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The recently synthesized first 4d transition-metal oxide-hydride LaSr3NiRuO4H4 with the unusual high H:O ratio surprisingly displays no magnetic order down to 1.8K.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 ～ 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 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 superexchange and thus accounts for the presence of no 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.