Manganese forms the highest number of oxidation states among the first-row transition metals. This t — Physical Chemistry — Electrochemistry Chemistry Question
Cyanido- and fluorido-complexes of manganese
Manganese forms the highest number of oxidation states among the first-row transition metals. This task deals with the synthesis and electronic structure of manganese cyanido- and fluorido-complexes in oxidation states +I to +IV.
Oxidation state +I
Metallic manganese reacts only slowly with water. It dissolves readily in deaerated solution of NaCN (c = 2 mol dm-3) to give colourless, diamagnetic Na5[Mn(CN)6] (1).
Write a balanced equation (1).
Model Answer
2 Mn + 12 NaCN + 2 H2O → 2 Na5[Mn(CN)6] + H2 + 2 NaOH
Draw the splitting diagram for the complex anion and fill in the electrons.
Model Answer
[VISUAL] Diagram – low-spin configuration d6
Oxidation state +II
Soluble manganese(+II) compounds (e.g. chloride, nitrate, sulfate) are common starting materials for the preparation of manganese complexes. Mn2+(aq) in deaerated aqueous solution reacts with excess CN− to give a blue ion [Mn(CN)6] 4− with a magnetic moment corresponding to one unpaired electron.
Mn2+(aq) ion can be considered as a high-spin hexaaqua-complex. Draw the splitting diagram, fill in the electrons and predict the number of unpaired electrons in the complex.
Model Answer
[VISUAL] Diagram – high-spin configuration d5
The complex has five unpaired electrons.
Draw the splitting diagram for [Mn(CN)6] 4− and fill in the electrons.
Model Answer
[VISUAL] Diagram – low-spin configuration d5
Oxidation state +III
Red [Mn(CN)6] 3− is an example of a rare low-spin manganese(+III) complex. It can be prepared by three different methods:
i. Stream of air is bubbled through the solution of a manganese(+II) salt and excess of cyanide (2).
ii. [Mn(CN)6] 4− is oxidized by 3% solution of hydrogen peroxide (3).
iii. Manganese(+II) chloride is oxidized by nitric acid in excess phosphoric acid (NO is formed) (4). The green-grey precipitate thus formed is filtered off and dissolved in potassium cyanide solution at 80 °C (non-redox reaction) (5).
Write balanced equations (2)–(5), (2) and (3) in an ionic form.
Model Answer
(2) 4 Mn2+ + O2 + 24 CN− + 2 H2O → 4 [Mn(CN)6] 3− + 4 OH−
(3) 2 [Mn(CN)6] 4− + H2O2 → 2 [Mn(CN)6] 3− + 2 OH−
(4) 3 MnCl2 + HNO3 + 3 H3PO4 → 3 MnPO4↓ + NO + 6 HCl + 2 H2O
(5) MnPO4 + 6 KCN → K3[Mn(CN)6] + K3PO4
Draw the splitting diagram for [Mn(CN)6] 3− ion and fill in the electrons.
Model Answer
[VISUAL] Diagram – low-spin configuration d4
Violet complex K3[MnF6] can be prepared by dissolving manganese dioxide in KHF2 aqueous solution (6).
Write a balanced equation (6).
Model Answer
(6) 4 MnO2 + 12 KHF2 → 4 K3[MnF6] + O2 + 6 H2O
Other manganese(+III) fluorido-complexes exist with seemingly different coordination numbers: Na2[MnF5], Cs[MnF4]. In reality, the coordination number of Mn atom is 6 in both cases. Octahedral units [MnF6] oct in the structure of these salts are interconnected by bridging F atoms.
Draw the splitting diagram for the octahedral species [MnF6] 3− and fill in the electrons.
Model Answer
[VISUAL] Diagram – high-spin configuration d4
Predict the structure of anionic 1D-chains present in Na2[MnF5].
Model Answer
[VISUAL] Sharing 1 bridging F atom between 2 neighbouring octahedral units corresponds to the stoichiometry [MnF5] 2−.
Predict the structure of anionic 2D-layers present in Cs[MnF4].
Model Answer
[VISUAL] The stoichiometry [MnF4] − could be achieved in a chain structure having 2 bridging F atoms between 2 neighbouring octahedral units. However, the structure is a 2D-anionic layer, so it is necessary to extend the structure to 2 dimensions – to have 4 bridging F atoms for each octahedral unit.
Oxidation state +IV
The oxidation of a [Mn(CN)6] 3− ion by nitrosyl chloride gives a [Mn(CN)6] 2− ion. When irradiated with sunlight, this ion undergoes reductive photolysis to give a tetrahedral [Mn(CN)4] 2− ion.
Draw the splitting diagram for [Mn(CN)6] 2− and fill in the electrons.
Model Answer
[VISUAL] Diagram – configuration d3
All the regular tetrahedral complexes are high-spin. Why? Draw the splitting diagram for [Mn(CN)4] 2− and fill in the electrons.
Model Answer
Since the magnitude of splitting in tetrahedral crystal field is about a half of octahedral (exactly Δtet = − 4/9 Δoct; the negative sign refers to the inverse order of split d-orbitals with respect to the octahedral crystal field), it is always lower than electron pairing energy (Δtet < P) which leads to high-spin configurations in tetrahedral complexes.
[VISUAL] Diagram – high-spin configuration d5
Yellow fluorido-complex K2[MnF6] can be prepared by reducing KMnO4 with hydrogen peroxide in the presence of KHF2 and HF (7).
Write a balanced equation (7).
Model Answer
(7) 2 KMnO4 + 3 H2O2 + 2 KHF2 + 8 HF → 2 K2[MnF6] + 3 O2 + 8 H2O
The electronic structure of [MnF6] 2− can be described qualitatively by the same splitting diagram as [Mn(CN)6] 2−. Why?
Model Answer
Manganese(IV) complexes have a configuration of d3. Since these three electrons occupy the t2g level only, low- and high-spin configurations cannot form regardless of the magnitude of the crystal field splitting.
Interestingly, complex K2[MnF6] can be used for non-electrolytic fluorine preparation. Upon heating, it reacts with SbF5 to give K[SbF6], MnF2 and fluorine (8).
Write a balanced equation (8).
Model Answer
(8) K2[MnF6] + 2 SbF5 → 2 K[SbF6] + MnF2 + F2