if i place C6H14 and KI in a test tube, then some metal salt (MNO3) is added...

If the M+ reacts with and oxidizes the I, What color is c6h14?

I think its purple.. but im not sure why it would be that color .. :S .. care to explain please ?

and if it doesnt react .. what color will it be.. is it colorless ? or still purple?

Oxidizing I^- forms I2 and molecular I2 dissolved in a non-polar solvent (such as hexane) with excess I^- gives I^-3 and that ion is a purple color.

so if it doesn't react.. is it colorless? because hexane is colorless.. ?

Ah, chemistry - where colors can be quite the confounding conundrum! Now let me clownishly clarify things for you. When C6H14, also known as hexane, is in its pure form, it's actually colorless. So, if it doesn't react with the metal salt (MNO3), it will remain colorless. It's like a stealthy little ninja, hiding in plain sight, or you know, like a transparent ghost haunting your test tube!

However, if the metal salt reacts and oxidizes the I (iodine) in KI (potassium iodide), things get a bit more colorful. The liberated iodine can form a complex with the hexane, resulting in a purple-ish color. So, if the reaction takes place successfully, you might just be blessed with a majestic, albeit unexpected, hue of purple in your test tube. Just remember, chemistry can be colorful and unpredictable like a whimsical paint palette!

The color of a substance is determined by the absorption and reflection of light by its molecules. In the case of C6H14 (hexane), since it is a hydrocarbon, it does not have any specific chromophores that would absorb visible light and produce color. Therefore, pure hexane is colorless.

If KI (potassium iodide) is added to the test tube, it will dissociate into K+ and I- ions. The I- ions will react with the metal salt MNO3, leading to the oxidation of I- to I2 (iodine). Iodine has a distinct purple color. Consequently, if I2 is formed, the color of the solution in the test tube containing C6H14 and KI will turn purple.

If the M+ ions do not react with and oxidize the I- ions, then no color change will occur in the solution. The C6H14 will remain colorless.

To determine the color of a compound, we need to consider its chemical structure and the presence of any conjugated systems or chromophores.

C6H14 is the chemical formula for hexane, which is a colorless liquid. In its pure form, hexane does not absorb visible light and therefore appears colorless.

When KI (potassium iodide) is added to the test tube, it dissociates into K+ and I- ions. These ions are generally colorless, so they do not contribute to the color of the solution.

However, when the metal salt MNO3 (e.g., Mn(NO3)2) is added, it is possible that a reaction occurs between the metal ion (M+) and the iodide ion (I-). This reaction can result in the oxidation of I- to iodine (I2), which is a brownish-purple solid.

So, if the M+ ion reacts and oxidizes the I- ion to form iodine, then the solution containing hexane and iodine would have a purple color due to the presence of iodine. It's important to note that the color intensity may vary depending on the concentration of iodine formed.

However, if the M+ ion does not react with and oxidize the I- ion, the color of the solution should remain colorless, as the individual components (hexane and KI) do not contribute to a visible color change.