3 Corrected electronegativity to reduction potential
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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
source | link

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.

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source | link

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.