Can you explain how we get gold(II) sulfide?
Brauer's Handbook of Preparative Inorganic Chemistry ([1, p. 1062] and references therein) lists the following method for gold(II) sulfide synthesis:


  $$\ce{$\underset{1/10}{}$\qquad$\underset{242.7}{\ce{8 AuCl3}}$ + $\underset{20.2\,l.}{\ce{9 H2S}}$ + $\underset{7.2}{\ce{4 H2O}}$ = $\underset{183.3}{\ce{8 AuS}}$ + $\underset{87.5}{\ce{24 HCl}}$ + $\underset{9.8}{\ce{H2SO4}}$}$$
  
  A neutral 1-3% solution of $\ce{AuCl3}$ is precipitated in the cold (the temperature must not exceed $\pu{40 ^\circ C}$) with $\ce{H2S}$ or an alkali sulfide. The precipitate is filtered, thoroughly washed with water, and
  treated with alcohol, anhydrous ether, $\ce{CS2}$ and again with ether.
  The product is dried at $\pu{130 ^\circ C}$ over $\ce{P2O5}$.
  
  Alternate method: A solution of $\ce{AuCl3}$ is added in drops to an
  aqueous solution of sodium dithiosulfatoaurate(I) [...]. 
  
  Properties: Formula weight 229.07. Deep black. Insoluble in water and acids; soluble in aqua regia and solutions of potassium cyanide and alkali polysulfides. Resistant to KOH in the cold, decomposes after prolonged boiling, liberating gold. Thermal decomposition begins at $\pu{140 ^\circ C}$.


Reference


Handbook of Preparative Inorganic Chemistry, 2nd ed.; Brauer, G., Ed.; Academic Press: New York; London, 1965; Vol. 2.