Samarium

Samarium dissolves readily in dilute sulphuric acid, forming a yellow solution of Sm(III) ions and hydrogen gas, H₂.

2 Sm(s) + 3 H₂SO₄(aq) 2 Sm³⁺(aq) + 3 SO₄²⁻(aq) + 3 H₂(g)

Samarium reacts slowly with oxygen, O₂, tarnishing the surface, under normal conditions. Samarium burns readily, forming Sm(III) oxide, Sm₂O₃:

4 Sm(s) + 3 O₂(g) 2 Sm₂O₃(s)

Samarium reacts with the halogens, forming the corresponding samarium(III) halides:

2 Sm(s) + 3 F₂(g) SmF₃(s) [white]
2 Sm(s) + 3 Cl₂(g) SmCl₃(s) [yellow]
2 Sm(s) + 3 Br₂(g) SmBr₃(s) [yellow]
2 Sm(s) + 3 I₂(g) SmI₃(s) [orange]

Samarium reacts slowly with cold water, and quickly with hot water, forming samarium hydroxide, Sm(OH)₃, and hydrogen gas, H₂.

2 Sm(s) + 6 H₂O(l) 2 Sm(OH)₃(aq) + 3 H₂(g)