  The recently synthesized first 4d transition-metal oxide-hydride
LaSr3NiRuO4H4 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+ - Ru2+ layered oxide LaSrNiRuO4 which has a rather high
ferromagnetic (FM) ordering Curie temperature TC about 250 K. In this work,
using density functional calculations with 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 Ru2+ ions (in addition to the common
Ni+ S=1/2 state but with a different orbital state): the Ru2+ S=0 state in
LaSr3NiRuO4H4, but the Ru2+ S=1 state in LaSrNiRuO4. The Ru2+ S=0 state has the
(xy)2(xz; yz)4 occupation due to the RuH4O2 octahedral coordination, and then
the nonmagnetic Ru2+ ions dilute the S=1/2 Ni+ sublattice which consequently
has a very weak antiferromagnetic (AF) superexchange and thus accounts for no
presence of magnetic order down to 1.8 K in LaSr3NiRuO4H4. In strong contrast,
the Ru2+ S=1 state in LaSrNiRuO4 has the (3z2-r2)2(xz; yz)3(xy)1 occupation due
to the planar square RuO4 coordination, and then the multi-orbital FM
superexchange between the S=1/2 Ni+ and S=1 Ru2+ ions gives rise to the high TC
in LaSrNiRuO4. This work highlights the importance of spin-orbital states in
determining the distinct magnetism.
