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As,

$$\ce{ZnCl2 + Na2CO3 -> ZnCO3 + 2 NaCl}$$ $$\ce{ZnCO3 -> ZnO + CO2}$$ $$\ce{Co(NO3)2 -> CoO + 2NO2 +\frac{1}{2}O2}$$

Hence, $\ce{CoO + ZnO -> CoZnO2}$

here, $\ce{Zn_{1-x}Co_xO2}$ (Rinmann's green)

Maybe, $\ce{CoO}$ being acidic in nature and $\ce{ZnO}$ being amphoteric reacts with each other, similarly in case of $\ce{CoAlO4}$ (Thénard's blue) but then why $\ce{CoO}$ react with $\ce{MgO}$ to form $\ce{CoO . MgO}$ (pink), even though they are both basic in nature?? Also, tell if it reacts ,what compound does this test form with anions such as $\ce{PO^{3-}_4}$ and $\ce{BO^{3-}_3}$ (as, it probably gives fusible blue colour with them) .

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    $\begingroup$ Chem+Math MathJax formatting: Basics / Expressions/formulas/equations / Upright vs Italics / Math SE Mathjax tutorial $\endgroup$
    – Poutnik
    Commented Apr 25 at 17:29
  • $\begingroup$ For future reference: for the body of questions, answers, and comments, chemistry.se offers to use mhchem as a comfortable method to add chemical equations and report numerical values (\pu{}) including a non-breakable space. $\endgroup$
    – Buttonwood
    Commented Apr 25 at 20:48
  • $\begingroup$ An element of why they form (typically from a melt) are the ionic radii, that the spinels offer «fitting pockets» to accommodate the relevant ions in place which (overall) is energetically favorable. As to why they may become colorful: electronic interactions and doping by the transition metal cations both electronically as well as spatially fitting. $\endgroup$
    – Buttonwood
    Commented Apr 25 at 20:53

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