One- and two-step spin-crossover behavior of [Fe(II)(isoxazole)(6)](2+) and the structure and magnetic properties of triangular [Fe(III)(3)O(OAc)(6)(isoxazole)(3)][ClO(4)].

The structure and spin-crossover magnetic behavior of [Fe(II)1(6)][BF(4)](2) (1 = isoxazole) and [Fe(II)1(6)][ClO(4)](2) have been studied. [Fe(II)1(6)][BF(4)](2) undergoes two reversible spin-crossover transitions at 91 and 192 K, and is the first two-step spin transition to undergo a simultaneous crystallographic phase transition, but does not exhibit thermal hysteresis. The single-crystal structure determinations at 260 [space group P3, a = 17.4387(4) A, c = 7.6847(2) A] and at 130 K [space group P1, a = 17.0901(2) A, b = 16.7481(2) A, c = 7.5413(1) A, alpha = 90.5309(6) degrees, beta = 91.5231(6) degrees, gamma = 117.8195(8) degrees ] reveal two different iron sites, Fe1 and Fe2, in a 1:2 ratio. The room-temperature magnetic moment of 5.0 mu(B) is consistent with high-spin Fe(II). A plateau in mu(T) having a moment of 3.3 mu(B) centered at 130 K suggests a mixed spin system of some high-spin and some low-spin Fe(II) molecules. On the basis of the Fe-N bond distances at the two temperatures, and the molar fraction of high-spin molecules at the transition plateau, Fe1 and Fe2 can be assigned to the 91 and 192 K transitions, respectively. [Fe(II)1(6)][ClO(4)](2) [space group P3, a = 17.5829(3) A, c = 7.8043(2) A, beta = 109.820 (3) degrees, T = 295 K] also possesses Fe1:Fe2 in a 1:2 ratio, and magnetic measurements show a single spin transition at 213 K, indicating that both Fe1 and Fe2 undergo a simultaneous spin transition. [Fe(II)1(6)][ClO(4)](2) slowly decomposes in solutions containing acetic anhydride to form [Fe(III)(3)O(OAc)(6)1(3)][ClO(4)] [space group I2, a = 10.1547(7) A, b = 16.5497(11) A, c = 10.3205(9) A, beta = 109.820 (3) degrees, T = 200 K]. The isosceles Fe(3) unit contains two Fe.Fe distances of 3.2844(1) A and a third Fe.Fe distance of 3.2857(1) A. The magnetic data can be fit to a trinuclear model with H = -2J(S(1)xS(2) + S(2)xS(3)) - 2J(13)(S(1)xS(3)), where J = -27.1 and J(13) = -32.5 cm(-1).