3 Corrected electronegativity to reduction potential edited Jun 26 '14 at 15:04 prooffreader 44122 silver badges77 bronze badges When you heat the copper wire in a flame, it is oxidized on the surface to copper (II) oxide: $$\ce{2Cu(s) + O2(g) -> 2CuO(s)}$$ Then when you mix it with the halide and heat it, the halide's higher electronegativityreduction potential makes it displace the oxygen. Here's an example with sodium chloride: $$\ce{CuO(s) + 2NaCl(s) -> CuCl2(g) + Na2O}$$ Copper halides are volatile (except for fluorides); the gas is hot enough to push some electrons into an excited state, and as they cool a tiny bit, the electrons drop back into their rest state and emit the excess energy in the form of a photon with a wavelength corresponding to the color green. When you heat the copper wire in a flame, it is oxidized on the surface to copper (II) oxide: $$\ce{2Cu(s) + O2(g) -> 2CuO(s)}$$ Then when you mix it with the halide and heat it, the halide's higher electronegativity makes it displace the oxygen. Here's an example with sodium chloride: $$\ce{CuO(s) + 2NaCl(s) -> CuCl2(g) + Na2O}$$ Copper halides are volatile (except for fluorides); the gas is hot enough to push some electrons into an excited state, and as they cool a tiny bit, the electrons drop back into their rest state and emit the excess energy in the form of a photon with a wavelength corresponding to the color green. When you heat the copper wire in a flame, it is oxidized on the surface to copper (II) oxide: $$\ce{2Cu(s) + O2(g) -> 2CuO(s)}$$ Then when you mix it with the halide and heat it, the higher reduction potential makes it displace the oxygen. Here's an example with sodium chloride: $$\ce{CuO(s) + 2NaCl(s) -> CuCl2(g) + Na2O}$$ Copper halides are volatile (except for fluorides); the gas is hot enough to push some electrons into an excited state, and as they cool a tiny bit, the electrons drop back into their rest state and emit the excess energy in the form of a photon with a wavelength corresponding to the color green. 2 added 16 characters in body edited Jun 26 '14 at 6:00 jonsca♦ 1,80477 gold badges2626 silver badges5252 bronze badges When you heat the copper wire in a flame, it is oxidized on the surface to copper (II) oxide: 2Cu(s) + O2(g) --> 2CuO(s)$$\ce{2Cu(s) + O2(g) -> 2CuO(s)}$$ Then when you mix it with the halide and heat it, the halide's higher electronegativity makes it displace the oxygen. Here's an example with sodium chloride: CuO(s) + 2NaCl(s) --> CuCl2(g) + Na2O$$\ce{CuO(s) + 2NaCl(s) -> CuCl2(g) + Na2O}$$ Copper halides are volatile (except for fluorides); the gas is hot enough to push some electrons into an excited state, and as they cool a tiny bit, the electrons drop back into their rest state and emit the excess energy in the form of a photon with a wavelength corresponding to the color green. When you heat the copper wire in a flame, it is oxidized on the surface to copper (II) oxide: 2Cu(s) + O2(g) --> 2CuO(s) Then when you mix it with the halide and heat it, the halide's higher electronegativity makes it displace the oxygen. Here's an example with sodium chloride: CuO(s) + 2NaCl(s) --> CuCl2(g) + Na2O Copper halides are volatile (except for fluorides); the gas is hot enough to push some electrons into an excited state, and as they cool a tiny bit, the electrons drop back into their rest state and emit the excess energy in the form of a photon with a wavelength corresponding to the color green. When you heat the copper wire in a flame, it is oxidized on the surface to copper (II) oxide: $$\ce{2Cu(s) + O2(g) -> 2CuO(s)}$$ Then when you mix it with the halide and heat it, the halide's higher electronegativity makes it displace the oxygen. Here's an example with sodium chloride: $$\ce{CuO(s) + 2NaCl(s) -> CuCl2(g) + Na2O}$$ Copper halides are volatile (except for fluorides); the gas is hot enough to push some electrons into an excited state, and as they cool a tiny bit, the electrons drop back into their rest state and emit the excess energy in the form of a photon with a wavelength corresponding to the color green. 1 answered Jun 26 '14 at 4:15 prooffreader 44122 silver badges77 bronze badges When you heat the copper wire in a flame, it is oxidized on the surface to copper (II) oxide: 2Cu(s) + O2(g) --> 2CuO(s) Then when you mix it with the halide and heat it, the halide's higher electronegativity makes it displace the oxygen. Here's an example with sodium chloride: CuO(s) + 2NaCl(s) --> CuCl2(g) + Na2O Copper halides are volatile (except for fluorides); the gas is hot enough to push some electrons into an excited state, and as they cool a tiny bit, the electrons drop back into their rest state and emit the excess energy in the form of a photon with a wavelength corresponding to the color green.