑹]𝟎𝟎 𝒕𝟏⁄ ⁄ 𝟐 𝟐 2.303 2.303 𝑥 0.3010 Or, t1/2 = log 2 = k 𝑘 𝟎.𝟔𝟗𝟑 Or, t1/2 = 𝒌 +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 27  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® Thus for a first order reaction, half-life period is independent of initial concentration of the reactants. Or, it is a constant. Pseudo first order reaction These are reactions which appear to follow higher order but actually follow first order kinetics. In these reactions the concentration of one of the reactants is large excess and so its change in concentration does not affect the rate of the reaction. E.g. 1. Hydrolysis of ester (ethyl acetate) H+ CH3COOC2H5 + H2O → CH3COOH + C2H5OH; r = k [CH3COOC2H5] Here the concentration of water does not change during the reaction. 2. Inversion of cane sugar H+ C12H22O11 + H2O → C6H12O6 + C6H12O6; r = k [C12H22O11] Cane sugar Glucose Fructose Rate of reaction and Temperature Most of the chemical reactions are accelerated by increase in temperature. It has been found that for a chemical reaction, when the temperature is increased by 10°, the rate of the reaction and the rate constant is nearly doubled. The ratio of the rate constants of a reaction at two temperatures differing by 100 is called temperature coefficient. Rate constant of the reaction at (T + 10)K i.e., Temperature coefficient = Rate constant of the reaction at T K The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation. The equation is: −𝑬𝒂⁄ k = A. 𝒆 𝑹𝑻 Where A is a constant called the Arrhenius parameter or the frequency factor or the pre- exponential factor. It is constant for a particular reaction. R is the universal gas constant, T is the absolute temperature and Ea is activation energy measured in joules/mole (J mol –1). [The above equation was first proposed by J H van’t Hoff, but its physical justification and interpretation was given by Arrhenius.] According to Arrhenius, a chemical reaction occurs by the collision of reactant molecules. All the molecular collisions are not effective. For effective collision, the colliding molecules should have a minimum kinetic energy called activation energy. When such molecules collide, an unstable intermediate called activated complex is formed, which decomposes to form products. For e.g. the reaction between H2 and I2 to form HI takes place in the following steps. H2 + I2 → 2HI If we plot a graph between potential energy and the progress of reaction, we get the following graph. +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 28  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® [Reaction coordinate represents the progress of reaction] In a reaction, all the molecules in the reacting species do not have the same kinetic energy. Ludwig Boltzmann and James Clark Maxwell calculated the distribution of kinetic energy among molecules. They plotted a graph between the fraction of molecules with a given kinetic energy against kinetic energy. This plot is known as Maxwell- Boltzmann distribution curve of molecular energies. Maxwell-Boltzmann Distribution curve Distribution curves at 2 different temperatures The peak of the curve corresponds to the most probable kinetic energy. It is the kinetic energy possessed by maximum fraction of molecules. When the temperature is raised, the maximum of the curve moves to the higher energy value and the curve spreads to the right. That is the fraction of molecules with activation energy increases (almost doubled). At normal temperature the fraction of molecules having energy equal to or greater than activation energy is very low. As the temperature increases, this fraction increases and hence the rate of reaction also increases. In the Arrhenius equation the factor e– Ea/RT corresponds to the fraction of molecules that have kinetic energy greater than Ea. 𝑬𝒂 The logarithmic form of Arrhenius equation is: ln k = – + ln A 𝑹𝑻 This equation is of the form y = mx + c, equation for a straight line. So if we plot a graph between ln k against 1/T, we get a straight line graph as follows: 𝐸 The slope of the graph = – 𝑎⁄𝑅 and the y-intercept = ln A. +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 29  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® Calculation of Activation energy by knowing the rate constants at two different temperatures Consider a reaction R → P. If k1 is the rate constant of the reaction at temperature T1 and k2 is the rate constant at temperature T2, then: −E ln k1 = RTa + ln A ………………. (1) and 1 −E ln k2 = RTa + ln A ……………… (2) 2 Equation (2) – (1) gives, −𝐸𝑎 −𝐸𝑎 ln k2 – ln k1 = [ + 𝑙𝑛𝐴] – [ + 𝑙𝑛𝐴] 𝑅𝑇2 𝑅𝑇1 𝑘2 𝐸𝑎 𝐸𝑎 Or, ln =– + 𝑘1 𝑅𝑇2 𝑅𝑇1 𝑘2 𝐸𝑎 1 1 ln = [ − ] 𝑘1 𝑅 𝑇1 𝑇2 On changing the base of logarithm, we get: 𝑘2 𝐸𝑎 1 1 2.303 log = [ − ] 𝑘1 𝑅 𝑇1 𝑇2 𝑘2 𝐸𝑎 1 1 Or, log = [ − ] 𝑘1 2.303 𝑅 𝑇1 𝑇2 From this equation we can calculate the value of activation energy (E a), by knowing all other values. Effect of Catalyst A catalyst is a substance that increases the rate of a reaction without itself undergoing any permanent chemical change. The action of the catalyst can be explained by intermediate complex theory. According to this theory, a catalyst combines with reactant to form an unstable intermediate complex, which decomposes to form products and the catalyst. A catalyst increases the rate of a chemical reaction by providing an alternate path with low activation energy between reactants and products. [A substance that decreases the rate of a reaction is called inhibitor]. The energy profile diagram for a reaction with catalyst and without catalyst is shown below: +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 30  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® The important characteristics of a catalyst are: 1. A small amount of the catalyst can catalyse a large amount of reactants. 2. A catalyst does not change the Gibbs energy of a reaction (ΔrG). It catalyses the spontaneous reactions but does not catalyse non-spontaneous reactions. 3. A catalyst does not change the equilibrium constant of a reaction, but it helps to attain the equilibrium faster by increasing the rate of both forward as well as the backward reactions. Collision Theory This theory was developed by Max Trautz and William Lewis. It is based on kinetic theory of gases. According to this theory, the reactant molecules are assumed to be hard spheres and reaction is occurred when molecules collide with each other. The number of collisions per second per unit volume of the reaction mixture is known as collision frequency (Z). Another factor which affects the rate of chemical reactions is activation energy. For a bimolecular elementary reaction A + B → Products Rate of reaction can be expressed as −𝑬𝒂⁄ Rate (r) = ZAB 𝒆 𝑹𝑻 −𝑬𝒂 Where ZAB represents the collision frequency of reactants, A and B and 𝒆 ⁄𝑹𝑻 represents the fraction of molecules with energies equal to or greater than Ea. Comparing with Arrhenius equation, we can see that A is related to collision frequency. A third factor which affects the rate of a chemical reaction is the proper orientation. The proper orientation of reactant molecules lead to bond formation. While, improper orientation makes them bounce back and no products are formed. To account for this, probability or steric factor (P) is introduced. So the above equation becomes: −𝑬𝒂 Rate (r) = P.ZAB. 𝒆 ⁄𝑹𝑻 Thus, in collision theory activation energy and proper orientation of the molecules together determine the criteria for an effective collision and hence the rate of a chemical reaction. +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 31  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® 4. The d and f Block Elements Elements from 3rd group to 12th group in the Modern Periodic table are called d-block elements. In these elements their last electron enters in the penultimate d- sub shell. They are placed in between s-block and p-block elements. They show a regular transition from the highly electropositive metals of s- block elements to the less electropositive p-block elements. So they are called transition elements. Transition elements can be defined as elements which contain partially filled d orbitals in their atomic state or in any of their oxidation state. The elements Zn, Cd and Hg contain only completely filled d-orbitals. So they are not regarded as transition elements. Or, they are called pseudo transition elements. There are four series of transition elements. 1) 3d series [from Scandium (Sc, z = 21) to zinc (Zn, z = 30)] 2) 4d series [from Yttrium(Y, z = 39) to cadmium, (Cd, z = 48)] 3) 5d series [from Lanthanum, (La, z =57), Hafnium, (Hf, z = 72) to Mercury, (Hg, z=80)] 4) 6d series [from Actinium, (Ac, z=89), Rutherfordium, (Rf, z=104) to copernicium, (Cn, z=112)] Electronic Configuration General outer electronic configuration of d-block elements is (n-1) d1-10 ns1-2. There is only a small difference in energy between (n-1)d orbital and ns orbital. So in some cases ns electrons are also transferred to (n-1)d level. The electronic configurations of Cr and Cu in the 3d series show some exceptions. 5 1 24Cr – [Ar] 3d 4s 10 1 29Cu – [Ar] 3d 4s This is due to the extra stability of half-filled (d5) and completely-filled electronic configurations (d10). The electronic configurations of Zn, Cd and Hg are represented by the general formula (n-1)d10 ns2. The orbitals in these elements are completely filled in the ground state as well as in their common oxidation states. So they are not regarded as transition elements. Electronic configuration of first row transition series elements: Atomic number Element Symbol Electronic configuration 21 Scandium Sc [Ar] 3d1 4s2 22 Titanium Ti [Ar] 3d2 4s2 23 Vanadium V [Ar] 3d3 4s2 24 Chromium Cr [Ar] 3d5 4s1 25 Manganese Mn [Ar] 3d5 4s2 26 Iron Fe [Ar] 3d6 4s2 27 Cobalt Co [Ar] 3d7 4s2 28 Nickel Ni [Ar] 3d8 4s2 29 Copper Cu [Ar] 3d10 4s1 30 Zinc Zn [Ar] 3d10 4s2 +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 32  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® General characteristics of transition elements 1. Atomic and ionic radii In a transition series, the atomic and ionic radii first decrease, then become constant and increases towards the end of the series. This is because in transition elements, the new electron enters in a penultimate d-orbital. Initially since there is a few numbers of d electrons, the shielding effect is very poor. As the atomic number increases, the nuclear charge also increases, so the atomic radius decreases. Towards the middle of the series, the increase in nuclear charge is balanced by the shielding effect and so the atomic radius becomes constant. Towards the end of the series, as the e- - e- repulsion increases the atomic radius also increases The atomic and ionic radii of 2nd and 3rd row transition metals are quite similar. This is due to the Lanthanide contraction. In between the 2nd and 3rd row transition elements, 4f electrons are present. The 4f electrons have very poor shielding effect and as a result the atomic and ionic radii of Lanthanides decrease from left to right (Lanthanide contraction). 2. Melting and boiling points Transition elements are generally very hard and have high melting and boiling points. This is due to the participation of (n-1)d electrons (penultimate d electrons) in addition to the ns electrons (valence s electrons) in metallic bonding. In a transition series, the melting and boiling points 1st increases up to the middle and then decreases. This can be explained in terms of metallic bond strength and heat of atomization which depend on the number of unpaired electrons. As the number of unpaired electrons increases, the metallic bond strength increases and hence the heat of atomization. In a transition series, the number of unpaired electrons increases up to the middle and then decreases. Hence the melting point also increases first and then decreases. Manganese (Mn) and Technetium (Tc) have low melting point even though they have d5 configuration. This is because of their low heat of atomization. The m.ps of second and third row transition series are higher than that of the first row due to their higher enthalpies of atomization. 3. Ionisation enthalpy The ionisation enthalpy of transition elements generally increases from left to right. This is due to increase in the nuclear charge from left to right. But the increase is not regular. This can be explained as follows. After the removal of one electron, the relative energies of 4s and 3d orbitals get changed. Hence the remaining electron in the 4s level is transferred to 3d level. So the unipositive ions have dn configuration with no 4s electrons. During this re-organisation of electrons, some energy is released and it is known as exchange energy. So the net energy required to remove the 1st electrons is equal to the sum of ionisation enthalpy and exchange energy. The first ionisation enthalpies of Cr and Cu are low. This is because the removal of one electron does not change their d configuration. Similarly first ionisation enthalpy of Zn is high because after the removal of one electron there is no change in the d configuration. +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 33  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® Zn → +Zn+ e- 10 3d 4s 2 3d 4s1 10 The 2nd I.E of Cr and Cu are very high. This is because the removal of one more e- from these metals disrupted their stable configuration (d5 or d10) The 2nd ionisation enthalpies of Mn and Zn are low, this is because after the removal of one more electron, they attain the stable half filled or completely filled electronic configuration. 4. Oxidation State Transition metals show variable oxidation states. This is because in these elements d and s electrons have comparable energies. So in chemical reaction along with s-electrons, d-electrons also participate. In a given transition series, the maximum oxidation state increases up to the middle and then decreases. This is due to the half-filled or fully filled electronic configuration (noble gas configuration). The common oxidation state of 1st row transition elements is +2. The maximum oxidation state increases from top to bottom in a group. In lower oxidation state, the transition element mainly forms ionic compounds. Scandium generally shows +3 oxidation state because after the removal of 3 electrons, it gets a stable noble gas configuration (d0). The oxidation state of Zinc is +2, because of the completely filled configuration of Zn2+. 5. Electrode Potential The electrode potential values of first row transition series elements generally increases from left to right with some exceptions. The E0(Cu2+|Cu) is positive (+0.34V), while the E0 values of all the other first row transition elements are –ve. This is because the electrode potential depends on heat of atomization, ionization enthalpies and hydration enthalpy. For copper, the high energy to transform solid Cu to aqueous Cu2+ is not balanced by its hydration enthalpy. So, copper does not easily react with acid and liberate H2. Only oxidizing acids [e.g. HNO3 and hot conc. H2SO4] react with Cu and the acid get reduced. Along the series the E0 values become less –ve due to the increase in the sum of 1st and 2nd ionisation enthalpies. The E0 values of Mn2+ and Zn2+ are more –ve, this is because of the half-filled configuration of Mn2+ (d5) and completely filled configuration of Zn2+ (d10). E0(M3+|M2+) value of scandium (Sc) is very low and that for zinc (Zn) is very high. This is because of their stable electronic configuration. E0 (Mn3+|Mn2+) is high because of the stable half-filled configuration of Mn2+. Similarly E0(Fe2+|Fe3+) is low, this is because after the removal of one electron from Fe2+, it gets a stable electronic configuration. Fe2+ → Fe3+ + e- 3d6 3d5 Q. Cu2+ is more stable than Cu+ in aqueous solution. Why? This is due to the greater –ve hydration enthalpy of Cu2+(aq) than Cu+(aq). In the case of Cu2+, the hydration enthalpy is compensated by second ionisation enthalpy. 6. Magnetic Properties Transition metals show mainly two types of magnetic properties - paramagnetism and diamagnetism. Some transition metals also show ferromagnetism which is an extreme case of paramagnetism. Paramagnetism arises from the presence of unpaired electrons. Each unpaired e- is associated with a spin magnetic moment and an orbital magnetic moment. For the compounds of 1 st row transition elements, the contribution of orbital magnetic moment is effectively cancelled and so only spin magnetic moment is considered. It is determined by the no. of unpaired e-s and is calculated by the spin only formula: +2 Chemistry Notes – 2024-25 by ANIL KUMAR K L, PHSS VANDIPERIYAR, IDUKKI Page 34  Join Now: https://join.hsslive.in Downloaded from https://www.hsslive.in ® µs = √n(n + 2) where n is the no. of unpaired electrons and µs is the spin only magnetic moment in the unit of Bohr Magneton (B.M). The magnetic moment increases with increase in no. of unpaired e-s. Thus the observed magnetic moment gives an idea about the no. of unpaired e-s present in the atom or ion. 7. Formation of coloured ions or compounds Most of the Transition metals ions or compounds are coloured. This is because of the presence of partially filled d orbitals. When an electron from a lower energy d orbital is exited to higher d level, it absorbs energy and this is equal to the energy of certain colours in visible region. So the colour observed is the complementary colour of the light absorbed. In aqueous solution most of the transition metal ions are coloured since water molecules act as the ligands. Among Ti2+ and Ti4+, Ti2+ is coloured while Ti4+ is colourless. This is because Ti4+ has no partially filled d orbitals. Ti2+ - [Ar] 3d2 Ti4+ - [Ar] 3d0 Similarly among Cu+ and Cu2+, Cu2+ is coloured while Cu+ is colourless. This is due to the lack (absence) of partially filled d orbitals in Cu+. 8. Formation of Complexes Transition metals form a large no. of complexes. This is due to: 1. Comparatively smaller size 2. High ionic charge 3. Presence of partially filled d orbitals 4. Ability to show variable oxidation state Eg: K4[Fe(CN)6], K3[Fe(CN)6], [Ni(CO)4] etc. 9. Catalytic Property Transition metals act as catalysts in a large no. of chemical reactions. This is due to their large surface area and their ability to show variable oxidation state. 10. Interstitial Compound Formation These are formed when smaller atoms like H, N, C, B etc. are trapped inside the crystal lattice of the metal. They are usually non-stoichiometric and neither typically ionic nor covalent. E.g.: Fe3H, Mn4N, TiC, VH0.56, TiH1.7 etc. Some the properties