Contrasting Magnetism in Isovalent Layered LaSr_3NiRuO_4H_4 and LaSrNiRuO_4 due to Distinct Spin-Orbital States

The recently synthesized first 4d transition-metal oxide-hydride LaSr_3NiRuO_4H_4 with the unusual high H:O ratio surprisingly displays no magnetic order down to 1.8 K. This is in sharp contrast to the similar unusual low-valent Ni^+-Ru^2+ layered oxide LaSrNiRuO_4 which has a rather high ferromagnetic (FM) ordering Curie temperature T_\rm C\sim 250 K. Using density functional calculations with the aid of crystal field level diagrams and superexchange pictures, we find that the contrasting magnetism is due to the distinct spin-orbital states of the Ru^2+ ions (in addition to the common Ni^+ S=1/2 state but with a different orbital state): the Ru^2+ S=0 state in LaSr_3NiRuO_4H_4, but the Ru^2+ S=1 state in LaSrNiRuO_4. The Ru^2+ S=0 state has the (xy)^2(xz,yz)^4 occupation due to the RuH_4O_2 octahedral coordination, and then the nonmagnetic Ru^2+ ions dilute the S=1/2 Ni^+ sublattice which consequently has a very weak antiferromagnetic superexchange and thus accounts for the presence of no magnetic order down to 1.8 K in LaSr_3NiRuO_4H_4. In strong contrast, the Ru^2+ S=1 state in LaSrNiRuO_4 has the (3z^2-r^2)^2(xz,yz)^3(xy)^1 occupation due to the planar square RuO_4 coordination, and then the multi-orbital FM superexchange between the S=1/2 Ni^+ and S=1 Ru^2+ ions gives rise to the high T_\rm C in LaSrNiRuO_4. This work highlights the importance of spin-orbital states in determining the distinct magnetism.