Hydride neutron diffraction studies

Aluminum thorium hydride, neutron diffraction studies on, 8 238-239... 

Cerium diiodide, 20 4 Cerium dioxide, 20 67, 69, 73, 74 Cerium hydride, neutron diffraction studies on, 8 238... 

Neutron diffraction studies have shown that in both systems Pd-H (17) and Ni-H (18) the hydrogen atoms during the process of hydride phase formation occupy octahedral positions inside the metal lattice. It is a process of ordering of the dissolved hydrogen in the a-solid solution leading to a hydride precipitation. In common with all other transition metal hydrides these also are of nonstoichiometric composition. As the respective atomic ratios of the components amount to approximately H/Me = 0.6, the hydrogen atoms thus occupy only some of the crystallographic positions available to them. 

Neutron Diffraction Studies of Tetrahedral Cluster Transition Metal Hydride Complexes HFeCo3(CO)9-(P(OCH3)3)3 and H3Ni4(C5H5)4... 

Proton NMR cannot be used as an indication of the position occupied by hydrides in HNCC in solution as the range of chemical shifts observed is enormous (441). For example, all fully characterized carbonyl clusters that contain interstitial hydrides are listed in Table II, with chemical shifts from 23.2 6 in [Co6(CO)isH]- to -24(5 in [Nii2(CO)2iH] -. The octahedral [Rue(CO).8H]- was the first reported cluster for which an interstitial hydride was assigned on the basis of X-ray (30, 31) and solid-state infrared spectroscopy studies (33). However, because of the extremely low field position of the NMR signal (16.4 S), it was suspected to be of the formyl type (417). Its interstitial position was later unequivocally established by neutron diffraction studies (32). The observation of the satellites in the... 

X-ray and neutron diffraction studies show that in these hydrides the H ion has a crystallographic radius between those of F and Cl". Thus the electrostatic lattice energies of the hydride and the fluoride and chloride of a given metal will be similar. These facts and a consideration of the Bom-Haber cycles lead us to conclude that only the most electropositive metals can form ionic hydrides, since in these cases relatively little energy is required to form the metal ion. 

---