The Activation of N2 and the Nitrogenases Flashcards
The Iron-molybdenum co-factor is the active site in nitrogenase
What is the overall reaction which occurs in the system
N₂ + 8H⁺ + 8e⁻ → 2NH₃ +H₂
Not really sure what the nitrogen is used for so can simplify down to:
N₂ + 6H⁺ 6e⁻ → 2NH₃
Why does s-p mixing occur for N₂ but not O₂ or F₂
Because due to higher electronegativity the energy gap between 2s and 2p is greater in O & F
(greater energy gap between ‘σs’ and ‘σp’
What does the Molecular Orbital diagram look like for nitrogen?
What is the bond order
Bond order = 8(0.5) - 2(0.5) = 3
How many electrons are needed to break the dinitrogen bond?
6 electrons are needed
The HOMO on dinitrogen is the σ3 (σs star) orbtial
what is the shape of this orbital
The LUMO on dinitrogen is the degenerate πp star orbtials
What is the shape of these orbtials
How does dinitrogen interact with a metal centre as a ligand
- It uses the electron pair in its HOMO to form a sigma bond with the metal using an empty d-orbital on the x-axis
- AND
- π interaction with LP in dxy, dyz, dxz orbtials allows for backbonding onto the nitrogen
dinitrogen is a weaker pi-acceptor than carbon monoxide or cyanide
Why?
- Because carbon is less electronegative than nitrogen so it donates more electrons to the metal
- And because oxygen is very electronegative it is a better acceptor
Nature uses the pi-backbonding in nitrogen to break its bond
How
Pi-backbonding occupies the π star orbitals on N₂ unit
This lowers bond order and weakens bond, hence potentially breaking it
Why is dinitrogen Raman active but not IR active
And why when nitrogen has bound to a metal is now IR active
- Dinitrogen is Raman active only as there is no change in dipole but there is a change in polarisation
- Nitrogen bond to a metal results in a dipole (taking electron of alpha nitrogen resulting in it becoming positively charged and the beta nitrogen becoming more negative due to pi-backbonding
- The coordination of dinitrogen weakens the bond which causes a decrease in the vibrational frequency
How does the change in nitrogens dipole when binding to a metal activative it?
Building up the negative charge on the rear nitrogen, means it can be attacked by H⁺
(Coordinated dinitrogen is susceptible to electrophilic attack - e.g. by H⁺ at the terminal N atoms)
The following complex can be reacted to add a dinitrogen ligand
How is it prepared
Reacting it with zinc and water, then with nitrogen
Goes from Ru(III) to Ru(II)
Why is Ruthenium a good metal to bind to nitrogen?
- Ru(II) is a d⁶
- And a 4d metal, hence is low spin
- maximises the amount of dxy, dxz and dyz electrons that are available to be backdonated to the dinitrogen
Why is Molybdenum in this centre a good metal to bind to nitrogen
Also a d⁶ centre
(typically a max of 2 dinitrogen molecules can bind to a d-transition metal centre due to being such a weak pi acceptor ligand)
What are the changes in oxidation states in the following reaction?
- Mo(0) → Mo(IV) + 4e-
- 2H⁺ + 4e- + N₂ → N-NH₂
- The 4e- have gone into the π star orbtials in nitrogen and bond order has been lowered to 3 in dinitrogen to 1 in N-NH₂ (hydrazido)
What does protonation of coordinated N₂ result in?
Break the dinitrogen bond
However very energy intensive reaction and forms lots of Mo(IV) salts
The H⁺/e⁻ addition to coordinated N₂ may be relevent to the function of the nitrogenases; however, this would imply that intermediate NₘHₙ species would be formed
However the function of nitrogenases results in reduced N₂→2NH₃ with no detectable intermediates
What is one way that N₂ can be cleaved in one step?
- 3e- have come from one Mo and 3e- come from the other to fully occupy all the antibonding orbitals on the dinitrogen unit - this breaks the bond
- Then can acid to last compound to form ammonia
- Breaks bond in 1 step
- But not catalytic - hence not cyclical
This is the Vandulov-Schrock Cycle for N₂ reduction
Sterically demanding ligands sets prevent dimerization but allow N₂ to bind
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