Ammonia molecular orbitals

The tetramethylammonium salt [Me4N][NSO] is obtained by cation exchange between M[NSO] (M = Rb, Cs) and tetramethylammonium chloride in liquid ammonia. An X-ray structural determination reveals approximately equal bond lengths of 1.43 and 1.44 A for the S-N and S-O bonds, respectively, and a bond angle characteristic bands in the IR spectrum at ca. 1270-1280, 985-1000 and 505-530 cm , corresponding to o(S-N), o(S-O) and (5(NSO), respectively. Ab initio molecular orbital calculations, including a correlation energy correction, indicate that the [NSO] anion is more stable than the isomer [SNO] by at least 9.1 kcal mol . ... 

Amination of heterocyclic nitro compounds such as nitroquinolines, nitroisoquinolines, or nitropyridines can be carried out by means of a liquid ammonia-KMn04 system, which has been studied by Wozniak and coworkers (Eq. 9.44). Frontier molecular orbital calculation can predict the reactivity and regioselectivity of this amination.75 In a similar way, nitroquinolines are methylaminated with a liquid methylamine solution of KMn04.76... 

Reduction of benzenoid hydrocarbons with solvated electrons generated by the solution of an alkali metal in liquid ammonia, the Birch reaction [34], involves homogeneous electron addition to the lowest unoccupied 7t-molecular orbital. Protonation of the radical-anion leads to a radical intermediate, which accepts a further electron. Protonation of the delocalised carbanion then occurs at the point of highest charge density and a non-conjugated cyclohexadiene 6 is formed by reduction of the benzene ring. An alcohol is usually added to the reaction mixture and acts as a proton source. The non-conjugated cyclohexadiene is stable in the presence of... 

It is assumed that the reader has previously learned, in undergraduate inorganic or physical chemistry classes, how symmetry arises in molecular shapes and structures and what symmetry elements are (e.g., planes, axes of rotation, centers of inversion, etc.). For the reader who feels, after reading this appendix, that additional background is needed, the texts by Cotton and EWK, as well as most physical chemistry texts can be consulted. We review and teach here only that material that is of direct application to symmetry analysis of molecular orbitals and vibrations and rotations of molecules. We use a specific example, the ammonia molecule, to introduce and illustrate the important aspects of point group symmetry. 

The photoelectron spectrum of the ammonia molecule is consistent with the molecular orbital conclusions there are three ionization energies at... 

The final molecule of this series is methane, the tetrahedral structure of which follows if a fourth unit positive charge is removed from the nucleus in the ammonia lone-pair direction. There are now four equivalent bonding orbitals, which may be represented approximately as linear combinations of carbon s-p hybrid and hydrogen Is functions. The transformation from molecular orbitals into equivalent orbitals or vice versa is exactly the same as for the neon atom. 

But how do we account for the bond angles in water (104°) and ammonia (107°) when the only atomic orbitals are at 90° to each other All the covalent compounds of elements in the row Li to Ne raise this difficulty. Water (H2O) and ammonia (NH3) have angles between their bonds that are roughly tetrahedral and methane (CH4) is exactly tetrahedral but how can the atomic orbitals combine to rationalize this shape The carbon atom has electrons only in the first and second shells, and the Is orbital is too low in energy to contribute to any molecular orbitals, which leaves only the 2s and 2p orbitals. The problem is that the 2p orbitals are at right angles to each other and methane does not have any 90° bonds. (So don t draw any either Remember Chapter 2.). Let us consider exactly where the atoms are in methane and see if we can combine the AOs in such a way as to make satisfactory molecular orbitals. 

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