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Caie a2 level chemistry 9701 theory v1
Chemistry of the Living World (CHEM 110)
University of Auckland
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ZNOTES
SUMMARIZED NOTES ON THE THEORY SYLLABUS
CHEMISTRY (9701)
CAIE A2 LEVEL
UPDATED TO 2022 SYLLABUS
1. Chemical Energetics
1. Enthalpy change of Atomization
Enthalpy change of atomization ( ): enthalpy change when 1 mole of gaseous atoms is formed from its elements under standard conditions e: Na (s) Na (g) e: 1/2Cl 2 (g) Cl (g) Note: specic value may not be given and you must use combination of enthalpies e.
1. Lattice Energy
Lattice Energy ( ): enthalpy change when one mole of an ionic compound is formed from its gaseous ions under standard conditions. e: Na+(g) + Cl-(g) NaCl (s) More –ve stronger ionic bonding more stable ionic compound If theoretical and experimental value similar, bonding is pure ionic, otherwise it is intermediate between ionic and covalent
1. Factors Aecting Lattice Energy
Radius of ion: the smaller the radius/smaller the size of ion, greater the charge density Charge on ion: Greater charge density, greater electrostatic attraction between ions, more exothermic, more stable compound
1. Electron Anity
First electron Anity ( ): enthalpy change when 1 mole of electrons is added to 1 mole of gaseous atoms to form 1 mole of gaseous anions under standard conditions e: Cl (g) + e- Cl- (g) 1 st electron anity is exothermic 2 nd, 3rd... electron anities are endothermic because when electron added to –ve ion, increased repulsion present therefore requires input of energy
1. Born-Haber Cycle
of an element under standard conditions = 0 Endothermic Processes: Bottom to top: Exothermic Processes: Top to bottom:
1. Dissolving Salts
Enthalpy change of solution ( ): enthalpy change when one mole of ionic solid is dissolved in sucient water to form a very dilute solution under standard conditions (can be +ve or –ve) e. NaCl(s) + aq NaCl(aq) Enthalpy change of hydration ( ): enthalpy change when one mole of specied gaseous ions dissolves in sucient water to form a very dilute solution under standard conditions e. Na+(g) + aq Na+(aq) Factors aecting factors aecting Solubility of ionic salts depend on value of : the more –ve = more soluble
1. Ion Polarization
Ion polarization: distortion of the electron cloud on an anion by a neighbouring cation
####### ΔHat θ
####### →
####### →
####### ΔHlat θ
####### →
####### ΔH lat θ → →
####### ΔHlat θ
####### ΔH lat θ ∝
####### Radius of Ion
####### Charge on Ion
####### ΔHlat θ
####### ΔHea θ
####### →
####### ΔHfθ
####### ΔHiθ
####### ΔHat θ
####### ΔH ea θ 2 nd and 3rd
####### ΔHfθ
####### ΔHea θ
####### ΔHlatt θ
####### ΔHhyd θ
####### ΔHsol θ
####### →
####### ΔHhyd θ
####### →
####### Δ Hθ hyd = Δ Hθ lat
####### ΔHθ sol
1 faraday (F): quantity of electric charge carried by 1 mol of electrons = 96,500 C Relationship: = Faraday’s Constant = Avagadro’s Constant = charge on 1 electron Example: To liberate 1 mol of H2(g) 2H+(aq) + 2e- H2(g) requires 2 faradays of electricity to form 1 mole
2. Standard Electrode Potentials
Standard electrode (redox) potential : the electrode potential of a half cell measured under standard conditions using a standard hydrogen electrode as the other half cell Standard cell potential : the dierence in standard electrode potential between two half cells Standard conditions: Temperature: 298 K Pressure: 1 atm (101 kPa) Concentration: 1 mol dm-
2. Electrochemical Cell
ANODE -ve electrode/half cell CATHODE +ve electrode/half cell Eθ of half-cell is more -ve (less +ve) Eθ of half-cell is more +ve (less -ve) Metal loses e-, forms +ve ions = oxidation occurs Metal ion gains e-, solid metal deposited = reduction occurs Cell Notation Zn(s) | Zn2+(aq) Cu2+(aq) | Cu(s) The more –ve (less +ve) the Eθ, the more easily it gets oxidized ∴ stronger reducing agent The more +ve (less -ve) the Eθ, the more easily it gets reduced ∴ stronger oxidizing agent Eθcell = EθRed - EθOx Eθcell = 0 - -0 = 1.
2. Salt Bridge
Used to complete the electrical circuit allowing the movement of ions between two half cells so that ionic balance is maintained; electrically neutral A strip of lter paper soaked in KNO3(aq) Salt used should not react with ions forming a ppt. e. KCl in a cell containing Ag2+ would form insoluble ppt.
2. Standard Hydrogen Electrode
(S.H.)
It is a reference cell, used to measure of any other half cell Consists of H2(g) at 298K and 1atm bubbling around a Pt electrode in contact with an aq. solution of H+ ions at 1. mol dm-3 conc. (e. HCl(aq) or ½ H 2 SO4(aq)) Pt electrode is inert, allows conduction and is coated with nely divided Pt serving as a catalyst. Being porous, retains large amounts of H2(g)
2. Measuring of a Half Cell using
S.H.
Metals in contact with their aq. solution
####### F = Le
####### F
####### L
####### e
####### →
####### ∴
####### Eθ
####### Ecell θ
####### Eθ
E
####### θ
Non-metals in contact with their aq. solution Pairs of ions with dierent oxidation states
2. Feasibility of a Reaction
For a spontaneous (i. feasible) reaction, must be positive Finding whether a reaction is feasible:Will chlorine oxidize ions to Fe^{3+}$ ions? Solution Identify which is the R.A./O. from questionChlorine is the O therefore is reduced Identify relevant equations from data booklet and place in correct direction. If using opposite direction, switch sign for value of Eθ Add Eθs together. If positive, reaction occurs and is feasible otherwise, does not occur spontaneously. Reaction occurs spontaneously and is feasible
2. Oxidizing Ability of Halogens (Gp.
17)
Electrode potential values can be used to determine ease of oxidation or reduction The more +ve , the easier it is to reduce therefore stronger oxidizing agent (ability). Relative reactivity of halogens can be determined by ordering them in descending order of , most reactive to least Reaction F 2 + 2e- 2F- +2. Reaction Cl 2 + 2e- 2Cl- +1. Br 2 + 2e- 2Br- +1. I 2 + 2e- 2I- +0.
2. Concentration of aq. Ions
Fe3+ + e- Fe2+ = 0 If increased or decreased, equi. favours forward reaction becomes more +ve (0) If increased or decreased, equi. favours backward reaction becomes more -ve (0) A given reaction will occur under non-standard conditions if the values of the two half reactions involved dier by more than 0 Changing conc. and non-standard conditions can also cause non-feasible standard reactions to occur as values can change greatly
2. Nernst Equation
Equation relating voltage of chemical cell to its standard potential and to concentrations of reactants & products is number of electrons transferred in reaction is the value of the Faraday constant 96,500 C mol- Simplied relationship by substituting standard values If conc. less than 1 mol dm-3, is –ve and is less than If conc. more than 1 mol dm-3, is +ve and is more than
2. Hydrogen Fuel Cell
Consists of: Two platinum coated porous electrodes that allow gases to pass through Electrolyte; either acidic or alkaline
####### Ecell θ
####### F e2+
####### Cl 2 + 2 e − →Cl−
####### E θ =+1.
####### F e 2+ →F e 3+ +e−
####### E θ =− +0( )
####### 1 − 0 = 0.
####### Eθ
####### Eθ
####### E
####### E
θ θ
####### →
####### →
####### →
####### →
####### ⇌ Eθ
####### F e3+ F e2+
####### ∴ E
####### F e2+ F e3+
####### ∴ E
####### Eθ
####### E
####### E = E θ + ln
####### zF
####### RT
####### [reduced f orm]
####### [ oxidized form ]
####### z
####### F
####### E = E θ + log
####### z
####### 0.
####### [reduced f orm]
####### [ oxidised form ]
####### log [oxidized f orm]
####### E Eθ
####### log [oxidized f orm]
####### E Eθ
pH of an alkali can be calculated using this formula Calculating pH of a strong base: Calculate the pH of a solution of sodium hydroxide of concentration 0 mol dm- Solution It is a strong base and dissociates completely the conc. of ions is equal to conc. of the solution Calculate pOH by logging Calculate pH by using the following
3. Buer Solutions
Buer solution: a solution in which pH doesn’t change signicantly when small amounts of acids or alkali added A buer solution can be: Acid: Weak acid + its salt e. CH 3 COH and CH 3 COONa Alkali: Weak alkali + its salt e. NH 3 and NH 4 Cl In both types, the salt provides more of its conjugate pair, for example HA(aq) H+(aq) + A-(aq) & MA(aq) M+(aq) + A-(aq)
3. Working of an Acid Buer
3. Controlling pH of Blood
Respiration in cells produces carbon dioxide and carbon dioxide combines with water in blood CO2(aq) + H 2 O(aq) H+(aq) + HCO3-(aq) The enzyme carbonic anhydrase is present to supply HCO3- ions and increase its conc. similar to acidic buers If H+ increases: If H+ decreases: Eq. shifts to the left Eq. shifts to the right Reduces H+ Increases H+ pH of blood unchanged
3. Calculating pH of Buer Solutions
Calculating pH of buer solution: A buer solution is made by adding 3 of CH 3 COONa to 1 dm 3 of 0 mol dm-3 CH 3 COOH 1. What is the pH of this buer (Ka ) What is the change in pH when 1 cm 3 of 1 mol dm-3 NaOH is added to 1 dm 3 of the buer? Solution Part (a): Calculate the moles and concentration of the salt Moles Concentration Use formula to calculate pH: Part (b): NaOH reacts with acid: NaOH(aq) + CH 3 COOH(aq) CH 3 COONa(aq) + H 2 O Calculate no. of moles of NaOH used: Moles Hence that much of acid has been reduced and the same amount has formed as salt. Calculate new concentrations of salt and acid. Conc. of Acid Conc. of Salt Use formula to calculate pH: Find the change in pH:
####### 14 = pH + pOH
####### [OH − ] =[NaOH = ] 0.
####### pOH = − log 0 = 1.
####### pH + pOH = 14
####### pH = 14 − pOH = 14 − 1 = 12.
####### ⇌ →
####### ⇌
####### pH = pK a +log (
####### [Acid ]
####### [ Salt ]
####### )
####### = 1 × 10−
####### = =
####### Molar Mass
####### Mass
####### =
####### 82
####### 3.
####### 0.
####### = =
####### Volume
####### Moles
####### =
####### 1
####### 0.
####### 0.
####### pH = − log (1 × 10 −5 ) + log ( =5.
####### 0.
####### 0.
####### )
####### = Concentration × V olume = 1 × 0 = 0.
####### = =
####### 1 + (1 × 10 −3 )
####### ( 0 × 1 ) – 0.
####### 8 × 10−
####### = =
####### 1 + (1 × 10 −3 )
####### 0 + 0.
####### 0.
####### pH = − log (1 × 10 −5 ) + log ( =5.
####### 8 × 10−
####### 0.
####### )
Calculating pH of buer solution with dierent volumes of acid and salts: What is the pH of the buer solution formed when 40cm 3 of 1 mol dm-3 of nitrous acid (pKa 3) is added to 20 cm 3 of 1 mol dm-3 of sodium hydroxide? Solution Part (a): Write out an equation for the reaction HNO2(aq) + NaOH(aq) NaNO2(aq) + H 2 O(l) Calculate the moles of each reactant Moles of HNO 2 Moles of NaOH NaOH is limiting reagent therefore moles of product and reactants are: NaNO 2 formed $= 0$ HNO 2 remaining $= 0 - 0 = 0$ Calculate concentration of salt and acid: Total Volume dm 3 Conc. of HNO 2 Conc. of NaNO 2 Use formula to calculate pH:
3. Solubility Product Ksp
When a sparingly soluble salt is shaken with distilled water and left to settle, the water contains aq. ions from the salt in very small conc. and a dynamic equilibrium is set up between ions and insoluble solid Solubility product (Ksp): product of conc. of each ion in a saturated solution of sparingly soluble salt at 298K raised to the power of moles in dissociation equation e. Fe 2 S3(s) 2Fe3+(aq) + 3S2-(aq) Ksp = 2Fe3+23S2-3 units: mol 5 dm- The conc. of ions is independent of amount of solid Ksp values change only with temperature Concentration of ions = solubility of salt Calculating solubility product from solubility: A saturated solution of magnesium uoride, MgF 2 , has a solubility of mol dm-3. Calculate the solubility product of magnesium uoride. Solution Write down the equilibrium equation 2F-(aq) Write down the equilibrium expression Solubility of salt = conc. of ions Substitute into equilibrium expression To calculate solubility from solubility product, apply the same method but place for conc. and nd it
3. Predicting Precipitation
Precipitation: when two aq. solutions of ionic substance are mixed, a ppt. will form provided the Ksp of the insoluble solid is less than ionic product of ions present Ionic Product > Ksp Predicting if insoluble salt will precipitate: Will a ppt. BaCO 3 (Ksp = ) form from solution containing 0 mol dm-3 Ba2+(aq) & 1 mol dm-3 CO32-(aq) Solution Calculate ionic product: Compare to Ksp value $1 > 8 \times 10^{- 9}\therefore$ ppt. will form
3. Common Ion Eect
Common ion eect: lowering of solubility of an ionic compound by addition of a common ion to the solution Calculating solubility with common ion: BaSO 4 is a sparingly soluble salt (Ksp mol 2 dm-6). Compare the solubility at 298K of BaSO 4 in an aq. solution of 0 mol dm-3 H 2 SO 4 to in pure water Solution Calculate conc. of Ba2+ in pure water Ksp = $\lbrack Ba^{2+}\rbrack \lbrack So_4^{2-} \rbrack$ and both conc. are equal $\therefore$ To calculate$Ba2+, rst nd conc. of common ion, SO42- H 2 SO 4 is a strong acid which dissociates completely
####### ΔpH = 5 − 5 = 0.
####### =
####### = 1 × (40 × 10 −3 ) =0.
####### = 1 × (20 × 10 −3 ) =0.
####### (20 + 40 ) × 10 −3 =0.
####### = 0 0 = 31
####### = 0 0 = 31
####### pH = 3+ log ( =3.
3 1 3 1
####### )
####### ⇌
####### 1 × 10−
####### M gF − 2(s) ⇌ M g(2+aq)
####### K sp =[M g 2+][2 F − 2]
####### ∴ [M g 2+ ] =[F − ] =1 × 10−
####### K sp =(1 ×
####### 10 −3 ) ×2 × (1 × 10 ) 2
2
####### K sp =7 × 10 −9 mol dm 3 −
####### x
####### 8 × 10−
####### 0 × 1 = 0.
####### = 1 × 10−
####### [Ba 2+ ] = 1 × 10−10 =1 × 10−
####### ∴ [H 2 SO 4 ] =[SO 4 ] =
2−
####### 0.
pc
3. Working of an Alkali Buer
4. Reaction Kinetics
4. Rate of Reaction
Rate of reaction is the change in concentration of products/reactants per unit time Unit: mol dm-3 s-
4. Rate Equation
Rate equations: states the relationship between the rate constant and the concentrations of those reactants Rate constant ( ): proportionality constant in rate eqn. Larger the rate constant, faster the reaction Depends on temperature Unit depends on overall order of reaction Order of reaction ( : power to which conc. of the reactant is raised in experimentally determined rate eqn. Overall order of reaction Half-life, of a reaction is the time taken for conc. of one of the reactants to fall by half
4. Order of Reaction
Order Half-life, t1/ Graph of Conc./Time Rate/Conc. Order Half-life, t1/ Graph of Decreases Constant Increases For zero order, rate of reaction not dependant on conc. of that reactant but it is need for completion of reaction. Reaction continues until all of that reactant is used up. Recognize shapes above in graphs given to work out order of reactant and construct rate equation. Calculate using data from graph(s). Constructing rate equation using conc. data: Using following data from the experiments construct a rate equation and calculate rate constant. Solution Use data and deduce orders by looking at changes in conc. and rates (arrows added above): Order of A = 1 Order of B = 1 Write as rate equation Calculate* * using data given from e. experiment 1
4. Calculating Rate Constant from t1/
Only used with rst order reactions
####### K = conc. of NH∗3(aq) =
conc. of NH∗3(org) 100×10− 9×10− 50×10− 6×10−
####### K pc =0.
####### R = t
[ ]
####### xA + yB → P roducts
####### R = k A[ ] B
m
####### [ ]
n
####### k
####### m, n)
####### = m + n
####### t 21
####### [A] 0
####### [A] 1
####### [A] 2
####### k
####### R = k A[ ][B]
####### k
####### k = =
####### [ A ][B]
####### R
####### =
####### 0 × 0.
####### 4 × 10−
####### 3
####### 1
####### ∴ R = A B
####### 3
####### 1 [
####### ] [ ]
Use the relationship:
4. Relationship of Temperature and k
Increasing the temperature, increases the value of When temp. is increased, the k. of reacting molecules increases resulting in more successful collisions Reactants change faster to products therefore conc. of reactants decreases. Using fraction above, numerator increases and denominator gets smaller increases.
4. Reaction Mechanism
Rate-determining step: the slowest step in a reaction mechanism In a multistep reaction: The rate of reaction is dependent on the slowest step that needs the highest activation energy. Rate equation includes only reactants that are present in the rate-determining step. The orders with respect to the reactants are the moles of the reactants in the rate determining step Hydrolysis of Alkyl Halide Primary alkyl halide mechanism: Tertiary alkyl halide mechanism: Constructing rate equation for multistep reaction The mechanism for the production of NO 2 F involves NO 2 + F 2 NF 2 + O 2 NF 2 + NO 2 NO 2 F + NF NF + O 2 NO 2 F a. What is the overall stoichiometric equation? b. Write a rate equation for the reaction. Solution Part (a) List all the reactants together and products together Reactants: NO 2 + F 2 + NF 2 + NO 2 + NF + O 2 Products: NF 2 + O 2 + NO 2 F + NF + NO 2 F Cancel the common things and form equation 2NO 2 + F 2 2NO 2 F Part (b) Rate equation will include only reactants of slowest step therefore:
4. Measuring Reaction Rates
Sampling: method that involves taking small sample of a reaction mixture at various times and then carrying out chemical analysis on sample. Chemical Analysis: C 4 H 9 Br + OH- C 4 H 9 OH + Br- Sampled removed at various times and quenched – stopping/slowing down reaction (e. cooling in ice) OH- conc. can be found using titration Plot graph and calculate rate of reaction Continuous: method that involves monitoring a physical property over a period of time Change in Volume of Gas Produced: Mg + 2HCl MgCl 2 + H 2 Measure change in volume of gas using a gas syringe Take down readings at regular intervals Plot graph and calculate rate of reaction Changes in Colour: CH 3 COCH 3 + I 2 CH 3 COCH 2 I + HI I 2 starts brown, fades through orange to yellow to colourless as iodine used up. Colorimeter measures amount of light absorbed as it passes through solution; recorded as absorbance. Before experiment, create calibration curve by nding absorbance of dierent conc. I 2 and plot a graph of
####### k =
####### t 21
####### 0.
####### k
####### k =
####### [A ][B]
####### R
####### ∴ k
####### C 3 H 7 CH 2 Br + OH −
slow
####### [C 3 H 7 BrOH]
####### [C 3 H 7 BrOH]
f ast
####### C 3 H 7 CH 2 OH +Br−
####### ∴ k[C 3 H 7 CH 2 Br][OH −]
####### (CH 3 ) 3 CBr
slow
####### (CH 3 )C + + Br−
####### (CH 3 )C + + OH −
f ast
####### (CH 3 )COH + Br−
####### ∴ R = k[(CH 3 ) 3 CBr]
slow f ast f ast
####### R = k[N O 2 ][F 2 ]
Honeycomb structure containing small beads coated with Pt, Pd and Rh Possible catalytic process: Adsorption of NOx and CO Weakening of covalent bonds within NOx and CO Formation of new bonds between Adjacent N atoms form N 2 CO and O atoms form CO 2 Desorption of N 2 and CO 2
5. Chemistry of Transition
Elements
Transition element: a d-block element that forms one or more stable ions with incomplete d orbitals
5. Electronic Congurations
Element Electronic Cong. Oxidation States Sc, scandium [Ar] + Ti, titanium [Ar] +3, + V, vanadium [Ar] +2, +3, +4, + Cr, chromium [Ar] +3, + Mn, manganese [Ar] +2, +4, +6, + Fe, iron [Ar] +2, + Co, cobalt [Ar] +2, + Ni, nickel [Ar] + Cu, copper [Ar] +1, + Zn, zinc [Ar] + Scandium and Zinc are not transition metals because: Sc3+ has no e-s in the d-orbital Zn2+ contains a full d-orbital Chromium and copper have anomalous congurations Cr: 4s electron demoted to half-ll 3d shell Cu: 4s electron demoted to full-full 3d shell creating a more stable conguration. When electrons added, ll 4s before 3d When electrons removed, remove from 4s before 3d
5. d-Orbitals
, and have d-orbitals in between axis and have d-orbitals along axis is formed by the merging of and
5. Physical Properties
5. Variable Oxidation States
Small energy dierence between 4s and 3d so electrons from both subshells can be removed to form a variety of oxidation states All transition metal exhibit two or more oxidation states Most common oxidation state +2 when 2e-s from 4s lost Transition elements show highest oxidation states when they combine with O or F (most electro-ve) When transition elements form compounds with high oxidation states above +4, they form large oxyanions and are covalent (acidic oxides) e. CrO4- or MnO4- When transition elements in lower oxidation states they form ionic compounds (basic oxides)
####### 3 d 1 4 s 2
####### 3 d 2 4 s 2
####### 3 d 3 4 s 2
####### 3 d 5 4 s 1
####### 3 d 5 4 s 2
####### 3 d 6 4 s 2
####### 3 d 7 4 s 2
####### 3 d 8 4 s 2
####### 3 d 10 4 s 1
####### 3 d 10 4 s 2
####### dxy dxz dyz
####### dx 2 −y 2 dz 2
####### dz 2 dx 2 −z 2 dy 2 −z 2
5. Complexes
Complex: is an ion or molecule formed by a central metal atom/ion surrounded by one or more ligands A complex consists of: Central transition metal ion (+ve) that can accept e-s Ligand (-ve): a species that contains a lone pair of e-s that forms a dative bond to a central metal atom/ion Coordination no.: number of coordinate or dative bonds to the central metal atom/ion Dierent metal ions show dierent coordination number with same ligands Transition metals form complexes because Ion are small in size so they have a strong electric eld around them which attract e--rich ligands They have empty 4s and 4p orbitals that are hybridised and can accept e-
5. Ligands
Monodentate ligands: forms only one coordinate bond with central metal ion (donates one pair of e-s) Anions Neutral Ligands Halide ions F-, Cl-, Br-, I- Water H2O Sulphide S2- Ammonia NH Nitrite NO22- Carbonyl CO Hydroxide OH- Cyanide CN- Thiocyanate SCN- Bidentate ligands: forms 2 coordinate bonds with central metal ion (donates 2 pairs of e-s per molecule) H 2 N – CH 2 – CH 2 – NH 2 - O – CO – CO – O- Ethylene diamine (en) (neutral) Oxalate ion (oxalato-) png" width=20%> Forms 6 coordinate bonds 4 from oxygen 2 from nitrogen
5. Writing Names of Complexes
To represent coordination number For monodentate ligands, use prexes mono-, di-, tri-, tetra-, hexa- For bi and polydentate ligands, use prexes mono-, bis-, tris-, tetrakis-, pentakis- Name the ligands If neutral and -ve ligands present, rst name -ve ligand If ligands all neutral/-ve or contain more than one of each, name in alphabetical order Anions Neutral Ligands Fluoro (F-), Chloro (Cl-) Amine (NH 3 ) Cyano (CN-) Aqua (H 2 O) Hydroxo (OH-) Name central metal ion If complex ion +ve/neutral – use normal name If complex ion –ve, use name ending –ate; special: Iron = Ferrate Lead = Plumbate Copper = Cuprate Oxidation no. of metal ion in Roman numeral and ‘ion’
5. Writing Formulae of Complexes
Metal ion written rst followed by ligand Place charge on formula in square brackets Total charge on complex = sum of charge on metal ions and charge on ligand If ligands all neutral, charge on complex = charge on metal ion
5. Shapes of Complexes
Coordination No. & Shape Diagram 6 Octahedral [Fe(H 2 O) 6 ]2+ 4 Tetrahedral [CoCl 4 ]2- 4 Square Planar
The greater the Kstab value, the more stable the complex Kstab values may be given on log 10 scale as values large Stability constants for complexes with bi/polydentate ligands very high
5. Colour of Complexes
The ve d-orbitals in isolated transition metal atoms/ions are degenerate; all at the same energy Coordinate bonding from ligands cause ve d-orbitals to split into two sets of non-degenerate orbitals at For octahedral complexes, ligands approach along axis and increases repulsion with and orbital causing them to be at higher energy e. an octahedral complex of Cu2+ Electrons from lower energy orbitals absorbed energy equal to from light and are excited to a higher energy level orbital Wavelengths of light equivalent to are absorbed and the rest are transmitted. Wavelengths transmitted merged together corresponds to the colour of the solution observed Conditions for complex to be coloured: At least one d-orbital must be occupied by an e- At least one d-orbital must not be fully occupied Hence, some compounds of d-block metals colourless: All Sc compounds colourless because Ar 3d 0 All Zn compound colourless because Ar 3d 10 Factors that determine colour of complex (which aects the splitting of d-orbitals): Magnitude of Strength of ligands Oxidation state of metal ion Geometry of complex Colour Spectrum R O Y G B I V Large = light absorbed from blue = complex red-ish Small = light absorbed from red = complex blue-ish
6. Arenes
6. Bonding in Benzene Ring
C – C bond length equal = 0 Benzene is a symmetrical at hexagonal planar molecule Consists of 6 C – C single bonds and remain electrons exist in delocalized system All C – C bonds are the same length (0) which are smaller than a single bond but greater than a double Each carbon forms 3 sigma bonds ( hybrid) 2 between carbon and carbon (on either side) 1 between carbon and hydrogen A p-orbital from each C atom overlaps above and below the plane of the ring forming two continuous loops; each C atom donates 1 electron into this bond (delocalized)
6. Properties of Benzene
Clear, colourless, non-polar liquid at r.t It is volatile and has a low b. Relatively unreactive due to strong bonding Burns with smoky ame
6. Naming Aromatic Compounds
Name with Benzene Name with Phenyl Chlorobenzene Phenol Methylbenzene Phenylethene Nitrobenzene Phenylamine Benzoic acid Phenyl ethanoate
####### ΔE
####### dx 2 −y 2 dz 2
####### ΔE
####### ΔE
####### [ ]
####### [ ]
####### ΔE
wavelength increases f requency and energy increases
####### ΔE
####### ΔE
####### sp 2
####### π
6. Electrophilic Substitution
Mechanism
An H-atom on the ring can be substituted by other atoms/groups All carbon atoms identical and benzene ring symmetrical hence any H-atom can be substituted An electrophile (E+) can attack the ring of e-s but benzene highly stable so E+ must be very strong. A +ve ion can be generated by heterolytic ssion of a covalent bond hence catalyst needed to generate E+ Mechanism: E+ attacks e- ring, accepting a pair of e- from benzene An intermediate +ve specie is formed in which +ve charge is delocalized around the ring (more stable than a normal carbocation) Proton eliminated from +ve intermediate to restore stability and system of delocalized e-s reformed
6. Electrophilic Substitution Reactions
Halogenation C 6 H6(l) + X2(g/l) C 6 H 5 X(l) + HX(g) Reagent: dry Cl 2 gas / pure Br 2 liquid Conditions: Temp./press.: r.t. Catalyst: anhydrous AlCl3(s) / FeBr3(s) Generating electrophile: catalysts are e- decient halogen carriers, form dative bond drawing e- from halogen molecule producing X+ AlCl 3 + Cl 2 AlCl4 - + Cl+ Regenerating catalyst: H+ + HCl + AlCl 3 Nitration C 6 H6(l) + H 2 SO 4 /HNO 3 C 6 H 5 NO2(l) + H 2 SO4(g) + H 2 O Reagent: nitrating mixture conc. HNO 3 / conc. H 2 SO 4 Condition: Temp.: 45-55oC under reux Generating electrophile: 2H 2 SO 4 + HNO 3 2HSO4- + NO2+ + H 3 O+ Regenerating catalyst: 2HSO4- + H 3 O+ + H+ 2H 2 SO 4 + H 2 O Alkylation / Acylation C 6 H6(l) + RCl/RCOCl C 6 H 5 R/C 6 H 5 RCO + HCl(g) Reagent: alkyl halide (RCl) / acyl chloride (RCOCl) Condition: Temp.: heat under reux Catalyst: anhydrous AlCl 3 Generating electrophile: AlCl 3 + RCl/RCOCl AlCl4 - + R+/RCO+ Regenerating catalyst: H+ + AlCl4 - HCl + AlCl 3
6. Electrophilic Addition Reaction
Hydrogenation C 6 H6(l) + 3H2(g) C 6 H 12 Type of reaction: electrophilic addition Conditions: Temp.: 150ºC Catalyst: nickel
6. Side Chain Reactions
Chloromethyl Benzene C 6 H 5 CH3(l) + Cl2(g) C 6 H 5 CH 2 Cl(l) + HCl Type of reaction: free radical substitution Reagent: Cl2(g) Condition: u. light With excess Cl 2 a mixture of di, tri chloromethyl benzene can be obtained Benzoic Acid C 6 H 5 CH3(l) + 3 O C 6 H 5 COOH(l) + 2H 2 O Type of reaction: oxidation Reagent: KMnO4(aq) or K 2 Cr 2 O7(aq) Condition: heat under reux (2 to 3 hours) Alkyl gp. always oxidized regardless of chain length If more than one alkyl gp., each one oxidized
6. Ring Activating/Deactivating
Groups
####### π
####### π
####### π
catalyst
####### → [ ]
####### [AICl 4 −] →
####### →
####### →
####### →
AlCl 3
####### [ ]
####### [ ]
N ickel
####### →
####### [ ] →
Reaction with Sodium C 6 H 5 OH(l) + Na(s) C 6 H 5 O-Na+(aq) + ½H2(g) Type of reaction: metal and acid, redox Condition: r.t. Reacts with Na liberating H 2 gas Note: phenol does not react with metal carbonate to liberate CO 2 hence shows its fairly weaker than other carboxylic acids
8. Reaction with Benzene Ring
Bromination Type of reaction: electrophilic substitution Reagent: aq. Bromine Condition: r.t. (no catalyst) Rapid reaction, forms a white ppt. No catalyst needed since Br 2 molecule easily polarized by increase e- density in ring Nitration Type of reaction: electrophilic substitution Reagent: dilute nitric acid Condition: r.t. If concentrated nitric acid used, then trinitrophenol produced
8. Relative Acidities
The stronger the acid: Higher the Ka value More easily H+ is donated More stable is the conjugate base Ethanol: C 2 H 5 OH + H 2 O C 2 H 5 O- + H 3 O+ Ka: 10-18 mol dm- Ability to donate H+ ions: C 2 H 5 is an e- donating gp. which increases charge density on O of OH More attraction between O–H so loss of H+ dicult Stability of conjugate base: In ethoxide ion, C 2 H 5 gp. increases –ve charge Makes ethoxide ion more basic than OH- Ability to accept H+ increases, moving equi. to left Water: H 2 O + H 2 O OH- + H 3 O+ Ka: 10-16 mol dm- No e- donating or withdrawing gp. present hence ability to donate and accept H+ ion is the same Phenol: C 6 H 5 OH + H 2 O C 6 H 5 O- + H 3 O+ Ka: 10-10 mol dm- Ability to donate H+ ions: OH is a ring activating gp. & the lone pair of e-s on O becomes part of delocalized e- system Decreases e- density on O of OH and attraction between O and H decreases so H+ lost more easily Stability of conjugate base: In phenoxide ion, -ve charge on O delocalized around ring and reduces tendency to attract H+ Conjugate base stable
8. Test for Phenol
r.t
####### ⇌
####### ⇌
####### ⇌
Test 1: Reagent: Iron(III) Chloride (FeCl3(aq)) – yellowish brown Observation: violet colour obtained Test 2: Reagent: bromine water (Br2(aq)) – orange Observation: white ppt. obtained
9. Acyl Chlorides
9. Preparation of Acyl Chlorides
These are derivatives of carboxylic acids; -OH gp. of acid replaced by Cl Named after corresponding acid using sux –oyl followed by chloride e. ethanoyl chloride
9. Reactivity of Acyl Chlorides
Acid chlorides are much more reactive than the carboxylic acid from which they are derived Polarity of both C O and C—Cl bond produces relatively large + charge on the carbon, making it e- decient and more susceptible to attack by a nucleophile
9. Nucleophilic Condensation
Mechanism
The carbonyl carbon being highly e- decient is attacked by nucleophile The Nu bonds to the carbon atom, the C O bond is broken leaving O with –ve charge Chloride is lost and carbonyl double bond is reformed
9. Nucleophilic Condensation
Reactions
Hydrolysis Forming Amides Primary amides: R—CO—Cl + NH3 →R—CO—NH2 + HCl Reagent: ammonia Secondary amides: R—CO—Cl + R’NH2 → R—CO—NHR’ + HCl Reagent: alkyl amine Note: If excess NH 3 used, the HCl formed is neutralized to NH 4 Cl Esterication R—CO—Cl + HO–R → R–CO–OR(aq) + HCl(aq) Reagent & Condition: Aliphatic alcohol at r.t Phenol dissolved in NaOH, warm it Hydrolysis of esters:
9. Relative Ease of Hydrolysis
####### =
####### δ
####### =
Caie a2 level chemistry 9701 theory v1
Course: Chemistry of the Living World (CHEM 110)
University: University of Auckland
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