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Arrange the following in order of decreasing stability. A blank molecular orbital diagram (Part A 1 figure) has been provided to help you.
Rank the fluorine species from most to least stable. To rank items as equivalent, overlap them. F2, F2+, F2-

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The correct order will be " F2-, F2, F2+ " because 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, F2+, F2 and the least stable F2-.
CintiaSalazar

Answer:

Then F- has greater stability, then F follows and the least stable is F +

Explanation:

Fluorine is an element of group 17 (VII A) that has 7 electrons in its last layer. This element tends to form anions (earning an electron). If the fluorine gains an electron, it acquires the electronic configuration of the neon (noble gas), forming the anion F-. This makes the flour acquire greater stability. and consider the F- as the most stable anion.

The F + ion is the most unstable element because in order to form it is called the Ionization Energy, which is the energy needed to start an electron to a gaseous atom, isolated and in a fundamental state. The electrons are attracted to the nucleus and it is necessary to provide energy to start them. In addition, the F +, having a positive charge, will have a greater tendency to acquire 2 electrons to complete its last layer and achieve stability, while the F has a tendency to acquire only 1 electron and form the F- anion to acquire stability.

Then F- has greater stability, then F follows and the least stable is F + .

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