00:01
Let's write down the three species that we are dealing with here.
00:06
F2 plus, f2 and f2 minus.
00:15
Now the number of valence electrons.
00:25
Since fluorine has seven electrons in its automorse shell, f2 will have 14 valence electrons.
00:34
F2 plus, since it's losing an electron, will have 13 electrons.
00:38
And f2 minuses it's gaining an electron will have 15 electrons so let's figure out the electronic configuration sigma ns sigma star ns sigma ns sigma np pi n p plus pi np so let's write x and y here and give this a denomination of set and pi star np x equals pi star np x equals pi star n p p y and last but not the least sigma n p z star so we have 13 electrons here so let's fill it up two three four five six seven eight nine ten eleven twelve thirteen so from this we can calculate the bond order which is a number of bonding orbitals minus the number of anti -bonding orbitals divided by two so we have one two 3, 4, 5, 6, 7, 8 bonding orbitals, bonding electrons, and then 1, 2, 3, 4, 5 anti -bonding electrons, which means that the bond order will be 1 .5.
02:04
Now let's do the same thing for f2.
02:10
Writing down the electronic configuration first.
02:38
Now let's fill up the electrons with 14, so 1, 2, 3, 4, 6, 7.
02:43
8, 9, 10, 11, 12, 13, 14.
02:46
The bond order will be 1, 2, 3, 4, 5, 6, 7, 8 minus 1, 2, 3, 4, 5, 6.
03:05
2 divided by 2 will be 1.
03:10
Now let's do the last one...
