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Chapter 11 Homework answers 3) C has only 2s and 2p orbitals, which formulates a maximum of 4 hybrid orbitals. Si has the 3d orbital in addition to the 3s and 3p, so it can forum up to 6 orbitals. 7) a) sp2 b)sp2 c)sp2 13) sp3d s + 3p + d sp2 s + 3p 15) I’ll upload this later , I have to figure out what I wrote 20) a) false, double bond = sigma bond + pi bond b) false, triple bond = sigma bond + 2 pi bond c) true d)true e) false, pi bond comes after sigma bond f) false, end to end electron sharing results in a bond with the electron density along the boundaries 23) (I can ’t draw on Wor d, so try and use the des cri pti on. If you really need it, I can go on WiZiQ and write it out on a white board) a) At the cent er is nitrogen. On one side of fluori ne with a single bond to nitrogen, and it has 6 electro ns. On the left side is oxygen with a double bond, and it has 4 electrons. Nitrogen has a sp2 hybridzation with 2 sigma bonds and a pi bind. b) At the center is two carbons, which are double bonded to each other. Each carbon is also attached to two fluorine atoms.  The fluorine atoms each h ave 6 el ectrons. The carbons have sp2 hybrization, with 3 sigma bonds and 1 pi bond. c) There are two carbons in the center, which are single bonded to each other . Each carbon i s also tripl e bonded to a nitrogen.  The nitrogen also has 2 electrons. The carbons have sp hybridzation, with 2 sigma bonds and 2 pi bonds. 25) H3C H C = C Single C – H bond and C – C bond are sigma bonds H CH3 Double C C bond is a sigma bond and a

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Page 1: Chapter 11 Homework Answers

8/14/2019 Chapter 11 Homework Answers

http://slidepdf.com/reader/full/chapter-11-homework-answers 1/2

Chapter 11 Homework answers

3) C has only 2s and 2p orbitals, which formulates a maximum of 4hybrid orbitals. Si has the 3d orbital in addition to the 3s and 3p, so itcan forum up to 6 orbitals.

7) a) sp2b)sp2c)sp2

13) sp3d s + 3p + d

sp2 s + 3p

15) I’ll upload this later, I have to figure out what I wrote

20) a) false, double bond = sigma bond + pi bondb) false, triple bond = sigma bond + 2 pi bondc) trued)truee) false, pi bond comes after sigma bondf) false, end to end electron sharing results in a bond with theelectron density along the boundaries

23) (I can’t draw on Word, so try and use the description. If youreally need it, I can go on WiZiQ and write it out on a whiteboard)

a) At the center is nitrogen. On one side of fluorine with a singlebond to nitrogen, and it has 6 electrons. On the left side isoxygen with a double bond, and it has 4 electrons. Nitrogen hasa sp2 hybridzation with 2 sigma bonds and a pi bind.b) At the center is two carbons, which are double bonded toeach other. Each carbon is also attached to two fluorine atoms. The fluorine atoms each have 6 electrons. The carbons have sp2hybrization, with 3 sigma bonds and 1 pi bond.c) There are two carbons in the center, which are single bondedto each other. Each carbon is also triple bonded to a nitrogen. The nitrogen also has 2 electrons. The carbons have sp

hybridzation, with 2 sigma bonds and 2 pi bonds.

25)

H3C HC = C Single C – H bond and C – C bond are

sigma bonds

H CH3 Double C – C bond is a sigma bond and a

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pi bond Trans

(there is a double bond between the Cs and single bonds from thedouble bonds to the four on the sides)

H3C CH3C = C Single C – H bond and C – C bond are

sigma bonds

H C Double C – C bond is a sigma bond and api bond

cis(same thing here)