Preparation,crystal structure,and properties of the mixed-valence compound Nb3Se5Cl7

The new compound Nb₃Se₅Cl₇ was prepared by heating 2NbSe₂Cl₂ + 1NbCl₄ at 530°C for 2–3 weeks. The compound is monoclinic with a = 7.599, b = 12.675, c = 8.051Å; β = 106.27°; space group $\text{P2}{}_1\text{m}$. The corresponding bromide, Nb₃Se₅Br₇ (obtained by decomposition of NbSe₂Br₂ under NbSeBr₃), is isotypic with a = 7.621, b = 12.833, c = 8.069Å; β = 106.21°. from the crystal structure and XPS spectra it follows that Nb₃Se₅Cl₇ can be formulated as: Nb⁴⁺₂Nb⁵⁺₁(Se₂)^2?₂Se^2?₁Cl^?₇]. The structure consists of chains of composition Nb⁴⁺₂(Se₂)^2?₂Cl^?₅], to which side chains Nb⁵⁺Se^2?Cl^?₂] are attached. The Nb⁴⁺ atoms form pairs (NbNb = 2.94 Å) which explains that Nb₃Se₅Cl₇ is a diamagnetic semiconductor with a band gap (1.59 eV at 5°K, 1.49 eV at 300°K) very similar to that of NbSe₂Cl₂.