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You can tell if you need a double bond if the number of valence electrons needed for each atom to form an octet exceeds the number of valence electrons actually available. One way to determine which atoms are double bonded is by looking at formal charge, and the "best" case would be where the formal charges of each atom in a lewis structure is closest to 0.
Nicole Jakiel 4F wrote:When drawing Lewis Structures, how do we know if we need to draw a double bond, and how do you determine which element gets the double bond?
To elaborate on Joonsoo's post, formal charge is determined by the formula: FC=V-(L+S/2), where FC is formal charge, V is the number of valence electrons of the atom, and L is the number of lone pairs.
For example, if we're trying to determine the Lewis structure for ONF we can try "testing" configurations to see which one is most stable/has more formal charges equal to 0. So I can try placing a double bond between O and N and then fill in the octets, like so: (https://www2.southeastern.edu/Academics ... re/onf.jpg). Yet I'm still not sure if this is correct, so I use the formula and calculate each atom's formal charge (it helps to look at the structure and see which variables to contribute to which numbers in this case).
For O: FC=4-(4+2/2)=0
For N: FC=6-(2+6/2)=1
For F: FC=7-(6+2/2)=0
From these calculations, I can confirm that this Lewis structure is most ideal.
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