Question

Lewis Structures

CHM113 Chapter 8 Resource Worksheets Resource 4: Lewis Structures It is recommended that you complete this worksheet on a separate sheet of paper. There are multiple methods of calculating formal charge. Any of the following formulas will work FC - VE - BE-NBE F C - VE - (NBE+ BE) FC - VE - NBE - Bonds 1. Draw Lewis structures (including lone pairs) for the following molecules and calculate formal charge for each atom a. NCI b. COS c. PBR d. HCN e. HS f. CH 0 h. SF i. CH g. XeF j. N:O: (N's bonded) 2. Draw Lewis structures (including lone pairs) for the following molecules with charges. Calculate the individual charges on each atom and be sure to indicate the overall formal charge on the molecule. a. H:0 b. BF, c. ICI d. CIF: e. C.02 f. SF 8 ОН hN; (try N-N-N) 3. For each of the following, draw all the possible Lewis structures, showing resonance Indicate all formal charges a. N.O b. NO; 4. Consider SO; a. Draw the Lewis dot structure that minimizes the formal charges and expands the octet. b. Draw the Lewis dot structures that have resonance and do not expand the octet. 5. Consider IO. a. Draw the Lewis dot structure that minimizes the formal charge and places the negative charge on the more EN atom, and expands the octet.

Answer 1

Solution.

hewis Structure - piso known as hemis dof structure These stoue tuves represents the bandling cin atoms and Shows all lone pairs present. Electrons are represented by clot & Count the total number of Valence electrons can a molecule → SJ o charoge add le; 94 o Change cleaned Subtract Yes Central atom would be one with less electronegativity [H always stays as terminal ] son [Single atom : o usually " as terminal would be at C always as central atom centre] 3) Docw the one bond between each atom each bond would have 2 eis : 4) Left eis are assigned to ators to fulfill Octet Total valence electrons = 5+7(3) = 26 N ce :☺:N:Ü: 7 l ice: b) COS Tolal valence electrons = 4+6+6 → 16 Ś

C) PB83 Total valence electrons → 5+7(3) = 26 es. Bo: P : 35:7 08 B-P-Bis:) d) HCN Total valence electrons > I+4+5 = 10es. y te v [H:C:::N:) or 4-ČEN: B Total Valence electrons = 2(1) +6 = 8

C) PB83 Total valence electrons → 5+7(3) = 26 es. Bo: P : 35:7 08 B-P-Bis:) d) HCN Total valence electrons > I+4+5 = 10es. y te v [H:C:::N:) or 4-ČEN: B Total Valence electrons = 2(1) +6 = 8