E12 10004 2021 Student Report Template PDF

Title E12 10004 2021 Student Report Template
Author Jane Doe
Course Chemistry 2
Institution University of Melbourne
Pages 6
File Size 193.5 KB
File Type PDF
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REPORT COVER SHEET AND DECLARATION School of Chemistry The University of Melbourne Laboratory Report Cover Sheet Student Name: Student Number: Subject Name & Code: Demonstrator: Experiment Title:

Meraal Zaib 1268766 CHEM10004 Jacob Rowan Experiment 12: Coordination Chemistry – Synthesis of Hexaamminecobalt(III) Chloride.

Due Date: _________13th October 2021____________________ By submitting work for assessment, I hereby declare that I understand the University’s policy on academic integrity and I declare that: • This laboratory report is my own original work and does not involve plagiarism or unauthorised

collusion, except where due credit is given to the work of others. The report is based on results and spectra obtained by me during my laboratory session. • This laboratory report has not previously been submitted for assessment in this or any other subject. For the purposes of assessment, I give the assessor of this assignment the permission to: • Reproduce this laboratory report and provide a copy to another member of staff; and • Take steps to authenticate the assignment/laboratory report, including communicating a copy of this assignment to a checking service (which may retain a copy of the assignment on its database for future plagiarism checking). Feedback on Report: Feedback on your report and the mark you received will be available on the Online Practical Assignments page on Canvas. Plagiarism: Plagiarism is the act of representing as one's own original work the creative works of another, without appropriate acknowledgment of the author or source. Collusion: Collusion is the presentation by a student of an assignment as his or her own work, but which is in fact the result in whole or in part of unauthorised collaboration with another person or persons. Collusion involves the cooperation of two or more students in plagiarism or other forms of academic misconduct. Both collusion and plagiarism can even occur in group work. For examples of plagiarism, collusion and academic misconduct in group work please see the University’s policy on Academic Honesty and Plagiarism: https://academichonesty.unimelb.edu.au Plagiarism and collusion constitute cheating. Disciplinary action will be taken against students who engage in plagiarism and collusion as outlined in University policy. Proven involvement in plagiarism or collusion may be recorded on your academic file in accordance with Statute 13.1.18.

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Experiment 12: Coordination Chemistry – Synthesis of Hexaamminecobalt(III) Chloride. Author(s): Meraal Zaib Day/Time/Group number: Tues/2pm/21 Abstract: (Summary of what you did and what you found out) In this experiment, a typical coordination complex, hexaamminecobalt(III) chloride, [Co(NH3)6]Cl3 was synthesised.

Introduction and Aim: (What is a coordination complex? Why are they of interest? Which metals are most commonly involved? What is the most common geometry?) A molecule coordination complex is a New materials, biological applications

Experimental: (How did you perform your experiment?)

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Results and Discussion: REMEMBER TO SHOW ALL WORKING FOR CALCULATIONS AND TO SUPPORT YOUR ANSWERS WITH EXPERIMENTAL EVIDENCE OR THEORY Part A and B. Synthesis of hexaamminecobalt(II) chloride. Question 1 What characteristic is essential for a molecule to act as a ligand? Must have basicity and have at least one donor atom with an electron pair used to form covalent bonds with the central atom.

Yield Mass of hexamminecobalt(III) chloride: 0.22 g a) Calculate the theoretical yield of hexaamminecobalt(III) chloride based on the no. of mole of CoSO4 7H2O used. m(CoSO47H2O) = n*M = 2.66*10-3 mol* 267.48 g mol-1 = 0.711 g = 711 mg b) Express your actual yield as a percentage of the theoretical yield. Actual yield/theoretical yield*100% = 0.22 g / 0.711 g* 100% = 31%

Part C: Test Tube Reactions of Co(III) and Co(II) and Ni(II) Complexes Record your observations and write an ionic equation for any overall reaction, or No Change, as required. Test 1 Observations Overall Reaction equation Test 2 Observations

Overall Reaction equation

Reactions of Co(III) complex

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[Co(NH3)6]Cl3, add AgNO3

Before the silver nitrate is added, the solution is yellow, after addition of AgNO3, the solution turns darker yellow and a faint white precipitate can be seen forming. The solution becomes more opaque over time. Co3+ (aq) + Ag (s) + Cl- (aq)  Ag2Co (s) Reactions of Co(III) complex  [Co(NH3)6]Cl3, add i)heat and ii)NaOH i)heat No change to colour of litmus paper. The vapours are neutral in pH, could be water vapour. ii)add NaOH Overtime, the solution turned dark brown, and developed a black precipitate. i)heat No reaction.

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ii)add NaOH Co3+ (aq) + 2NaOH- (aq)  Co(OH)3 (s) + NaCl (aq)

Test 3 Observations

Explanation

Test 4 Observations

Reactions of Co(II) complex  [Co(OH 2)6]2+(aq) ion: stabilization energy 2[Co(OH2)6]2+ (aq) + 4Cl- (aq)[CoCl ⇋ (aq) + 6H2O (l) 4] Add HCl: Colour of cobalt went from orange to a rose, mauve colour. Adding more HCl, the solution then turned blue/purple. Cool: The solution turned more transparent and lighter purple in colour. Heat: The solution turned bright blue (colour it was when adding HCl) (Effect of heat, identify species formed on heating) The effect of heating the solution shifted the equilibrium to the right. By favouring the formation of the product ([CoCl 4]2-). The colour of the solution becomes what it was when adding HCl. Heating will favour the tetrahedral complex Reactions of Co(II) complex  [Co(OH2)6]2+(aq) ion: concentration effect Add HCl: Pink solution turns light blue Add water: The dilution with water resulted in the solution turning pink.

Explanation

Test 6 Observations

Overall Reaction equation

(Effect of HCl, identify species formed) The increase in chlorine ions will mean the eqiulbiruim will try to decrease chlorine concentration by pushing the reaction to the left. Le Chaterlier’s principle explains the shift of the equilibrium to the left as an increase in chlorine ions will favour the production of [Co(OH2)6]2+, or Co(II) which is blue in dry form. Reactions of Ni(II) complex, add NH3 Add water: A cyan solution forms. Add NH3: The solution turns cyan to a bright blue. A clear layer is visible at the top of the test tube, indicating the volatility of the ammonia. Ammonia replaces (displacement), which pushes the reaction forward. Add ethylenediamine: (Passive diffusion) The solution turns purple/violet after shaking. Heat: While a layers formed indicating competition between species, after mixing, test tube 1 contained a deeper purple solution, while test tube 2 was lighter purple. Add NH3: Ni2+ (aq) + NH3 (aq)  [Ni(NH3)6]2+ (aq) Add ethylenediamine: Ni2+ (aq) + (en) (aq)  tris(ethylenediamine)nickel(II) perchlorate (s) + Heat:

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Reactions of Ni(II) complex, add ethylenediamine

Test 7 Observations

Add water: Add ethylenediamine: Add NH3: Heat:

Overall Reaction equation

Add ethylenediamine: Add NH3: Heat:

Question 2 For Test 3, equation below, label the forward direction as exothermic or endothermic. Explain your answer using Le Chatelier’s Principle. 2[Co(OH2)6]2+ (aq) + 4Cl- (aq) [CoCl ⇋ (aq) + 6H2O (l) 4] Forward reaction is

Question 3 Draw the 3 coordination complexes formed in Test 6, beginning with the Ni(II) complex in water.

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Question 4 Provide an explanation for the results of Tests 6 and 7: i.compare the ligands H2O, NH3 and ethylenediamine – which one is most easily substituted?) ii. which is the most stable complex and why? Complex 3, as the number of ligands bound to it are greater. The denticity Enthalpy (tmep influecs how strongly entropy influences) Not spont as there is a small entropic barrier, as entropy is relatively small compared to enthalpy The bidente logand is stronger than monodentate (enthalpic) Theeqm constant more complex 3 is higher than complex 2 Complex 3 is relatively more stable than complex 2, we want a more negative enthalpy change Chelate effect Complex 2 chelating to metal is less stable

Conclusion: (What have you found out?)

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