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.
Barium is an element that is primarily used in fireworks because of its distinct green colour while on the other hand, lead is used in electrical industries due to its unique properties. The ions of these elements can simply be identified when its insoluble salts get precipitated by any process.
When you use the equation q = m x c x ∆T you will be able to find the energy gained or lost. The data for the water in this case is just there to distract you so ignore it. :D
