Explaining the magnetism of gold

2017-09-15T20:48:59Z (GMT) by Robson de Farias
<p>In the present work, a computational study is performed in order to clarify the possible magnetic nature of gold. For such purpose, gas phase Au<sub>2</sub> (zero charge) is modelled, in order to calculate its gas phase formation enthalpy. The calculated values were compared with the experimental value obtained by means of Knudsen effusion mass spectrometric studies [5]. Based on the obtained formation enthalpy values for Au<sub>2</sub>, the compound with two unpaired electrons is the most probable one. The calculated ionization energy of modelled Au<sub>2</sub> with two unpaired electrons is 8.94 eV and with zero unpaired electrons, 11.42 eV. The difference (11.42-8.94 = 2.48 eV = 239.29 kJmol<sup>-1</sup>), is in very good agreement with the experimental value of 226.2 ± 0.5 kJmol<sup>-1</sup> to the Au-Au bond<sup>7</sup>. So, as expected, in the specie with none unpaired electrons, the two 6s<sup>1</sup> (one of each gold atom) are paired, forming a chemical bond with bond order 1. On the other hand, in Au<sub>2</sub> with two unpaired electrons, the s-d hybridization prevails, because the relativistic contributions. A molecular orbital energy diagram for gas phase Au<sub>2</sub> is proposed, explaining its paramagnetism (and, by extension, the paramagnetism of gold clusters and nanoparticles).</p>