SOMEONE ASKED 👇
Arrange the following in order of decreasing stability : F2, F2-, F2+?
HERE THE ANSWERS 👇
F2-, F2, F2+
It’s a little hard to give a decent answer as you need to see a molecular orbital diagram. Each fluorine atom has 7 valence electrons, and in F2 there are 2 electrons in sigma(2S), 2 in sigma star, 2 in sigma2P, 4 in two pi bonding and 4 in two anti pi bonding. The bond order is 1/2(no. of bonding electrons – no. of antibonding electrons), so for F2 that is 1/2(8-6)=1 so F2 has a bond order of 1 (a single bond). Following the same for F2- you get 1/2(8-7)= 1/2 and for F2+ 1/2(8-5)= 1.5. So in order of stability you have; F2+, F2 and the least stable F2-.
F2- is the least stable, F2+ is the most stable, and F2 is in between.
From largest to smallest HCL NO HI F_2
Awesome except for the fact in the above answers they are using the greater than sign THE WRONG WAY
In general Fluorine is highest electronegative atom having electo negative value 4.2.
As the size of the decreases, the stability of that atom or molecule increases. Because the Nucleus has the capability of holding the outermost orbital electrons with much more attraction, as the size decreases.
loosing of electrons indicates, decreasinf the size and increasing the stability.
So here, F2+ lost one electron and hence decrease the size than F2.
F2- gained one elctron and hence increase the size than F2.
So the as per size the increasing order is
F2- > F2 > F2+
The stability orde is also,
F2- > F2 > F2+
F2-, F2+, F2
the answer is 1, single bondSource(s): mastering chemistry