The Crystal Structure of Potassium Metaborate, K3(B3O6)

In order to find the configuration of the metaborate group the crystal structure of potassium metaborate was determined. The crystals are rhombohedral a_r = 7.76A, α = 110° 36′ with six molecules KBO₂. The density is 2.348. The space group is ${\text{R3̄c(D}}_{\text{3d}}^6)$ . All atoms are lying on twofold axes at positions: ±(u, ½—u, ¼) $\mathrm{\circlearrowleft \hspace{0.17em}}$ . The parameters were determined by means of two‐dimensional Fourier analyses: $$\begin{array}{cccc}\mathrm{K}& \text{u=0.689±0.003,}& \mathrm{B}& \text{u=0.361±0.006,}\\ {\mathrm{O}}_{\mathrm{I}}& \text{0.465±0.006,}& {\mathrm{O}}_{\mathrm{II}}& \text{0.146±0.006.}\end{array}$$ The metaborate radical is (B₃O₆)^—3, a ring of three BO₃ triangles, rather than the endless chain of BO₃ triangles found in CaB₂O₄. The BO₃ groups are slightly distorted, the B–O distances being 1.33A, 1.38A and 1.38A, the O–O distances 2.30A, 2.38A and 2.38A. The nine particles of the radical lie in one plane. Potassium is surrounded by seven oxygen atoms at a distance 2.82A.