What is the product of this combination reaction? $A l (s) + I_{2} (s) \to$ $Al(s) + I_2(s) ->$ ?

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What is the product of this combination reaction? $A l (s) + I_{2} (s) \to$ $Al(s) + I_2(s) ->$ ?

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Answer 1 · 2017-03-01T17:53:33

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Explanation:

Aluminum is oxidized by iodine according to the following reaction:

$A l (s) + \frac{3}{2} I_{2} (s) \to A l I_{3} (s)$ $Al(s) + 3/2I_2(s) rarr AlI_3(s)$

While a subvalent aluminum iodide is known, the material you would put into a bottle is $aluminium(III) iodide$ $"aluminium(III) iodide"$ . Like the chloride, $A l I_{3}$ $AlI_3$ is dimeric in the solid state to give $A l_{2} I_{6}$ $Al_2I_6$ , i.e. ${I_{2} A l}_{2} {(μ - I)}_{2}$ ${I_2Al}_2(mu-I)_2$ .

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What is the product of this combination reaction? $A l (s) + I_{2} (s) \to$ $Al(s) + I_2(s) ->$ ?

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What is the product of this combination reaction? $A l (s) + I_{2} (s) \to$ $Al(s) + I_2(s) ->$ ?