Answers to Chapter 12

26 NSF is the National Science Foundation--the federal agency which funds lots of basic scientific research (including my own!)

9 e pairs. :NtripleS-F. Trigonal planar or sp2 hydridization around S.  The molecule is bent.

28 17 e pairs. The central S has one lone pair and 4 single bonds to F. Therefore it has a see-saw shape and sp3d hybridization. Sigma bonds are formed when an sp3d overlaps with a p orbital from F. Each sigma bond has 2 electrons.

30 a NO2- is trigonal planar for e and bent.  The hybridization around N is sp2
bI3- is linear with 3 e pairs around central I. sp3d hybridization around central I.  It is linear. sigma bonds form where sp3d orbitals overlap p orbital from terminal I's.
c Each C is bonded to one C and two O; one with a double bond and the other with a single bond.  The geometry around each C is sp2 or trigonal planar. sigma bonds are the result of overlaps of C's sp2 orbitals and sp2 from the other C or a p orbital from O.
pi bonds require overlaps of two like p orbitals from O's.
d 12 e pairs for HCO3-.
Trigonal planar, sp2. The central C makes single bonds to 2 O's and a double bond to one O. The H is singly bonded to a singly bonded O.   There is tetrahedral/bent geometry at this O and sp3 hybridization.

34
The end C's are sp2 hybridized and have trigonal geometry. The middle C has sp or linear geometry. The key observation is that the two pi bonds are perpendicular to each other so the two sets of hydrogens are in planes which are perpendicular to each other.

36
C2 has 8 valence electrons and the Lewis structure would predict a quadruple bond, BO=4.
Molecular orbital theory predicts sigma2s2, sigma*2s2, pi2p2, pi2p2.  This makes 4 net bonding electrons for a bond order of 2.

40 Not true. If the electron is lost from an antibonding orbital, the bond is strengthened. Ex. F2 F2+

44
a CO+ 9 valence e, BO 2.5 parmagnetic
b CN- 10 valence e BO 3 diamagnetic shorter, stronger bond.

48 Delocalization of electrons occurs if there are resonance forms
a HCO2- 2 forms Yes
b CO3- 3 forms Yes
c CH3+ 1 form No
64
Using atomic configurations, He is 1s2. Promote 1e to 2s and let a sigma bond form between the 2s orbitals of 2 He.
Alternately, use molecular orbital theory.
Normally the BO of He2 = 0 (1sigma s2, 1sigma*s2).  If we promote one of the antibonding e to the next orbital which is 2sigma s then we get BO=1
68
F2Cl- has 11 electron pairs so tbp e shape or AX2E3.  The 3 lone pairs around the central Cl are equitorial and the F's are axial.  The molecule is linear and the Cl has 5 sp3d orbitals.
F2Cl+ has 10 e pairs, 2 lone pairs, sp3 hybridization and tetrahedral/bent geometry. It is AX2E2.