Chem 115 Essay
If the blue balloon expands, the angle between red and green balloons decreases. c) Nonbonding (lone) electron pairs exert greater repulsive forces than bonding pairs, resulting in compression of adjacent bond angles. 9. 12 a) Both molecules would be symmetrical because all four surrounding atoms are the same. In a symmetrical tetrahedron, the four bond angles are equal to each other, with values of 109. 5°. The H-C-H angles in CH4 and the O-Cl-O angles in ClO4- will have values close to 109. 5°. ) Two 1) The H-N-H angle 2) The N-H distance In a trigonal pyramidal molecule, there are three bonding and one nonbonding electron domains. Since a nonbonding electron domain takes up more space (higher repulsion) it compresses bond angles, the H-N-H angles will be smaller than 109. 5°. The bond angle must be specified. N does not sit in the plane of the H atoms. The distance of N out of the plane is determined by the N-H distances, as well as the H-N-H angles. The N-H distance must also be specified. 9. 20 ) I can confidently predict the bond angles for H2S to be 180° because H2S electron domain geometry is linear with two bonding domains and zero non bonding domain. b) I can confidently predict the bond angles for BCl3 to 120° because I know BCl3 has three electron domains, all of them being bonding domain giving it a trigonal planar molecular geometry. c) I can confidently predict the bond angles for CH3I to be 109. 5 because CH3I molecular geometry is tetrahedral because it has four electron domains that are all bonding domains. d) I can confidently predict the bond angles of CBr4 to be 109. because CBr4 molecular geometry is tetrahedral, it has four electron domains that are all bonding domains e) I’m a bit uncertain about the bond angles of TeBr4 because Te has five electron domain around it, four from the Te—Br bonds and one nonbonding pair. Since nonbonding pairs take up more space and compress the other bond angles I don’t know the exact bond angles of TeBr4. I would guess the bond angles to be 186° and 116°. 9. 28 9. 62 0 +1 -1 -2 +1 -1 -1 +1 0 a.
The first structure would be the most important because the electronegative 0 atom has the negative formal charge. b. The more electrons that are being shared between two atoms the shorter the bond. Since the N-N bond length in N2O is longer than the typical N=N, the middle and right structures where N shares less than three electrons are contributors to the true structure. The middle structure where N and O share more than two electrons pairs does not contribute to the true structure because the N-O bond length is shorter than a typical N=O bond