LaSr₃NiRuO₄H₄: A 4d Transition‐Metal Oxide–Hydride Containing Metal Hydride Sheets

Abstract

The synthesis of the first 4d transition metal oxidehydride,LaSr 3 NiRuO 4 H 4 ,isprepared via topochemical anion exchange.Neutron diffraction data showthat the hydride ions occupyt he equatorial anion sites in the host lattice and as aresult the Ru and Ni cations are located in aplane containing only hydride ligands,au nique structural feature with obvious parallels to the CuO 2 sheets present in the superconducting cuprates.DFT calculations confirm the presence of S = 1 = 2 Ni + and S = 0, Ru 2+ centers,b ut neutron diffraction and mSR data show no evidence for long-range magnetic order between the Ni centers down to 1.8 K. The observed weak inter-cation magnetic coupling can be attributed to poor overlap between Ni 3d z 2 and H1si nthe super-exchange pathways.Complex transition-metal oxides continue to be the subject of extensive study because they exhibit aw ide variety of interesting physical and chemical properties.T hese range from magnetoresistance,high-temperature superconductivity, and collective magnetism, to ferroelectricity,ionic conductivity,a nd unusual catalytic and photocatalytic behavior. [1] Ty pically the properties of complex oxides are tuned via cation substitutions,b ut modifications to the anion lattice, either by the introduction of anion vacancies or by substituting non-oxide heteroanions,c an also be used to modify the chemical and physical behavior of oxides.F or example,anion doping allows metal oxidation states to be adjusted, the onsite electronic configuration of transition metal centers to be modified (through ligand-field interactions) and inter-cation couplings to be tuned (though cation-anion-cation linkages), providing access to novel electronic states.Thecontrasting features of oxide (O 2À )and hydride (H À ) anions offer many attractive opportunities to modify the electronic properties of host oxide phases by anion substitution. Themost obvious difference between oxide and hydride ions is their charge,a saresult of which hydride-for-oxide substitution necessarily involves reduction, allowing access to unusually low transition-metal oxidation states.T he conversion of insulating A II TiO 3 oxides into metallic A II TiO 3Àx H y oxide-hydrides is aclassic example of the use of hydride-foroxide substitution to modify materials properties. [2] Thelower electronegativity of hydride compared to oxide also implies ah igher degree of covalency and orbital mixing in M À H bonds compared to M À Oanalogues,and as aresult the band structures of oxide-hydrides will be qualitatively different from their parent oxides.F urthermore,m agnetic coupling strengths can be strongly enhanced in oxide-hydride phases, resulting in the high magnetic ordering temperatures observed for LaSrCoO 3 H 0.7 ,S rVO 2 H, and SrCrO 2 H. [3] A final, more subtle difference between oxide and hydride ions is the absence of p-symmetry valence orbitals on the H À anion. Theo rbital connectivity of ap hase can therefore be altered dramatically,e specially if the oxide and hydride anions adopt an ordered arrangement....

Contrasting Magnetism in Isovalent Layered LaSr3NiRuO4H4 and LaSrNiRuO4 due to Distinct Spin-Orbital States*

“…Note that our PBE+U calculations give almost the same results to the LSDA+U ones, with the corresponding FM-AF energy difference of 0.14 meV/fu and the Ni + spin moment of 0.91 µ B (the same as in Ref. [10]) for the S=1/2 Ni + and S=0 Ru 2+ ground state. Thus, these results well account for the experimental observation that LaSr 3 NiRuO 4 H 4 is not magnetically ordered down to 1.8 K and it could well be PM [10].…”

“…[10]) for the S=1/2 Ni + and S=0 Ru 2+ ground state. Thus, these results well account for the experimental observation that LaSr 3 NiRuO 4 H 4 is not magnetically ordered down to 1.8 K and it could well be PM [10]. The PM behavior of LaSr 3 NiRuO 4 H 4 is exactly due to the dilution of the magnetic S=1/2 Ni + sublattice by the nonmagnetic S=0 Ru 2+ ions.…”

“…Thus, the H corner-shared (Ni,Ru)H 4 O 2 octahedra form the (Ni,Ru)H 2 square planar sheets similar to the CuO 2 sheets in the superconducting cuprates. LaSr 3 NiRuO 4 H 4 does not display a long-range magnetic order down to 1.8 K and could thus well be paramagnetic (PM), according to the neutron diffraction and muon spin resonance measurements [10]. The absence of magnetic order was ascribed to the nonmagnetic Ru 2+ S=0 state, which dilutes the S=1/2 Ni + magnetic lattice and suppresses the superexchange interactions therein.…”

“…LaSrCoO 3 H 0.7 , SrVO 2 H, and SrCrO 2 H display high-temperature magnetic ordering [7][8][9]. Very recently, the first 4d TM oxide-hydride, LaSr 3 NiRuO 4 H 4 , was synthesized from the Ruddlesden-Popper LaSr 3 NiRuO 8 via topochemical anion exchange [10]. Its layered structure has the -(La,Sr)O-(Ni,Ru)H 2 -(La,Sr)O-stacking sequence along the c-axis, see FIG.1(a).…”

“…Its layered structure has the -(La,Sr)O-(Ni,Ru)H 2 -(La,Sr)O-stacking sequence along the c-axis, see FIG.1(a). The Ni/Ru cations are bonded, along the c-axis, to two O anions of the neighboring (La,Sr)O sheets, and in the ab plane, they are coordinated completely by the H anions after the topochemical anion exchange [10]. Thus, the H corner-shared (Ni,Ru)H 4 O 2 octahedra form the (Ni,Ru)H 2 square planar sheets similar to the CuO 2 sheets in the superconducting cuprates.…”