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Prepared By Dr M R Choudhary K V SAC AHMEDABAD SOLUTION Concentration of Solutions:- Molarity [M]: No. of moles of solute present per liter of solution in called ‘molarity of solution’.. It is temperature depend...

Prepared By Dr M R Choudhary K V SAC AHMEDABAD SOLUTION Concentration of Solutions:- Molarity [M]: No. of moles of solute present per liter of solution in called ‘molarity of solution’.. It is temperature dependent. M= No. of Moles of solute / Volume of solution in litre Molality [m]: No. of moles of solute present per kg. of solvent is called molality. It is temperature independent. m= No. of Moles of solute / Mass of Solvent in Kg Mole Fraction: Ratio of moles of components to total no. of moles of all the components of solution is called mole fraction (x) of the component. Henry’s law :- “The partial pressure of the gas in vapour phase p is proportional to the mole fraction of the gas x in the solution.” P=KH x Applications of Henery Law(1) In Packing of soda/ Soft drinks(2) In Deep see diving( He = 11.7 % , N2 = 56.2 % and O2 =32.1% ) (3) Functions of lungs (4)At high altitudes pressure is low there for breathing problems lead to ANOXIA (unable to think and weak) Vapour Pressure: The pressure exerted by vapours over the liquid surface at equilibrium is called vapour pressure of the liquid. Raoult’s Law:“The V.P. of any volatile component in the solution is directly proportional to its mole fraction”. Raoult’s Law for Solutions Containing Non-Volatile Solute Raoult’s Law for Solution Containing Volatile Solute Ideal and Non-Ideal Solutions: Ideal Solution Non Ideal Solution Follows Raoult’s law at all temperature and Does not follow Raoult’s law at all temperature and concentrations. P = PA + PB pressure. P ≠ PA + PB Intermolecular forces in resulting solution are Intermolecular forces in resulting solution are different same as in pure components. from the inter molecular force of pure components. A A–B=A–A=B–B – B ≠ A – A, B – B No change in volume while mixing components. Change in volume while mixing components. Δ V mix = 0 Δ V mix ≠ 0 No heat change take place while mixing the Heat changes take place while mixing the components. Δ H mix = 0 components. Δ H mix ≠ 0 Eg: n–hexane+ n–heptanes & benzene + toluene Eg: Acetone + Water &Acetone + CHCl3 Non Ideal Solution Showing Positive elevation from Raoult’s Law Showing Negative Deviation from Raoult’s Law PT > P A + PB Resulting intermolecular force is PT < PA + PB Resulting intermolecular force is weaker than pure components. stronger than pure components. Δ V mix = +ve Δ V mix = –ve Δ H mix = +ve Endothermic mixing process Δ H mix = –ve Exothermic process Acetone + Water , Alcohol + Water Carboxylic Acid + Water  Azeothrpes: (Constant Boiling Mixture) :-Solution in which components are present in a fixed proportion, boils at a constant temperature irrespective of boiling point of pure components ๏ Minimum Boiling Azeothrpes: Boils at a temperature lower than b.ps. of pure components. [95% Alcohol] ๏ Maximum Boiling Azeothrpes: Boils at a temperature higher than b.p.s. of pure components. [68% HNO3] Colligative Properties:- The properties of dilute solutions which depend only on number particles of solute present in the solution and not on their identity are called colligative properties 1-Relative Lowering of 2- Elevation of Boiling 3-Depression of Freezing 4- Osmotic Pressure Point : Point Vapour Pressure ΔTb = Kb m V = nRT = Osmotic pressure R =Gas constant Π =CRT (C= Molarity) Osmosis:- The phenomenon of the passage of pure solven from a region of lower concentration (of the solution) to a region of its higher concentration through a semi-permeable membrane. Osmotic Pressure: Excess pressure which must be applied to a solution in order to prevent flow of solvent into the solution through the semi-permeable membrane. Reverse Osmosis: If pressure greater than osmotic pressure is applied then, flow of solvent molecules is reversed, i.e. from a higher concentration solution to lower concentrated solution. This phenomenon is called “Reverse Osmosis”. It is used in water purification and desalination of water.  UNIT-3 ELECTROCHEMISTRY SrNo Terminology 1. 2 3 4 5 Variation of molar conductivity with concentration For strong electrolytes λ decreases slightly with increase in concentration due to increase in number of ions per unit volume. e.g KCl For weak electrolytes it increases sharply with decrease in concentration as ionization of weak electrolytes increases on dilution e.g CH3COOH 6 Kohlrausch’s Law : According to this law, molar conductivity of an electrolyte, at infinite dilution can be expressed   2 as the sum of contributions from its individual ions e.g.  CaCl 2   (Ca )  2  (Cl  ) 7 Faraday – First law of electrolysis : The amount of substance deposited during electrolysis is directly proportional to quantity of electricity passed. W  zit and W=EIt/96500 8 Nernst equation Equilibrium constant Kc At equilibrium Q = Kc and Ecell = 0 0.059 [ product ] 0.059 Ecell  E cell  log E  cell  log Kc n [react.] n 9. Electrochemical Cell and Gibbs energy of the reaction Products of electrolysis NaCl (molten) Cathode : Na+(l) + e– → Na(s) Anode : Cl–→ ½Cl2+e– NaCl (aq) Cathode : H2O (l ) + e– → ½H2(g) + OH– Anode : Cl–→ ½Cl2+e– AgNO3(aq)-Ag electrodes Cathode : Ag+(aq) + e- Ag(s) Anode: Ag(s)  Ag+(aq) + e- AgNO3(aq)- Pt electrodes Cathode : Ag+(aq) + e- Ag(s) Anode: 2H2O(l )→ O2(g) + 4H+(aq) + 4e– CuCl(aq)- Pt electrodes Cathode : Cu+(aq) + e- Cu(s) Anode: 2H2O(l )→ O2(g) + 4H+(aq) + 4e– Electrochemical Cell Electrolytic Cell 1. It is device to convert chemical energy to electrical energy 1. It is device to convert electrical energy into chemical energy 2. It is based upon redox reaction which is spontaneous. 2. The redox reaction is non-spontaneous and takes place only when electrical energy is supplied. SNO Name of cell Electrolyte Chemical Reaction Remark 1 Mercury Cell Paste of Anode:- zinc-mercury amalgam Provide constant voltage and does   KOH & Zn( Hg )  2OH ZnO( s)  H 2O  2e not corroded due to absence of Zn(OH)2 Cathode:- HgO and carbon ion in overall reaction.  HgO  H O  2e Hg (1)  2OH  Voltage 1.35V 2 Overall reaction:- Zn(Hg) + HgO  ZnO + H2 2 Dry Cell : ZnCl2 and Anode :- Zinc rod Zn(s) → Zn2++2e- Corroded due to the presence of NH4Cl Cathode:-Carbon rod surrounded by MnO2 ion in overall reaction. 2MnO2 + 2NH4+ + 2e- Mn2O3 +2NH3 ZnCl2 is used to make a complex Overall Reaction:- [Zn(NH3)2Cl2] it saves the bursting Zn + 2MnO2 + 2NH4+ Zn2+ + Mn2O3 + 2 NH3 of cell, Voltage 1.5V 3 Lead Storage 38% H2SO4 Anode:- Pb grid Can be Recharged by reversing the (Secondary PbSO4 ( s)  2e  Pb( s)  SO4 2 ( aq) electrodes Reactant become Batteries) : products and products become Cathode:-Pb grid filled with PbO2 reactants PbO2(S) + SO42- + 4H+(aq) +2e- PbSO4(s) + 2H2O(l) Overall Cell Reaction:- Pb(s)  PbO2 (s)  2H 2 SO4 2PbSO4 (s)  2H 2O(1) 4 Fuel Cell : KOH Anode :- 2 H 2  4OH  4 H 2O  4e  Eco friendly, solution Efficiency 60-70% Cathode O2 ( g )  2 H 2O(l )  4e  4OH  (aq) Overall Reaction:- 2 H 2 ( g )  O2 ( g ) 2 H 2O(l ) 2 5. Corrosion H2CO3 Anode (oxidation) :- Fe( s)  Fe (aq)  2e Formula of rust Fe2O3. x H2O Cathode (Reduction) :- O2 ( g )  4 H  (ag )  4e  2 H 2O(l ) Atmospheric Oxidation:- 2Fe2+(aq) + 2H2O(l)+1/2O2 Fe2O3(s)+4H+(aq) By covering the surface with paint or by some chemicals (e.g. bisphenol). / Cover the surface by other Prevention of metals (Sn, Zn, etc.) that are inert or react to save the object. An electrochemical method (sacrificial Corrosion electrode like Mg, Zn, etc.) which corrodes itself but saves the object. Observe the graph (a)The curve ‘Y’ is for KCl or CH3COOH? Ans:- It is for CH3COOH. (b)What is intercept on Λm axis for ‘X’ equal to? Ans:- equal Λm shown in figure (c) Give mathematical equation representing straight line. between Λm(molar conductivity) Vs C Ans: (Molar- (d) What is slope equal to? Ans: Slope = –A concentration) and (e) What happens to molar conductivity on dilution in case of weak electrolyte and why? answer the questions Ans: Λm for weak electrolyte increases sharply on dilution because based on graph. both number of ions as well as mobility of ions increases.  CHAPTER-4 CHEMICAL KINETICS RATE OF The change in concentration of either reactant or product per unit time. REACTION Example : N2 + 3H2 - 2NH3 Unit of rate Mol L-1time-1 or mol L-1 S-1 Rate Law The expression of rate of reaction in term of concentration of reactant aA + bB Product Rate=K[A]x[B]y Rate Constant Specific Rate of reaction :-Rate of reaction when concentration are taken as unity Rate=K[A]x[B]y [A]=1 [B]=1 Rate=K Unit of rate constatnt N - Order of reaction Order of The sum of powerof concentration of reactant in rate law expression reaction aA + bB Product Rate=K[A]x[B]y Order of reaction = x+y It may be Zero,Positive or Negative or fraction.It is experimental quantity. Mechanism of Elementary reaction:-Chemical reaction complete in single step and have exponant in rate law Reaction expression equal to their stochiometric coefficient.Zero order reaction can not be an elementry reaction. Complex Reaction:- For the complex reaction,the overall reaction is controlled by the slowest step. Difference Order of Reaction Molecularity between  1.The sum of the exponents (powers) by  1.The no. atoms , ions or molecule that must Order of which the concentration terms are raised collide with one another simultaneously so that Reaction and in rate law. chemical reaction take place Molecularity  2.Determined Experimently by rate law  2.Theoritical concept ,determined by slowest step  3.May be Zero and negetive  3.Never be Zero and negetive  4.May be Fraction  4.Neverbe Fraction Pseudo first A chemical reaction which seems to be higher order but actually they are of first order order reaction  For these reaction,Order of reaction is one but molecularity is more than one Example :-Hydrolysis of ester in acedic medium, Inversion of cane sugar in acidic medium METHOD OF Consider the following data for the reaction A + B = Products and Determine the order of the reaction DETERMINING with respect to A and w.r.t. B and the overall order of the reaction. Also calculate rate constant. ORDER OF Exp. Initial conc. (A) Initial conc. (B) Initial rate (mol L-1s-1) REACTION 1 0.10 M 1.00 M 2.1 x 10-3 2 0.20 M 1.00 M 8.4 x 10-3 3 0.20 M 2.00 M 8.4 x 10-3 ANSWER: Exp. Rate = K [A]x [B]y Eq.( i)/(ii) Then(1/2)x = (1/4) 1 Rate = K [0.10 ]x [1.00 ]y = 2.1 x 10-3 …………………….(i) Therefore x = 2 2 Rate = K [0.20 ]x [1.00 ]y = 8.4 x 10-3 …………………….(ii) Eq.(ii)/(iii) Then(1/2)y = (1) 3 Rate = K [0.20 ]x [2.00 ]y = 8.4 x 10-3 …………………….(iii) Therefore y = 0 Rate law=K[A]2[B]0 K =2.1 x 10-3 / [0.1]2[1.00]0 Differential and Integrated Rate Laws: Zero Order Reactions: First Order Reactions: [A] = -K0t + [A0] that is Y = -mx + C ln[A] = -K1t + ln[A0] that is Y = -mx + C [A]0-[A]t = k0t or k1=( 2.303/ t)log ([A]0 / [A]).Units of k1 = s-1EXAMPLES Unit of rate constant = mol litre-1s-1 Examples:  Enzyme catalyzed reactions  Half Life: Time required to consume the half of initial Half Life:Time required to consume the half of initial concentration of reactant is called half life period. concentration of reactant is called half life period.At t = At t = t1/2[A] = [A0]/2 t1/2[A] = [A0]/2 K0t1/2 = [A]0-[A0]/2 t1/2 = [A]0/2k k1=( 2.303/ t1/2 )log ([A]0 / [A0/2]) t1/2 = 0.693/k1 Half life is directly proportion of the initial concentration Half-life is independent of the initial concentration of the of the reactant for a zero order reaction. reactant for a first order reaction The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume. SO2Cl2 (g) → SO2(g) + Cl2(g). Calculate the rate constant. THRESHOLD ENERGY The Minimum extra amt of energy which the reactants molecules must have sothat reactant molecule reacts and overcome the energy barriers and converts into products. ACTIVATION ENERGY The Minimum extra amt of energy absorbed by the reactant molecule so that their energy becomes equal to threshold energy.  d and f- BLOCK ELEMENTS (By Dr M R CHOUDHARY 846913570) 1 d-block elements are called transition elements incomplete d-orbitals in neutral or most stable oxidation state. 2 Zn, Hg, Cd, are not included as transition as they have filled d-orbitals in neutral and most stable +2 elements and why? oxidation state. 3 Sc3+ has d0 configuration but it is transition metal Because of partially filled d1 configuration of Sc 4 First series transition elements shows irregular Due to shielding effect of inner 3d electrons which overcome trend of atomic radii ENC. 5 Atomic radii of Zr to Hf are almost same. due to Lanthenoid Contraction. 6 First series transition elements shows irregular It is because of mutual exchange of energy between d and f trend of IE orbitals after removal of one electron. 7 Transition metals have high enthalpy of due to presence of unpaired electrons form strong metallic atomization and melting point. bonding 8 Zn, Cd, Hg are not very hard and have low M.P. Due to presence of all paired electrons they have least enthalpy and B.P. of atomization, very weak metallic bond & are not very hard with low M.P. and B.P. 9 d-block elements shows a large no. of oxidation due to presence of unpaired electrons or partially filled d- state orbitals. 10 Which first series transition metal shows highest Mn shows highest number of oxidation state due presence of number of oxidation states and why? highest number of unpaired electrons. 11 Which first series transition metal do not show Sc , due to achieving noble gas configuration in +3 O.S variable oxidation states and why? 12 Cr3+ is more stable as compared to Cr2+ due to stable half filled t2g3 configuration. 13 Cu2+ is more stable in solution than Cu+ due to high hydration enthalpy of Cu2+ ion in solution. 14 Which first series transition metal shows +1 Cu shows +1 oxidation state in gaseous form due to stable full oxidation states and why? filled 3d10 configuration. 15 Higher oxidation states of transition metals because higher oxidation states stabilized by highly found in their oxides and fluorides electronegative elements such as O and F 16 higher oxidation states are more common in due to formation of dπ-pπ bond between metal and oxygen. oxides rather than fluorides of transition metals. 17 +2 O.S becomes more stable in first series as number of unpaired electrons decreases. transition elements moving from left to right 18 Write oxoanions of first series transition metals MnO4-, Cr2O72-, VO3-, CrO42- in which group number and O.N. of transition metal are same- 19 Moving from left to right E0 for M2+/M becomes trends due to irregular variation of IE1 + IE2. less negative but it shows irregular trend 20 For copper E0 Cu2+/Cu is +ve (+0.34 V) it is due to low hydration enthalpy of Cu2+ which cannot compensate ∆subH0 + IE1 + IE2. 21 For Mn , E0 Mn3+/Mn2+ is +ve because by converting into Mn2+ it achieves half filled configuration. 22 For Cr , E Cr /Cr is −ve 0 3+ 2+ because Cr3+ is more stable due to half filled t2g configuration. 23 Most of the compounds of transition metal are It is due to d-d transition coloured in solid state as well as in their solution 24 (Sc3+, Ti4+) does not exhibit any colour Due to absent of unpaired e- as d-d transition is not possible 25 (Zn2+, Cu+) does not exhibit any colour full filled d-orbitals (d10) as d-d transition is not possible 26 Transition metals or ions are paramagnetic in Due to presence of unpaired electrons (d1-9) nature 27 Transition metal form large number of High charge by mass ratio, availability of vacant d-orbital & coordination complexes Variable oxidation state 28 What are Interstitial compounds? Compounds in which small elements like H, B, C, N trapped in the interstitial space of metal lattice. 29 Transition metals form Interstitial compounds. due to availability of interstitial space in metal lattice. 30 Why interstitial compounds are found to be more These compounds have high M.P as compared to pure metals. useful than pure metal? These are very hard and are Chemically inert. 31 Transition metals and their compounds are Due to large surface area, variable oxidation state & Availability widely used as catalyst of vacant d-orbitals 32 transition metals form alloy. Due to comparable metallic radii 33 Why oxides of Higher oxidation states of Higher oxidation states are more polarizing in nature. hence transition metals are acidic in nature whereas are acidic in nature whereas lower O.S. are less polarizing lower O.S. are basic in nature? hence are basic in nature. 34 Actinoids shows greater number of oxidation Due to comparable energy of 5f, 6d, and 7s orbitals. states then lanthanoids 35 It is difficult to study chemistry of actinoids. Due to radio active nature and small half life 36 Ce(IV) is a good oxidizing agent in aq. soln because in aq. solution its E0Ce4+/ Ce3+ is +1.74 V 37 Actinoid contraction is greater from element to because of poor shielding by 5f orbitals due to its superficial element as compared to lanthanoids. position. Potassium Dichromate [K2Cr2O7] Potassium Permanganate [KMnO4] Preparation: Preparation: from pyrolucite ore [MnO2] It is prepared from chromite ore : FeCr2O4/ FeO Cr2O3 Step 1: Step 1: 2MnO2 (Black)+4KOH+O2 2K2MnO4 (Green)+ 2H2O 4FeCr2O4 + 8Na2CO3 +7O2 8Na2CrO4 + 2Fe2O3 + 8CO2 Step 2: Step 2: 2Na2CrO4 (Yellow)+ H2SO4 3MnO42- (Green)+4H+ 2MnO4-( Purple)+MnO2+ 2H2O Na2Cr2O7(Orange)+ Na2SO4 + H2O Oxidizing Character: (Oxidation in acidic medium) Step 3: Na2Cr2O7 + 2KCl K2Cr2O7 + 2NaCl MnO4- + 8H+ + 5e- Mn2+ + 4H2O Orange Orange Eq. mass of KMnO4 = 158/5 = 31.6 Effect of pH: MnO4-+5Fe2++8H+ Mn2++5Fe3++ 4H2O Cr2O72- (Orange) + 2OH- 2CrO42- (Yellow) + H2O 2MnO4- + 10Cl- + 16H+ 2Mn2+ + 8H2O + 5Cl2 2CrO42- (Yellow) + 2H+ Cr2O72- (Orange) + H2O 2MnO4- +5C2O42-+ 16H+ 2Mn2++10CO2+ 8H2O Oxidizing Character: Dichromate ion acts as strong 2MnO4-+5SO32-+6H+ 2Mn2+ + 5SO42- + 3H2O oxidizing agent in acidic medium. 2MnO4 +5NO2 +6H - - + 2Mn2+ + 5NO3- + 3H2O Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O 2MnO4-+5S2-+16H+ 2Mn2+ + 5S + 8H2O Cr2O72- + 14H+ + 6I- 2Cr3+ + 7H2O + 3I2 2MnO4- +3H2O+ I- 2MnO2 +IO3- + 2OH- Cr2O7 + 14H + 6Fe 2- + 2+ 2Cr3+ 7H2O + 6Fe3+ 8MnO4 +3S2O3 +H2O - 2- 8MnO2 + 6SO42- + 2OH- Cr2O72- + 14H+ + 3Sn2+ 2Cr3+ 7H2O + 3Sn4+ 2MnO4- +3 Mn2++2 H2O 5MnO2 +4H+ Cr2O72- + 8H+ + 3H2S 2Cr3+ + 7H2O + 3S f-Block Elements Consequences: 1. Resemblance in the properties of second and General configuration (n – 2) f 1-14 (n – 1)d 0-1 ns2 third transition series from Zr & Hf) atomic size of 2nd and 3rd 57La [Xe] 5d1 6s2 89Ac [Rn] 6d1 7s2 - transition series become same. 58Ce [Xe] 4f1 5d1 6s2 91Pa [Rn] 5f2 6d1 7s2 2. Similarity in the properties of Lanthanoids:- The decrease in 63Eu [Xe] 4f7 5d0 6s2 92U [Rn] 5f3 6d1 7s2 atomic radii from Lanthenoid to Lanthenoid is very less hence 64Gd [Xe] 4f 5d 6s 7 1 2 94Pu [Rn] 5f6 6d0 7s2 it is very difficult to separate these elements in pure state. 70Yb [Xe] 4f14 5d0 6s2 3. Basic strength of hydroxides of Lanthanoids decreases from Lanthanoid Contraction- Regular decrease in atomic Lanthenium (La) to Lutesium (Lu). or ionic radii of Lanthenoid with increase in atomic Misch Metal – It is alloy of Lanthanoids(95%) with iron number is called Lanthenoid contraction. (5%)and Trace of S, C, Al ,Ca Lanthenoid – 95% Metal + Trace Cause: It is due to imperfect shielding of one 4f orbital of S, C, Al ,Ca + Iron – 5% by another 4f orbital due to which ENC regularly Uses- making bullet shells and flinter of lighters. increases with increase in atomic number. Lanthanoids Actinoids Last electron enters in 4f orbital Last electron enters in 5f orbital Are natural elements and non radioactive in nature Are radioactive in nature and after Np they are synthetic elements Shows less number of oxidation states [+2,+3,+4] Shows greater number of oxidation states [+3,+4,+5,+6,+7] Element to element lanthanoid contraction is less Element to element actinoid contraction is more  Coordination Compounds Ligand: Provide pair of electron for formation of Coordination Number: Number of coordinate bonds coordinate bond to central metal atom is called ligand. formed by ligands with central metal atom in the complex (Lewis Base) is called coordination number of complex. Types of ligands:- eg. - [Mg(EDTA)]2- [Fe(CN)6]4- [Co(en)3]3+ (a) Monodentate :-One donar atom    (b) Bidentate :- Two donar atom C.N. 6 6 6 (c) Ambidentate:- Two donar atom but at a time 5) Homoleptic and Hetroleptic Complexes: form only one bond Eg: SCN- and NCS- A complex with same type of ligands surrounding central F -Fluorido, Cl- - Chlorido, Br- -Bromido - metal atom is called homoleptic complex. E.g.- I- - Iodido, H- -Hydrido, OH- - Hydroxido [Cu(NH3)4]2+, [Pt(NH3)4]2+ CN- -Cyanido, SCN- - Thiocyanato A complex with different types of ligands surrounding NCS- - Isothiocyanato, CH3COO- -Acetato central metal atom is called hetroleptic complex. E.g. SO42- - Sulphato, CO32- - Carbonato [Pt(NH3)2 Cl2] (COO-)2 or C2O42- - Oxalato (ox), NO3- - Nitrato 6. Coordination Polyhedron: Spatial arrangement of NO2- - Nitrito(N), PO43- - Phosphato ligands around central metal atom or ion is called Neutral ligands- H2O- Aqua , NH3 - Ammine coordination polyhedron. R NH2 -Alkanamine, CO –Carbonyl, Writing Formula of Coordination Complex: H2NCH2CH2NH2 - ethane-1,2–diamine 1. Cation is written first followed by anion. Ethylenediaminetetraacetate ion (EDTA4–) 2. In coordination sphere central metal atom is written Ligand which has two different donor atoms and first followed by ligands. either of the two ligetes in the complex is called 3. If more than one type of ligands are present then ambidentate ligand. Examples NO2 – and SCN– alphabet is followed. Na2[Fe(H2O)4 (NO) Br] Nomenclature of Coordination Compounds: (5) [Pt (NH3)2 Cl (NO2)] (1) Na2 [Fe (H2O) Br4 (NO)] Diamminechloridonitritoplatinum (II) Sodium tetrabromidonitrosylferrate (II) (6) [Ni (CO)4] Tetracarbonyl Nickel (O) (2) [Fe (H2O)4 Br (NO)] SO4 (7) CO (NH3)5 CO3] Cl Tetraquabromidonitrosyliron (III) sulphate Pentamminecarbonato cobalt (III) chloride (3) K4 [Fe (CN)6 ] (8) K3 [Cr (C2O4)3] Potassium hexacyanoferrate (II) Potassium trioxalate chromate (III) (4) [Cr (NH3)3 (H2O)3 ] Cl3 (9) Hg [Co (SCN)4] Triamminetriaquachromium (III) Chloride Mercury tetrathiocyanato cobaltate (III) Isomerism in Coordination Compounds (b) Stereoisomerism (a) Structural isomerism (i) Geometrical isomerism- (i) Linkage isomerism- ambidentate ligand. M–SCN, Square planner complexes with formula MA2B2- two M–NCS & (–NO2 ), (–ONO) geometrical isomers, MABCD- Three isomers (ii) Coordination isomerism-Both complex Octahedral complexes- MA4B2, M(aa)2 A2 – Cis & trans [Co(NH3 )6 ][Cr(CN)6 ] & [Cr(NH3)6 ][Co(CN)6 ] MA3B3- Fac & mer isomers (iii) Ionisation isomerism- (ii) Optical isomerism- Optical isomers are mirror [Co(NH3)5(SO4)]Br & [Co(NH3 )5Br]SO4. images that cannot be superimposed on one another. (iv) Solvate isomerism- exchange of H2O as ligand These are called as enantiomers. and counter anion [Cr(H2O)6]Cl3&[Cr(H2O)5Cl]Cl2.H2O [PtCl2 (en)2 ] 2+ Bonding in coordination compounds- (I) Werner’s Coordination Theory- Primary valence is ionizable and refers charge present on complex ion. Secondary valence is non-ionizable and refers coordination number of complex. The ions/groups bound by the secondary linkages to the metal have characteristic spatial arrangements corresponding to different coordination numbers. 2) Valence Bond Theory or VBT Complex Oxidation Hybridi Geometry Magnetic nature unpaired Outer sphere / state of CMA -zation electrons Inner sphere [Co(NH3)6] 3+ +3 d sp 2 3 Octahedral Diamagnetic 0 Inner sphere [CoF6 ]3- +3 sp3 d2 Octahedral Paramagnetic 0 Outer sphere [Co(C2O4)3]3- +3 d2sp3 Octahedral Diamagnetic 0 Inner sphere [Mn(CN)6]4- +2 d2sp3 Octahedral Paramagnetic 1 Inner sphere [Fe(CN)6]3- +3 d2sp3 Octahedral Paramagnetic 1 Inner sphere [Fe(H2O)6]3+ +3 sp3 d2 Octahedral Paramagnetic 5 Outer sphere [Ni(NH3)6]2+ +2 sp3 d2 Octahedral Paramagnetic 2 Outer sphere [NiCl4]2- +2 sp3 Tetrahedral Paramagnetic 2 ------------------ [Ni(CO)4] 0 sp3 Tetrahedral Diamagnetic 0 ---------------- [Ni(CN)4 ]2– +2 dsp2 Square planner Diamagnetic 0 ----------------- 3) Crystal Field Theory [CFT] Splitting of d-orbitals in octahedral field: Splitting of d-orbitals in tetrahedral field: Metal Carbonyls: Homoleptic complex of transition b) Metallurgy: metal with carbonyl ligand are called metal carbonyls. Extraction of Au, Ag, by cyanide process. E.g. [Ni(CO)4], [Fe(CO)5], [Cr(CO)6],[CO2(CO)8], Refining of Ni by Mond’s process [CO3(CO)12], [Mn2(CO)10] Refining of Zr by Van Arkel process. The metal-carbon bond in metal carbonyls possess c) Quantitative Analysis: Estimation of hardness of water both σ and π character. The M–C σ bond is formed by Ca2+ and Mg2+ are estimated by complexomatric titration the donation of lone pair of electrons on the carbonyl using EDTA. carbon into a vacant orbital of the metal. The M–C π d) Qualitative Analysis: Cu2+, Fe2+, Zn2+, Ni2+, NO-3, etc ions bond is formed by the donation of a pair of electrons are confirmed by reactions involving complex formation. from a filled d orbital of metal into the vacant e) Industrial Catalyst: [(Ph3P)3RhCl], a Wilkinson catalyst, antibonding π* orbital of carbon monoxide. The metal is used for the hydrogenation of alkenes to ligand bonding creates a synergic effect which f) Black & White Photography: strengthens the bond between CO and the metal. In black and white photography, the developed film is fixed by washing with hypo solution which dissolves the undecomposed AgBr to form a complex ion, [Ag(S2O3 )2 ]3–. g) As medicines: Cisplatin [Cancer Treatment], EDTA [Used to remove heavy metal poison using chelate Application of Coordination Complexes: therapy. a) Biological System: Chlorophyll – complex of Mg Haemoglobin – complex of Fe, Vit. B-12 – complex of Co Prepared By Dr M R Choudhary K V SAC AHMEDABAD 8469113570  NAME REACTION A. HALOALKANES AND HALOARENES 1. Wurtz reaction : Alkyl halides react with sodium in dry 2R – X +2Na D ry e t her R – R + 2NaX ether to give hydrocarbons(Alkanes) containing double the number of carbon atoms present in the halide. 2CH3 –Br +2Na D  CH3 – CH3 + 2Na Br ryether 2. Fittig reaction: Aryl halides(Haloarenes) when treated 2Ar – X +2NaD  Ar – Ar + 2NaX ryether with sodium in dry ether gives in which two aryl groups are joined together. Wurtz-Fittig reaction : A mixture of an alkyl halide Ar – X + Na + R – X D  Ar – R + 2NaX ryether (Haloalkanes) and aryl halide(Haloarenes.) gives alkylarene when treated with sodium in dry ether. 3. Finkelstein reaction: Alkyl chlorides/ bromides reacts R – Cl +NaI a  R – I + NaCl cet one with NaI in dry acetone to give Alkyl iodides. R – Br +NaI a  R – I + NaBr cet one Swarts reaction Heating of alkyl chloride/bromide in the R – X AgF, Hg 2F2 , CoF2 or SbF3       R–F presence of a metallic fluoride such as AgF,Hg2F2, CoF2 or CH3 –Br + AgF CH3 –F + AgBr SbF3 to give alkyl fluorides 12. ALCOHOLS,PHENOLS AND ETHERS Hydroboration –oxidation reaction : The alcohol CH3-CH=CH2( i ) B2 H 6 ( ii ) H2O2/O H     CH3-CH2-CH2-OH obtained through anti- Markownikov ’s addition of H2O Reimer-Tiemann reaction- Phenol reacts with chloroform and NaOH to give salicylaldehyde Kolbe’s reaction:- Phenol reacts with NaOH followed by CO2 in acidic mediumto give salicylaldehyde Williamson’s synthesis :In this method an primary CH3Cl + CH3 ONa → CH3OCH3 + NaCl alkyl halide reacts with a sodium alkoxide to form symmetrical or unsymmetrical ethers. 12. ALDEHYDES, KETONES AND CARBOXYLIC ACID Aldol Condensation : Aldehydes and ketones having at least one α – hydrogen undergo a condensation reaction in the presence of dilute alkali (NaOH ,KOH etc.)as catalyst to form β- hydroxyaldehyde (aldol)or β – hydroxyketone (ketol) respectively Cross Aldol Condensation: When aldol condensation is carried out between two different aldehydesand / or ketones, it is called cross aldol condensation. If both of them contain α-hydrogen atoms, it gives a mixture of four products. Cannizzaro Reaction: Aldehydes, which do not have an α-hydrogen atom, undergo self oxidation and reduction reaction on treatment with concentrated alkali.In this reaction one molecule of the aldehydeis reduced to alcohol and another is oxidized to carboxylic acid salt. Clemmensen Reaction: The carbonyl group of CH3CHO n  H Z g ,Conc.H Cl CH3CH3 aldehydes and Ketones are reduced to CH2 group on treatment with zinc amalgam(Zn/Hg) and concentrated CH3CO CH3 Z   CH3CH2CH3 nH g,H Cl hydrochloric acid.(Conc.HCl) Wolf Kishner Reaction:- reaction used to convert carbonyl group (-CO-) into methylene groups through reaction with hydrazine and KOH- ethylene glycol Rosenmund Reaction: Acyl chloride (acid chloride) is hydrogenated over catalyst, palladium on BaSO4 Etard Reaction- oxidation of toluene with chromyl chloride (CrO2Cl2) in CCl4 to give aromatic aldehyde Stephen Reaction: Alkyl nitriles on reduction with SnCl2 and Conc. HCl, give corresponding aldehydes. Gatterman-Koch reaction: benzene is treated with carbon monoxide in the acidic medium in presence of anhydrous aluminium chloride to give benzaldehyde Ozonolysis of alkenes involves the addition of ozone molecule to alkene to form ozonide, and thencleavage of the ozonide by Zn-H2O to Aldehyde and/or Ketones. Decarboxylation : Sodium salts of acids when heated with soda lime, alkanes are formed. Hell Volhard Zelinsky: (HVZ) Carboxylic acids having an α-hydrogen are halogenated at the α-position on treatment with halogenin the presence of small amount of red phosphorus to give α- halocarboxylic acids 13. AMINES Hoffmann bromamide degradation reaction: primary RCONH2 + 4NaOH + Br2 →RNH2 + 2 NaBr + Na2CO3 + 2H2O amides are treated with bromine in thepresence of an CH3CONH2  Br2  4 KOH CH3 NH2  K2CO3  alkali, a primary amine containing one carbon less than 2KBr  2H2O the amide is formed. Gabriel phthalimide synthesis: Phthalimide on treatment with KOH which on heating with alkyl halidefollowed by alkaline hydrolysis produces the corresponding primary amine. Carbylamine reaction(Isocyanide test):Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides Isocyanides or carbylamines which are foul smelling Diazotization reaction- Aromatic amine reacts with nitrous acid and Conc HCl to form diazonium salt Sandmeyers reaction: The Cl–, Br– and CN– nucleophiles can easily be introduced in the benzene ring in the presence of Cu(I) ion [CuCl] Gatterman reaction: The Cl–, Br– and CN– nucleophiles can easily be introduced in the benzene ring in the presence of copper powder  DISTINGUISH TEST S.No Test Reagent Inference 1. Lucas test :To distinguishbetween ZnCl2/HCl 0 (3 ) Alcohols gives Turbidity Primary (10), Secondary (20), & (immediately), 20 Turbidity after some Tertiary (30) Alcohols) time (5-10 min) 10 does not give Turbidity at room temp. 2. Iodoform test (Alcohols I2 / NaOH Yellow Ppt of CHI3 is formed containing CH3-CH(OH)-linkage) and CH3CO- in Ald/Ketone 3. Neutral ferric chloridetest Neutral FeCl3 Phenols give Violet colouration (Phenol) 4. Tollens test [Aliphatic Ammonical AgNO3 Bright silver mirror [Ag] is produced due Aldehydes(e.g.Ethanal,Pro panal etc) & to the formation of silver metal. Aromatic Aldehydes (Benzaldehyde) (HCOOH is gives this test) 5. Fehling’s test [OnlyAliphatic Fehling solution A Reddish brown precipitate of [Cu2O] is Aldehydes] (aqueous CuSO4 & obtained. **Aromatic aldehyde do not Fehling solution B (alkaline sodium potassium tartarate is give this test alkaline sodium also known as Rochelle salt) potassium tartarate) 6. Sodium bicarbonate test (Aliphatic NaHCO3 Effervescence due to evolution of CO2 & AromaticCarboxylic acids) Sodium gas. Hydrogencarbonate 7. Isocyanide test Primary Aliphatic Chloroform(CHCl3) Unpleasent odur (foul smelling) of & Aromatic amines. + Alcoholic KOH isocyanides or carbylamines. 8. Heinsberg test(To distinguish Benzenesulphonyl Product of 1o Amines soluble in alkali. between (10), (20), & (30) Amines. chloride C6H5SO2Cl Product of 2o Amines are insoluble in alkali 3o amines do not react. 9. Azo dye test(Aniline) (NaNO2 + HCl) Reaction with NaNO2 + HCl at 273- [Nitrous acid] followed 278 K gives BDC which forms a by β -napthol brilliant orange Azo dye with β- napthol in sodium hydroxide REASONING TYPE OF QUESTIONS 10. HALOALKANES AND HALOARENES QUESTION-REASONING ANSWER- REASON 1 Benzyl chloride is highly reactive towards the SN1 Due to the stability of benzyl carbocation due to reaction. resonance 2 2-bromobutane is optically active but 1- Because 2-Bromobutane has a chiral centre so mirror bromobutane is optically inactive image are nonsuperimposable 3 Electrophilic substitution reactions in haloarenes Due to – I effect of halogen predominant over +R occur slowly. effect 4 Which would undergo SN1 reaction faster because 3o carbocation is more stable than 1ocarbocation 5 Why haloarenes are less reactive than haloalkanes In haloarenes C—X bond acquires a partial double towards nucleophilic substitution reactions bond character due to resonance 6 Which compound in each of the following pairs will (i) iodine is a better leaving group because of its larger react faster in SN2 reaction? Why? size. (ii) three bulky methyl group hinder the (i) CH3Br or CH3I (ii) (CH3)3C-Cl or CH3-Cl approaching nucleophile 7 Compound (I) reacts faster in SN1 reaction as it is a 2° alkyl halide 8 A solution of KOH hydrolyses CH3CH(Cl)CH2CH3 and CH3CH2ClCHCH3 more easily hydrolysed as it forms 20 CH3CH2CH2CH2Cl. Which one of these is more easily carbocation which is more stable than 10 carbocation hydrolysed and why? 9 State one use each of DDT and iodoform DDT: It is used as insecticide Iodoform: Iodoform is used as an antiseptic. 10 What is known as a racemic mixture? Give an An equi-molar mixture of d- and l- isomers(50:50 d+l) example. For example, butan-2-ol. A racemic mixture is optically inactive due to external compensation 11 Although chlorine is an electron withdrawing Through resonance effect, chlorine tends to stabilize group, yet it is ortho-, para-directing in the carbocation and the effect is onlyapplicable at electrophilic aromatic substitution reactions. ortho and para-positions. Explain why it is so? 12 Grignard’s reagents should be prepared under This is because Grignard reagent forms alkanes by anhydrous conditions, why? reacting with moisture. 13 the dipole moment of Chlorobenzene is lower than due to resonance ,Chlorobenzene shorter C—Cl that of Cyclohexyl chloride bond(sp2)than cyclohexyl chloride C—Cl bond(sp3) 14 Chloroform is stored in closed dark brown bottles chloroform is slowly oxidised by air in the presence of light to form poisonous gas phosgene.COCl2 11. ALCOHOL, PHENOL AND ETHER 1 p-nitro phenol is more acidic than p-methyl phenol Due to –I/–R effect of –NO2 group & +I /+R effect of CH3 2 p-nitrophenol is more acidic than o-nitrophenol OR p-nitro phenol has intermolecular H-bond while o-nitro O-nitrophenol is steam volatile ,not p-nitrophenol phenol has intramolecular H-bond. 3 Phenol is more acidic than ethanol. phenoxide ion stabilised by resonance) 4 Boiling point of ethanol is higher in comparison to Because of hydrogen bonding in ethanol methoxymethane. 5 The C – O – H bond angle in alcohols is slightly less Due to lone pair- lone pair repulsion on oxygen than the tetrahedral angle (109o28’). 6 Although phenoxide ion has more number of carboxylate ion stabilized through equivalent resonance resonating structures than carboxylate ion, whereas phenoxide ion through non equivalent carboxylic acid is a stronger acid than phenol. resonance 12. ALDEHYDES KETONES AND CARBOXYLIC ACIDS 1 Increasing order reactivity towards nucleophilic , addition OR (reactivity towards HCN) Butanone < Propanone < Propanal < Ethanal Methyl tert-butyl ketone < Acetone < Acetaldehyde 2 Acidic Character =(1/ pKa) (CH3)2CHCOOH < CH3CH(Br) CH2COOH < CH3COOH < Cl-CH2-COOH < F-CH2-COOH CH3CH2CH(Br)COOH 4-Methoxy benzoic acid Aniline > tr-nitroaniline 10 Aromatic primary amines cannot be prepared by aryl halides do not undergo nuleophilic substitution Gabriel phthalimide synthesis. easily due to presence of partial double character 11 Although amino group is o– and p– directing in In presence of conc HNO3 ,aniline gets protonated to electrophilic substitution rean, aniline on nitration form the anilinium ion which is meta directing gives a substantial amount of m-nitroaniline.  MECHANISM 1. SN1 mechanism  It is Two step reactions.  Step I: In the first step slow dissociation of alkyl halide takes place by reversible reaction forming a carbocation.  Step II: The carbocation at once combines with the nucleophile to form final product (racemic mixture) 2. SN2 mechanism  It is One step reaction 3. Mechanism for the Hydration of alkenes to alcohol Mechanism- The mechanism of the reaction involves the following three steps: Step 1: Protonation of alkene to form carbocation by electrophilic attack of H3O+. Step 2: Nucleophilic attack of water on carbocation Step 3: Deprotonation to form an alcohol 4. Mechanism for the acidic Dehydration of alcohols to give alkenes (At 443 K) Step 1: Formation of protonated alcohol. Step 2: Formation of carbocation: It is the slowest step and hence, the rate determining step of the reaction. Step 3: Formation of ethene by elimination of a proton 5. Mechanism for the acidic Dehydration of alcohols to give ethers (At 413K) 6. Nucleophilic addition reaction in carbonyl compounds.  BIOMOLECULES Carbohydrates are classified on the basis of their behavior on hydrolysis : Monosaccharides: A carbohydrate that cannot be hydrolyzed further to give simpler unit of examples are glucose, fructose, ribose, etc. Oligosaccharides: Carbohydrates that yield two to ten monosaccharide units, on hydrolysis. For example, sucrose , Maltose and Lactose Polysaccharides: Carbohydrates which yield more than 10 of monosaccharide units on hydrolysis are called polysaccharides. Some common examples are starch, cellulose, glycogen, gums, etc. Reducing sugars: All those carbohydrates which reduce Fehling’s solution and Tollens’ reagent are referred to as reducing sugars. Examples: All monosaccharides, Maltose and Lactose. Non-reducing In disaccharides, if the reducing groups of monosaccharides i.e., aldehydes or sugars: ketone groups are bonded, these are non-reducing sugars e.g. Starch, Sucrose Anomers: α & β - Glucose , which differ in the orientation of − OH group at C1. Proteins: proteins are polymer of α- amino acids ,joined by peptide bonds. They also known as polyamides. Types of Proteins: (i) Fibrous proteins:- The polypeptide chains run parallel and are held by H- bond or disulphide linkage , Insoluble in water Eg : Keratin ,Myocin (ii) Globular proteins:- Polypeptides coil around to give a spherical shape, Soluble in water Eg : Insulin ,Albumins Structure and shape 1) Primary structure : It is a specific sequence of amino acids of Proteins: 2) Secondary structure: It represent shape ie. α - halix and β- pleated sheet. i) α - halix : polypeptide chain twisted in to a right handed screw by forming H-bonds b/w NH group and >C=O grup. ii) β- pleated sheet : peptide chains laid side by side and held together by H-bonds 3) Tertiary structure: It represent further folding of the secondary structure. It gives rise to two major molecular shapes viz. fibrous and globular. 4) Quaternary structure of proteins: composed of two or more polypeptide chains referred to as sub-units. The spatial arrangement of these subunits with respect to each other. Denaturation of Disturbing the 20 and 30 structures of proteins by heating or changing pH. eg: proteins coagulation of egg white on boiling. Nucleic acids Polymer of nucleotides linked by phosphodiester linkage -3’ 5’ linkage Nucleotide Each nucleotide contains N-base, Sugar and Phosphate. Nucleoside Contains N-base & Sugar. Double helical Two strand of DNA coiled around each other and held together by H-bonds structure of DNA b/w pairs of bases. Such as − C≡G−, and −A =T− Purines : Adinine & Guanine And Pyrimidines : Cytosine, Uracil & Thymine DNA N-Bases : A,G ,C &T, Double helix, Contains 2-deoxy ribose sugar, Transfer heredity characters. RNA N-Bases: A,G C & U, Single helix, Contains Ribose sugar, Helps in proteins synthesis. Preparation of Glucose  Structure of Glucose Reactions of Glucose Reactions that prove cyclic i. Aldehyde group present but glucose does not react with NaHSO3& NH3. structure of Glucose ii. Glucose does not give the Schiff’s Test & 2,4-DNP test for aldehyde. iii. Glucose penta-acetate does not react hydroxyl amine, which shows that aldehyde group is absent in glucose. iv. Glucose exist in two stereo-isomeric forms ( &).  All observations indicate that free aldehydic group is not present in glucose. Cyclic Structure of Glucose Structure of Nucleotide Prepared By:- Dr M R Choudhary K V SAC AHMEDABAD 8469113570