Procedures for the synthesis of several compounds of transition metal X are given below. A solution — Physical Chemistry — Thermodynamics Chemistry Question
Transition metal compounds
Procedures for the synthesis of several compounds of transition metal X are given below.
A solution of 2 g of very fine powder A in 50 cm-3 of 28% sodium hydroxide is triturated in a small Erlenmeyer flask with 3.5 g of finely ground Na2SO3 · 7 H2O; the flask stands in an ice bath. The trituration requires about 10 minutes, that is, until a light-blue crystalline slurry is obtained. The mixture is then transported under vacuum onto an ice-cooled glass filter, and the product washed thoroughly with 28% sodium hydroxide at 0 °C. The wet preparation is rapidly spread in a thin layer on fresh clay and stored at 0 °C in an evacuated desiccator (no drying agent)… The preparative procedure should be designed so as to avoid contamination by silicates or aluminates … Product B, in the form of well-crystallized sky-blue rods, remains stable at 0 °C if kept free of H2O and CO2… A solution of B in 50% potassium hydroxide turns grassy green upon heating or dilution; simultaneously, C is precipitated.
In a pure form, salt D, which is a main constituent of B, is prepared according to the following procedure: «NaOH is entirely dehydrated by heating in silver pot at 400 ºС and mixed with C in a such way that Na : X molar ratio is 3 : 1. Mixture is heated to 800 ºС in a silver pot and kept under oxygen flow for 5 h. The formed product D is rapidly cooled to room temperature». Salt D is a dark-green compound inert to CO2.
A solution of 30 g of KOH in 50 cm-3 of water is prepared; 10 g of А is added and the mixture is boiled in an open 250 cm-3 Erlenmeyer flask until a pure green solution is obtained. The water lost by evaporation is then replaced and the flask set in ice. The precipitated black-green crystals, which show a purplish luster, are collected on a Pyrex glass filter, washed (high suction) with some 1 M potassium hydroxide, and dried over P2O5. The formed compound E can be recrystallized by dissolving in dil. KOH and evaporated in vacuum».
Determine the element Х and molecular formulae of A - Е using the following data: a) sodium weight content in В is 18.1 %; b) the weight content of the element Х in А, В, С, D, and Е is 34.8, 13.3, 63.2, 29.3, and 27.9 %, respectively.
Model Answer
Anhydrous salt D is the main constituent of compound B. We may suppose that B is a hydrate of D.The Na : X molar ratio in D is 3 : 1. D is not a binary compound Na3X as in this case МХ = (29.3 × 69 / 70.7) = 28.6. There is no such element. So, D contains some other element(s) too. Oxygen is the most probable element, i.e., D is Na3XOn (salt D cannot have formulae of Na3НmXOn type as all volatiles should be removed under reaction conditions used for synthesis of D (heating at 800 ºС). High content of Х in compound С allows one to suppose that C is a binary compound, i.e., it is an oxide of Х. Now we can determine Х.
Therefore, Х is Mn and С is MnO2. From the content of Mn in D we derive its formula Na3MnO4. The manganese oxidation state in this compound is V. Under heating or cooling, the alkaline solution of D disproportionates, giving solid MnO2 and a green solution. Solutions of manganese(VII) derivatives are usually purple but not green. Therefore, the solution contains a salt of manganese(VI). The analogous green solution is formed in the last procedure. We may conclude that this procedure leads to manganate, K2MnO4. Indeed, the content of Mn in K2MnO4 (compound E) is 27.9 %. Compound В (a Mn(V) derivative) is obtained by the reaction of А with sodium sulfite which is a well-known reducing agent. Heating of the alkaline solution of A affords K2MnO4. It is possible only if A is a Mn(VII) derivative. Indeed, the Mn content in A corresponds to the formula of KMnO4. The remaining unknown compound is B. Above we supposed that B is a hydrate of D. Calculations using the formula of Na3MnO4·nH2O lead to Мr(В) = 413.5. It corresponds to n = 12.5. However, Мr(В) = 381.2 from the Na content. In other words, Na : Mn ratio in В is not 3 : 1 but 3.25 : 1. This additional sodium appears due to the presence of some other Na compound(s) in the solvate. To determine this compound, the analysis of the synthetic procedure is required. During the synthesis of B solvate is washed with NaOH solution. So, the possible formula of B is Na3MnO4 · 0.25 NaOH · n H2O. From Na and Mn content we conclude that n = 12. Finally, В is [4 Na3MnO4 · NaOH · 48 H2O].
Write all the reaction equations.
Model Answer
Four reactions are discussed in the text. They are:
i. 4 KMnO4 + 4 Na2SO3 · 7 H2O + 13 NaOH + 16 H2O → [4 Na3MnO4· NaOH · 48 H2O]↓ + 4 Na2SO4 + 4 KOH
or (4 KMnO4 + 4 Na2SO3 + 13 NaOH + 44 H2O → [4 Na3MnO4 · NaOH · 48 H2O]↓ + 4 Na2SO4 + 4 KOH)
ii. 2 Na3MnO4+ 2 H2O → Na2MnO4 + MnO2 + 4 NaOH
iii. 12 NaOH + 4 MnO2 + O2 → 4 Na3MnO4 + 6 H2O
iv. 4 KMnO4 + 4 KOH → 4 K2MnO4 + O2 + 2 H2O