# I know that Bond Order tell us the stability of the bond of the two atoms. So the more bond order ## reach closer to zero, is it more stable?

No, it’s actually the other way around, the larger the value of the bond order, the more stable the bond.
Bond order is calculated by using the number of electrons that reside in and in .
##”B.O.” = 1/2 (“bonding electrons” – “antibonding electrons”)##
The bigger the difference between how many electrons reside in bonding orbitals and how many reside in antibonding orbitals, the higher the bond order, and thus the stronger the bond.
Keep in mind that triple bonds are more stable than double bonds, which in turn are more stable than single bonds, so a higher bond order indicates greater stability.
A bond order equal to 1 implies that you have 2 more electrons located in bonding orbitals than in antibonding orbitals, which means that the atoms form a single bond.
A bond order of 2 implies that you have 4 more electrons in bonding orbitals than in antibonding orbitals, so the atoms will form a double bond.
Likewise, a bond order of 3 implies that you have 6 more electrons in bonding orbitals than in antibonding orbitals ##->## the atoms form a triple bond.
Similarly, a bond over of 1.5 is more stable than a bond order of 1, a bond order of 2.5 is more stable than a bond order of 2, and so on.
Check out this example of how adding/removing electrons from the oxygen molecule affects bond order and stability:

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I know that Bond Order tell us the stability of the bond of the two atoms. So the more bond order ## reach closer to zero, is it more stable?
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