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a) Draw the Lewis structure of formamide, HCONH2, with a tetravalent carbon, divalent oxygen, and trivalent nitrogen bearing zero formal charges. Show all atoms, bonds, and lone pairs. b) This amide has an important resonance structure with a double C-N bond and a negative formal charge at oxygen. Draw this resonance Lewis structure. Show all atoms, bonds, lone pairs, and formal charges. c) Assuming that the hybridization of C, O, and N atoms in formamide is sp2 for all three atoms, draw the valence bond diagram representing chemical bonding in the molecule of HCONH2 based on the Lewis structure from part a, showing atomic and hybridized orbitals as boxes, valence electrons as arrows in the boxes, and chemical bonds as lines connecting the boxes. Note that the lone pair on N occupies an unhybridized 2p orbital due to the contribution of resonance structure b. (Use Lecture 25 materials for examples of such VB diagrams.) d) Draw another VB diagram for the molecule of HCONH2 showing the shapes of the valence orbitals involved and their mutual overlap. Note that the 2p-orbital on N partially overlaps with the 2p-orbital of O, creating a delocalized π system (in addition to the expected C-O σ bond).
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a) The Lewis structure of formamide (HCONH2) with a tetravalent carbon, divalent oxygen, and trivalent nitrogen bearing zero formal charges is: $$\text{H}\text{-}\text{C}\text{=}\text{O}\text{-}\text{N}\text{-}\text{H}_2$$ Show more…
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a) Draw the Lewis structure of formamide, HCONH2, with a tetravalent carbon, divalent oxygen, and trivalent nitrogen bearing zero formal charges. Show all atoms, bonds, and lone pairs. b) This amide has an important resonance structure with a double C-N bond and a negative formal charge at oxygen. Draw this resonance Lewis structure: Show all atoms, bonds, lone pairs, and formal charges. c) Assuming that the hybridization of C, O, and N atoms in formamide is sp2 for all three atoms, draw the valence bond diagram representing chemical bonding in the molecule of HCONH2 based on the Lewis structure from part a, showing atomic and hybridized orbitals as boxes, valence electrons as arrows in the boxes, and chemical bonds as lines connecting the boxes. Note that the lone pair on N occupies an unhybridized 2p orbital due to the contribution of resonance structure b. (Use Lecture 25 materials for examples of such VB diagrams) d) Draw another VB diagram for the molecule of HCONH2 showing the shapes of the valence orbitals involved and their mutual overlap. Note that the 2p-orbital on N partially overlaps with the 2p-orbital of C creating a delocalized π system (in addition to the expected C-O π bond)
Qbs E.
a. Draw the Lewis structure of $\mathrm{H}_{2} \mathrm{CO}$. b. Describe the orbitals used by the carbon atom in bonding and indicate the approximate bond angles. SOLUTION TO 31 a. Our first attempt at a Lewis structure (drawing the atoms with the hydrogens on the outside of the molecule) shows that carbon is the only atom that does not form the needed number of bonds. If we place a double bond between carbon and oxygen and move the $\mathrm{H}$ from $\mathrm{O}$ to $\mathrm{C}$ (which still keeps it on the outside of the molecule), then all the atoms end up with the correct number of bonds. Lone-pair electrons are used to give oxygen a filled outer shell. When we check to see if any atom needs to be assigned a formal charge, we find that none of them does. SOLUTION TO 31 b. Because carbon forms a double bond, we know that it uses $s p^{2}$ orbitals (as it does in ethene) to bond to the two hydrogens and the oxygen. It uses its "leftover" $p$ orbital to form the second bond to oxygen. Because carbon is $s p^{2}$ hybridized, the bond angles are approximately $120^{\circ}$.
B. Draw Lewis Structures for the following molecules (include all lone pair electrons). For each molecule, follow these steps: 1. Count the total number of valence electrons for each atom. 2. Predict the general connectivity of the molecule. 3. Use any remaining electrons as lone pairs to fill up unsatisfied octets. a. CH4 b. CH2Cl2 c. HI d. Br2 e. BF3 f. H2S g. N2H4 h. CH5N Typical number of bonds per charge-neutral atom: - Group 17 (F, Cl, Br, I) + Hydrogen: 1 - Group 16 (O, S, Se): 2 - Group 15 (N, P): 3 - Group 14 (C, Si): 4 Some molecules contain double/triple bonds, which require an additional rule: 4. If all electrons have been placed as bonds/lone pairs, but some atoms remain unfilled (subvalent), remove a lone pair from an adjacent atom and use it to form a double bond with the subvalent atom. If some atoms are still subvalent, remove additional lone pairs to make a triple bond or additional double bonds. a. O2 b. N2H2 c. CH2O d. C2H2 e. N2 f. C2Cl4 g. CO2 h. HCN 5. If the molecule has a charge, add electrons (for a negative charge) or subtract electrons (for a positive charge) from the total electron count in step 1. Some atoms will not have their typical number of bonds (as in the chart above). a. CH3CO2- b. CO3 2- c. NH4+ d. OH
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