Draw resonance structures for each the following compounds or ions, and use formal charges to indicate which resonance structure will make the largest contribution to the true structure, namely, which of the resonance structure most closely approximates the true structure of that compound or ion. Show your calculations. a. SCN-1 b. SO2 c. NO2 -1 d. OH e. BF3 f. CO2

I have attached an image file showing resonance structures. Equivalent resonance structures are not drawn in every case; I will leave it to you to recognize and add those as appropriate. I have left off the formal charges, but I will explain them.
a) thiocyanate
Numbers of valence electrons: S-6, C-4, N-5, plus 1 for the -1 charge. Sulfur can have more than an octet, but carbon and nitrogen must have 8 electrons total in any good Lewis structure.
Formal charges on the atoms are found by taking the atom's original number of valence electrons (which is equal to the atom's core charge) and subtracting 2 for each lone pair and 1 for each shared pair (covalent bond).
In the structure on the left, the formal charges are
S: 6-6-1=-1; C: 4-0-4=0; N: 5-2-3=0.
In the middle structure, we get
S: 6-4-2=0; C: 4-0-4=0; N: 5-4-2=-1
In the right-hand structure, we have
S: 6-6-1=-1; C 4-3=+1; N: 5-4-2=-2
The best structure normally has the least charge separation, and if there is a negative charge, it will be on the most electronegative atom. Nitrogen is the most electronegative of the three, so the center structure is expected to most closely approximate the "true" structure of the ion.
b) sulfur dioxide
Number of valence electrons = 6+6+6 = 18. Sulfur can accommodate more than an octet, but oxygen cannot (Period 2 of the periodic table; no close-lying d orbitals available).
Left structure:
O (left): 6-4-2=0; S: 6-2-4=0; O (right): 6-4-2=0
Middle structures (equivalent but calculation is for the upper one—reverse numbers for the lower one)
O (left): 6-6-1=-1; S: 6-2-4=0; O (right): 6-2-3=+1
Right-hand structures (again equivalent—reverse numbers for the lower one)
O (left): 6-6-1=-1; S: 6-0-4=+2; O (right) 6-4-3=-1
The preferred structure here will be the left-hand structure with no charge separation.
c) nitrite
Total valence electrons = 6+5+6+1=18.
Here the left and middle structures are equivalent. The left has
O (left): 6-4-2=0; N: 5-2-3=0; O (right): 6-6-1=-1
Right-hand structure:
Both oxygens are the same, with 6-6-1=-1, while N has 5-2-2=+1. The equivalent left and middle structures are preferable; the ion has the negative charge distributed between the two oxygen atoms.
d) I presume you meant a hydroxide ion, OH-. This ion has 6+1+1 (8) valence electrons, and hydrogen will have a filled valence shell with only two.
Left-hand structure has
O: 6-6-1=-1; H: 1-0-1=0.
Right structure has
O: 6-8-0=-2; H: 1-0-0=+1.
The left-hand structure, with less charge separation, is preferable.
e) boron trifluoride
Total valence electrons = 3 + 7 + 7 + 7 = 24.
In the left-hand structure, all fluorines are equivalent. This structure, however, violates the octet rule, with only 6 valence electrons on boron. Formal charges are
B: 3-0-3=0; F: 7-6-1=0.
The right-hand structure is one of three equivalent structures having double bonds to different fluorines. It obeys the octet rule. Formal charges are
B: 3-0-4=-1; F (single bonded): 7-6-1=0; F (double bonded): 7-4-2=+1.
This would lead to partial positive charges on all three fluorines and a negative charge on boron—not likely considering the high electronegativity of fluorine. The left-hand structure, despite violating the octet rule, is preferable.
f) carbon dioxide
Total valence electrons are 6 + 4 + 6 = 16.
The left-hand structure has all partial charges equal to 0. The other two each have a partial charge of +1 on the carbon, -1 on the singly-bonded oxygen, and 0 on the doubly-bonded oxygen.
Because oxygen is significantly more electronegative than carbon, the structures showing negative charge on oxygen are considered to contribute significantly. If your assignment is to choose a single structure, however, we know that the molecule is nonpolar and the two oxygen atoms are equivalent, so the best choice would be the left-hand structure.