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Ch 10 Example Problems Answers

Ch 10 Example Problems Answers - HODSON CHE 1301 Petrucci...

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HODSON CHE 1301 Petrucci Ch10 Example Problems The following problems are taken from old exams ...... 1. In which of he following molecules would you expect the nitrogen to nitrogen bond to be the shortest (where appropriate, formulae are written as condensed structural formulae) ? a) NH 3 b) H 2 NNH 2 c)   O 2 NNO 2 d) N 2 e) N 2 O Answer: D Explanation: When considering bond lengths, we expect that a given bond will be shorter if there are more electrons involved, i.e. multiple bonds are stronger and hence shorter than single bonds (more electrons shared between the atoms meaning more electrostatic attraction to the nuclei, pulling the nuclei closer together). So, to find the strongest Nitrogen–Nitrogen bond above, we think about the presence of multiple bonds in our Lewis structures. Triple bonds are shorter and stronger than double bonds, which are shorter and stronger than single bonds. Remember: to draw a Lewis Structure, first find the number of valence electrons you have, then draw a skeletal diagram (see hints, slide 31 if necessary), then arrange the electrons to give the outermost atoms octets (duets for H), extra electrons go on central atoms as lone pairs, or if necessary include multiple bonds to satisfy octets for the central atoms. a) Ammonia, NH 3 . We could immediately eliminate this because there isn’t an N to N bond, but whatever …… V.e = 5 + (3 × 1) = 8. Skeletal diagram has N in center (H is always terminal), and three lines (bonds, 2 electrons each) are draw between the N and three H atoms; the remaining 2 electrons are placed on the N as a lone pair. (note: electron group geometry is tetrahedral, molecular geometry is trigonal pyramidal) b) H 2 NNH 2 v.e = (2 × 5) + (4 × 1) = 14. Skeletal diagram is trickier, but this is a condensed structural formula, so it shows connectivities where possible (read left to right in the formula to find the atomic framework). We know that H is always terminal, and that molecules tend to be symmetrical. Considering this we end up with the structure on the left below as our skeletal diagram, and then adding the remaining 4 electrons (10 electrons ‘used’ in 5 covalent bonds in the skeletal structure) as lone pair to the
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nitrogens comlpetes octets (notice the H atoms all have duets, and if you tried to have multiple bonds between the N atoms they’d go above their stable octets). c) Closely related to b) O 2 NNO 2 v.e = (2 × 5) + (4 × 6) = 34. Remember this is a condensed structural formula, so it shows connectivities where possible. We end up with the structure on the left below after adding the remaining 24 electrons (10 electrons were ‘used’ in 5 covalent bonds in the skeletal structure) as lone pairs to the oxygen atoms to complete octets. The nitrogen atoms still need another 2 electrons each, so we allow them to share lone pairs from the oxygen atoms. Note that resonance forms are possible
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