F2 is light yellow/colorless in hydrocarbon solvent. I2 is pink. If a student combines fluorine water with NaI (sodium iodide) in water, then adds pentane and observes a nearly colorless pentane layer, what can the student conclude? F is reduced from F0 to F - F is oxidized I - is oxidized to I0 I is reduced

Question

Abdullah Walter

Chemistry

1 August, 22:04

F2 is light yellow/colorless in hydrocarbon solvent. I2 is pink. If a student combines fluorine water with NaI (sodium iodide) in water, then adds pentane and observes a nearly colorless pentane layer, what can the student conclude? F is reduced from F0 to F - F is oxidized I - is oxidized to I0 I is reduced

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Robert Spence · Answer 1

F₂ + 2 NaI → 2 NaF + I₂

It is given that F₂ is light yellow / colorless in hydrocarbon solvent. The student combines Fluorine water with NaI in water. Then student adds pentane in the mixture of F₂ and NaI. After dissolution, solution was observed and a colorless pentane layer was seen. Alkanes are unreactive in nature. The C-H bond in alkane is difficult to break. whereas, F₂ is very reactive and reacts vigorously with alkanes in presence of light by free radical mechanism. It is given that the color of the solution is nearly colorless. F₂ when present in hydrocarbon solvent is light yellow / colorless / nearly colorless. Hence, F₂ is not reacting with hydrocarbon and there is no reaction taking place (No F ₂ is present)