LABORATORY EXPERIMENT REPORT, Assignments of Inorganic Chemistry

Download LABORATORY EXPERIMENT REPORT and more Assignments Inorganic Chemistry in PDF only on Docsity! Experiment 1 UV-Visible Spectroscopic Studies of hexaaquo-chromium (III) chloride, [Cr(H20)6]C13 and tetraaquo- manganese (II) chloride, [Mn(H20)4]C12 Introduction. Chromium compounds display a wide range of oxidation states and colors. Chromium is derived from the Greek word 'chromos' meaning mean color. Chromium shows the highest oxidation corresponding to the loss of all the outre electrons i.e. +6 but the oxidation +3 sites i.e. the most stable and most common. Compounds of Cr(IID) have the d3 configuration and are kinetically inert, undergoing slow ligand substitution reactions. The violet hexaquo ion [Cr(H20)6]3+ is a regular octahedral and exists in various salts such as the violet hydrate [Cr(H20)6]C13 and in extensive series of alums, MCr(SO4)2.12H20. The hexahydrate CrC13.6H20 exists in three isomeric forms. Manganese also displays a variety of oxidation states ranging from +2, +3, +4, +6, and +7 and all oxidation states from —3 to +7 have been observed. Mn2+ Compounds with oxidation states +5 (blue) and +6 (green) are strong oxidizing agents. The most stable oxidation state for manganese is +2, which has a pale pink color, and many manganese (II) compounds are known, such as manganese (II) sulfate (MnSO4) and manganese (II) chloride (MnC12). Manganese (II) most commonly exists with a high spin ground state because of the high pairing energy for manganese (II). However, a few examples of low-spin manganese (II) are known. The tetrahydrate MnCl2*4H20 is the most common form of "manganese (II) chloride" salts. Experiment Step 1: Preparation of reagents 1. Hexaaquachromium(I)chrolide, [Cr(H20)6]C13 was prepared by approximately weighing 0.2g of [Cr(H20)6]CI3 and dissolving the salt in a few mL of distilled water in 25mls volumetric flask and then fill to the flask mark. 2. Tetraaquo- manganese (II) chloride, [Mn(H20)4]Cl12 was prepared by approximately weighing 0.8g of [Mn(H20)4]Cl2 and dissolving the sail in few ml of distilled water in separate 25mls volumetric flask and then fill to the flask mark. Table 1. Reagents Weight Actual weight Diluted to [Cr(H20)6]C13 0.2g Approximated 25mls flask [Mn(H20)4]C12 0.8g Approximated 25mls flask Step 2: Instrument analysis UV-Visible Spectroscopic Set Up The instruments which was used in this experiment was Agilent Cary 60 Spectrophotometer. It was being operated in the scanning mode with the wavelength range of 200nm to 800nm. The analysis time was set at medium Analysis was done following the steps below; 1. The instrument was zeroed by filling the cuvette (1 cm) with distilled water as blank, and was inserted into the cell holder on the Agilent Cary 60 Spectrometer and press zero on the screen. The blank reading was recorded. 2. The cuvette was cleaned with distilled water and rinsed with prepared chromium solution 3. The cuvette was filled with chromium sample and placed in a cell holder. Scan was recorded by pressing start. 4. Step 2 and 3 three where repeated with prepared manganese sample. 5. After collecting the two spectra, the instrument analysis was stopped by placing finish. Results and Discussion Table 2. Absorbance reading of a blank Path Blank length(cm) Absorbance(nm) Distilled water 1 0 C=concentration=0.030mol/L l= cuvette (pathlength)=1.0cm. A= (Peak A) = 1.75, A= (Peak B) = 1.85. A (Blank) =0.001 PEAK A (Extinction coefficient) PEAK B(Extinction coefficient) £=A-0.001/C1 £= A-0.001/Cl1 =1.75 — 0.001/0.030M x 1.0 cm =1.85-0.001/0.030Mx1.0cm =58.3 M/I/cm/1 =61.6M/I/ cm/1 2. Compare the spectra of Cr and the Mn compounds and say why the Mn has such unnoticeable features why the Cr peaks are prominent Mn2+ are d5 ions and exhibit no spin-allowed transitions in either octahedral or tetrahedral environments. Transitions only occur via weak spin-orbit coupling leading to very low intensity bands and very pale colours. Spectra containing these ions is thus unoticeable. In the [Cr(H20)6] 3+ ion, the lower +3 charge on the metal means that the charge transfer bands occur in the ultraviolet and do not affect the colour. The charge transfer transitions involve excitation of an electron from an oxygenbased p-orbital (or lone pair orbital) into an empty metal d-orbital. They are fully orbitally allowed and hence the bands are intense and the colour is deep 3. Explain why compounds of Cr(III) under slow ligand substitution in their reactions. The Cr(IID) aqua ion has three electrons in the orbital of the d-shell, leading to considerable ligand field stabilization energy, and so undergoes ligand substitution reactions very slowly. Octahedral complexes with d° configurations, such as Cr** (d*), tend to be substitution- inert because of their high CFSE. 4. Give the structure and names of the three isomeric forms of hexahydrate CrCl3.6H20 OH2 Cl cl cl H2Ony,,, | ww wrOHe A201, | wwOHg 20/7, | aire! HOM, | wWOH2 —~ ; ~~ ; ~~ oN H0% | “ow, 4,0% | “ox, 10% | “oH, H,0% | “ou, OH2 OH2 OH, Cl (a) (b) (c) (d) (a) Hexaaquachromium(I])chloride (b) Pentaaquachlorochromium(III)chloride (c) Cis-Tetraaquadichlorochromium(IDchloride (d) Trans-Tetraaquadichrolochromium(III)chloride 5. Derive the spectroscopic (RS) terms of the highest spin multiplicities of the chromium ion in the complex [Cr(H20)«]Cls and use these to predict the expected number of UV peaks. For [Cr(H20)6]Cls Cr3+ is 3d? n=3 1=2. M= +2, +1, 0, -1,-2 tT Tt We know, d? = Then, L = 2+1 =3 1 1 1 S=Lmep+>+7=3 3 as+1=2(2)+1=4 Ground state symbol = 4, crystal filled spliting yields th states EE > ag) 4t2g 4tg4T19(P) With 4a, being the ground term Expected transition are: 4A2g > 4T2g 4A2g > 4T1g 4A2g > 4T1g(p) For d? electrons three different picks corresponding to three possible d-d transition should occur 6. Does the predicted number of peaks agree with the number of peaks observed on the spectra. Explain. No, the wavelength in UV/VIS only goes up to 200nm, but the energy for third transition is very high hence expecting to have a very low wavelength, less than 200nm which is out of uv/vis range thus only two peaks are observed. 7. Using the spectra and the appropriate TS diagram of this complex, determine the crystal field splitting parameter, o, From UV/VJS spectra we have V1 = 15873.0cm7+ V2 = 22727.3cm™+ 1. From the information given, the ratio v2 / vl = 22727.3 / 15873.0 = 1.43 2. Using a Tanabe-Sugano diagram for a d3 system this ratio is found at A/B=23.75 3. interpolation of the graph to find the Y-axis values for the spin-allowed transitions gives: v1/B=23.2 v2/B=33.2 v3/B= 52.4