Editing Samarium (section)

===Organometallic compounds===
Samarium forms a [[Cyclopentadiene|cyclopentadienide]] {{chem2|Sm(C5H5)3}} and its chloroderivatives {{chem2|Sm(C5H5)2Cl}} and {{chem2|Sm(C5H5)Cl2}}. They are prepared by reacting samarium trichloride with {{chem2|NaC5H5}} in [[tetrahydrofuran]]. Contrary to cyclopentadienides of most other lanthanides, in {{chem2|Sm(C5H5)3}} some {{chem2|C5H5}} rings bridge each other by forming ring vertexes η<sup>1</sup> or edges η<sup>2</sup> toward another neighboring samarium, thus creating polymeric chains.<ref name="g1248" /> The chloroderivative {{chem2|Sm(C5H5)2Cl}} has a dimer structure, which is more accurately expressed as {{chem2|(η(5)\sC5H5)2Sm(\m\sCl)2(\h(5)\sC5H5)2}}. There, the chlorine bridges can be replaced, for instance, by iodine, hydrogen or nitrogen atoms or by [[cyanide|CN]] groups.<ref name="g1249">[[#Greenwood|Greenwood]], p. 1249</ref>

The ({{chem2|C5H5}})<sup>−</sup> ion in samarium cyclopentadienides can be replaced by the indenide ({{chem2|C9H7}})<sup>−</sup> or [[Cyclooctatetraene|cyclooctatetraenide]] ({{chem2|C8H8}})<sup>2−</sup> ring, resulting in {{chem2|Sm(C9H7)3}} or {{chem2|KSm(\h(8)\sC8H8)2}}. The latter compound has a structure similar to [[uranocene]]. There is also a cyclopentadienide of divalent samarium, {{chem2|Sm(C5H5)2}} a solid that sublimates at about {{convert|85|C|F}}. Contrary to [[ferrocene]], the {{chem2|C5H5}} rings in {{chem2|Sm(C5H5)2}} are not parallel but are tilted by 40°.<ref name="g1249" /><ref>{{cite journal|last1=Evans|first1=William J.|last2=Hughes|first2=Laura A.|last3=Hanusa|first3=Timothy P.|title=Synthesis and x-ray crystal structure of bis(pentamethylcyclopentadienyl) complexes of samarium and europium: (C<sub>5</sub>Me<sub>5</sub>)<sub>2</sub>Sm and (C<sub>5</sub>Me<sub>5</sub>)<sub>2</sub>Eu|journal=Organometallics|volume=5|page=1285|date=1986|doi=10.1021/om00138a001|issue=7}}</ref>

A [[Salt metathesis reaction|metathesis reaction]] in tetrahydrofuran or [[diethyl ether|ether]] gives [[alkyl]]s and [[aryl]]s of samarium:<ref name="g1249" />
:{{chem2|SmCl3 + 3LiR → SmR3 + 3LiCl}}
:{{chem2|Sm(OR)3 + 3LiCH(SiMe3)2 → Sm{CH(SiMe3)2}3 + 3LiOR}}

Here R is a hydrocarbon group and Me = [[methyl]].