How do you separate a mixture of cis and trans potassium dioxalatodiaquochromate?

Answer 1

4 Cis- and trans potassium dioxalatodiaquachromato (III) Materials required:Oxalic acid, potassium dichromate a)Trans isomer of K{Cr(C2O4)2(H2O)2|3H2O Dissolve 12 g of oxalic acid monohydrate in a minimum amount of boiling water in a 250 ml beaker. Add in small portions a solution of 4 g of potassium oxalate in a minimum of hot water, covering the beaker while the violent reaction proceeds. Evaporate the solution to about one third of its original volume. Filter the crystals and wash with cold water and ethanol. A high yield cannot be expected since only a part of the product is isolated. Note that in solution there is an equilibrium between the cis- and thetrans isomer, but the low solubility of the trans isomer results in its initial deposition. Excessive spontaneous evaporation should be avoided since otherwise the product may be contaminated with the cis isomer. b) Cis isomer of K{CR(C2O4))(H2O)2}2H2O Prepare an intimate mixture of finely ground potassium dichromate(4 g) and oxalic acid dihydrate (12 g) and heap the powder in a 15 cm diameter evaporating dish. Place one drop of water in a small depression in the mixture and cover the dish with a watch glass. After a short induction period the reaction commences and soon becomes vigorous with the evolution of steam and carbon dioxide. A solution of the product cannot be formed, and hence an equilibrium mixture of cis- and trans-isomers is not formed. The outcome of this reaction is a purple viscous liquid to which is added 20 ml of ethanol. The mixture is stirred until the reaction product solidifies. If solidification is slow, decant the liquid and repeat with a second portion of ethanol until the product is entirely crystalline. Filter at the pump. ** Purity test of the isomers:Place a few crystals of the complex on a filter paper and add a few drops of diluted aqueous ammonia. The cis-isomer rapidly forms a deep green solution which spreads over the filter paper, and no solid remains. The trans-isomer forms a light brown solid which remains undissolved. These changes are due to the formation of cis- and trans- dioxalatohydroxoaquachromium(III) ions.

Answer 2

trans form: is less soluble than of cis form, and can be obtained by crystallizing the solution at ordinary temperatures.

the cis form: may be obtained by carrying out a preparation at high temperature in presence of EtOH.

Answer 3

One method to separate cis and trans potassium dioxalatodiaquochromate is through crystallization. By adjusting the temperature or adding a solvent, one form may crystallize out while the other remains in solution, allowing for physical separation. Additionally, chromatography techniques such as HPLC or column chromatography could also be used for separation based on differences in polarity or molecular size.