Shpend Mushkolaj Thank you very much for your comprehensive and insightful reply.

Zirconium diboride (ZrB2) is a highly covalent refractory ceramic material with a hexagonal crystal structure. ZrB2 is an ultra high temperature ceramic (UHTC) with a melting point of 3246 °C. This along with its relatively low density of ~6.09 g/cm3 (measured density may be higher due to hafnium impurities) and good high temperature strength makes it a candidate for high temperature aerospace applications such as hypersonic flight or rocket propulsion systems. It is an unusual ceramic, having relatively high thermal and electrical conductivities, properties it shares with isostructural titanium diboride and hafnium diboride.

ZrB2 parts are usually hot pressed (pressure applied to the heated powder) and then machined to shape. Sintering of ZrB2 is hindered by the material's covalent nature and presence of surface oxides which increase grain coarsening before densificationduring sintering. Pressureless sintering of ZrB2 is possible with sintering additives such as boron carbide and carbon which react with the surface oxides to increase the driving force for sintering but mechanical properties are degraded compared to hot pressed ZrB2.[2]

Additions of ~30 vol% SiC to ZrB2 is often added to ZrB2 to improve oxidation resistance through SiC creating a protective oxide layer - similar to aluminum's protective alumina layer

Mr. Asl,

If we have described the ZrB2 in analogy with a uranyl-like structure (attachment), where the initial state is ZrVIBIII2, that NBO analysis performed, has shown that in fact the final state is (delta-)BZr(2(delta+))B(delta-), due to a rather coivalent character of the Zr-B bonds.

You can compare the data of ZrB2 with thoise of isolated Zr6+ and B3-, which have been also aded in the attachment.

Zr in ZrB2 cannot have the oxidation state +6. This would require the ionisation of two electrons of the crypton shell. The maximum of Zr oxidation leads to +4, which is equal to the Kr shell.

As far as I know is ZrB2 isostructural to MgB2. If this is the case ZrB2 is a Zintl phase containing Zr2+ and (B2)2-. The latter is isoelectronic to C and thus (B2)2- formes graphite-like layers with hexagonal symmetry. The B-B bonds are thus completely covalent, the Zr-B bonds are more ionic from this simple structure description.

A more exhaustive description of the structure is only possible using MO. And here the nature of the 2 remaining electrons on Zr2+ is of interest, and of course, not clear. They will surely form bands.

Importantly, ZrB2 is SURELY NOT a molecule!

Mr. Klein,

ZrB2 is used as neutron burnable absorber in the nuclear industry (ref. [1]), having ability like uranium oxides to stabilize nonstoichometric compositions (refs. [2,3]) such as for example B2Zr0.75 and B2.34Zr. Furthermore there have studies employing experiments with Zr6+ ions or indicating stabilization of +6 oxidation state as well (refs. [4,5]). In addition "B" typically has stabilized oxidation state -3. Furthermore  the synthesis of ZrB2 has involved B2O3 (ZrO2 + B2O3 + 5C = ZrB2 + 5CO) as it has been reported in Ref [6] and presumably, the last process, if there has transition to BII should be realized with a change of multiplicity (attachment). Given that, there is fully possible to have ZrVIBIII2, Unfortunately there has no single crystal X-ray diffraction measurements, which to determine unambiguously the structure and geometry parameters such as Zr-B bonds.

But even if we accept your version for ZrIVBII2 as a true one, the NBO analysis performed (attachment) indicates a rather covalent character of Zr-B and (delta-)B-Zr(2delta+)-B(delta-) as well.

[1] P. Liu, P. Zhang, X. Pang, Q. Wang, T. Liu, A study on fabrication technique of ZrB2 target, Procedia Engineering 27 (2012) 1305-1312

[2] Knyshev, E.; Novgorodtsev, V.; Plyshevski, U.; Kobyakov, V.; Stepanova, Z. et al. Journal of the Less-Common Metals, 1976, 47, 273-278

[3] Russian Journal of Inorganic Chemistry, 2007, 52, 238-241, Effect of metal vacancies on the energy parameters of s-, p-, and d-metal diborides, I. Shein, A. Ivanovskii

[4] INFLUENCE OF OXYGEN CONTENT ON EVOLUTION OF THE STRUCTURE OF ALLOY Zr1%Nb UNDER ION IRRADIATION, O. Borodin, V. Bryk, R. Vasilenko, V. Voyevodin, I. Petelguzov, N. Rybalchenko, Questions on the Nuclear (Atomic?) Science and Technology, 2008; Series: Physics of the radiation damage and radiation materials research (?) (92), 53-61.

[5] Y. LI, J. G. HE, X. XUE, H. RU, X. HUANG, H. YANG, EXRACTION AND SEPARATION OF CERIUM(IV)/FLUORINE IN FLUORIDE-BEARING CERIUM SULFATE SOLUTION WITH FLUORIDE COORDINATION AGENT, METABK 53(3) 320-322 (2014)

[6] M. Thompson, W. Fahrenholtz, G. Hilmas, Effect of Starting Particle Size and Oxygen Content on Densification of ZrB2, J. Am. Ceram. Soc., 94, 429-435 (2011)

I don't want to be picky. But Ce is in group IV in the Periodic Table of the Elements. Thus, it has FOUR valence electrons. It is disputable what the nature of these FOUR electrons is, but there is no chance to remove further electrons than these four in a chemical process. Thus, there cannot be an oxidation state of +6 for Zr in a chemical compound.

It is one of the main principles of the periodic table of the elements, that the electron shells of the noble gases are very stable. As a consequence the number of electrons which might be removed or added to an atom in a chemical reaction is limited. E.g. Rb can only loose one electron, thus forming Rb+. Sr can loose one or two electrons, not more. Yttrium can loose one, two or three electrons to a more electronegative reaction partner, frequently, this is three for Y, thus Y3+ is the main oxidation state.

Secondly, the structure of ZrB2 is KNOWN: e.g. Zeng et al. Computational Materials Science 49 (2010) 814–819.

Mr. Klein,

First we are discussing about Zirkonium (Zr), not Cerium (Ce). For Zr there have reports as it was shown in my previouse message for oxidation state +6 (refs. [4,5]). Are these reports are true ones or not can be established after repeating the experiments of the authors. So that when you repeat their experiments and have unambigouse confirmation that at the shown experimental conditions Zr has oxidation state +4, not +6, than you could write an erratum explaining why their conclusions are wrong. 

Furthermore as I already has mentioned, if Zr is in +4 oxidation state, than "B" should be "-2". Or a change of the multiplicity should occur during the reaction described in ref. [6]. But it is well known that for a favourable transition there is no change in multiplicity. So that such as process leading to ZrIVBII2 under the shown reaction is, generally, unfavourable. The authors, however, have not provided assumption/disscusion about the oxidation state of the elements; about the mechanism of the reaction, experiments in this context and s.o.. They just have reported an obtaining of ZrB2 under a reaction, comparing with available powder XRD data....

Again the react: ZrO2 + B2O3 + 5C = ZrB2 + 5CO, ref. [6]

Second, you have provided a reference:

1. Crystal structure and elastic properties of ZrB compared with ZrB2: A first-principles study, Computational Materials Science, 49, 4, 2010, 814-819, H. Li, L. Zhang, Q. Zeng, J. Wang, L. Cheng, H. Ren, K. Kuan

But these authors have reported COMPUTATIONAL prediction of POSSIBLE structures on the base of known chemical compositions, using first principle MD calculations. The title is missleding. These are molecular dynamics simulations.

My comment was that there have not experimental, furthermore SINGLE CRYSTAL X-Ray diffraction data. There has only POWDER XRD data on the periodic translation as it was mentioned above, but there has no information about the geometry parameters.

Therefore, still the Zr-B distance is unknown....If somebody know it from experimental (not theoretical computational data) or there have somewhere report, which has included experimental SINGLR CRYSTAL XRD of ZrB2 could share in order to compute the level of covalency, even to make assumption about the oxidation state.

rfp

@Mehdi

You are right, the type of chemical bond of ZrB2 is mainly covalent, but you can easily estimate the percentage of covalent and ionic bonds. The smallest and highest electronegativity values have francium and fluorine, namely 0.7 and 3.98, respectively (in Pauling scale). One can define FrF with an electronegativity difference of 3.28 as 100% ionic and 0% covalent chemical bonding. Since the electronegativity difference in ZrB2 is 0.71, i.e. the bond is polar covalent. If we compare the electronegativity difference of FrF with ZrB2, one may estimate a percentage of 22% ionic type and 78% covalent type bond for ZrB2. Therefore, the total molecular quantum state Psi of ZrB2 is a superposition of two types of states, namely the main part double covalent and the smaller part ionic states.

PsiZrB2= 0.78* PsiB=Zr=B + 0.22* PsiB- -Zr++++B- -

I think it is very unlikely that a +6 oxidation state of Zr6+ might be present in ZrB2. The oxidation state Zr4+ and Hf4+ in ZrB2 and HfB2 may be easily proven from the superconductivity transition temperatures of Zr0.97V0.03B2 and Hf0.96V0.04B2. One additional hint for the presence of +4 oxidation state of Zr4+ and Hf4+ might result from the very large difference of the melting temperatures of 3520 K for ZrB2 and HfB2, and 1100 K for MgB2. The unit cells are hexagonal with almost identical lattice parameters, namely: a=3.086 Å, b=3.524 Å for MgB2 [1], a=3.17 Å, b=3.53 Å for ZrB2 [2], and a=3.14 Å, b=3.48 Å for HfB2 [2].

If we assume the presence of oxidation states for Mg2+, Zr4+ and Hf4+, then we can clarify the large difference in the melting points as a result of higher oxidation states of Zr and Hf and higher order covalent Zr=B and Hf=B bonding in ZrB2 and HfB2. The melting points of ZrB2 and HfB2 can be roughly estimated from the melting point of MgB2, namely 4*(1100K)*0.78=3430 K.

The superconducting transition temperatures for MgB2, Zr0.97V0.03B2 and Hf0.96V0.04B2 are: 39 K [1], 8.3K [3] and 7.3 K [3].

In the following I calculate the superconducting transition temperatures [4, 5] for MgB2, Zr0.97V0.03B2 and Hf0.96V0.04B2 to prove the above mentioned oxidation states. Tc calculations are in the attached file.

Shpend Mushkolaj Thank you very much for your comprehensive and insightful reply.

a late answer, but I am happy if my answer has helped you a little bit.

best regards