Fluorine is the most reactive halogen followed by chlorine, bromine and iodine
Explanation:
The reactivity of halogens decreases down the group, that is, fluorine is the most reactive halogen followed by chlorine and bromine. The least reactive halogen is iodine.
There are several reasons and viewpoints explaining this concept:
Fluorine is the smallest atom of all halogens, since: (a) it has the lowest number of electron shells and (b) it has the greatest electronegativity, so that the electrons are attracted with the greatest force towards the nucleus. Therefore, it has the lowest radius. This means when the protons in the nucleus attract the electrons from another atom, the distance between the nucleus of fluorine and the electrons of another atom will be the smallest of all halogens. The lower the distance, the greater the attraction force, so that we obtain the strongest bond with fluorine and the weakest with iodine, as iodine has the greatest radius of all. From this standpoint, fluorine is the most reactive.
Similarly, fluorine is least shielded atom, as it only has two valence shells. Going down the group, the number of shells increases, so iodine is the most shielded atom. Shielding increases the atomic radius, so that it's easier for fluorine to approach the electrons from the valence shell of another atom compared to more shielded chlorine, bromine and iodine.
I believe the correct answer is the second option. The type of decay that characterizes the change of nuclides to their respective daughter products would be exponential decay. This type of decay is characterized by the decrease of quantity of a material according to the equation y=ab^x.
A lit candle needs to draw oxygen from the air in order to continue burning. ... Thus, oxygen inside the glass jar will decrease and it gets filled with carbon dioxide, and eventually the candle's flame will Extinguish.
