1. Determine the hybridization scheme used by the underlined atom in each of the molecules below.
(a) CO2
2 bonding groups, 0 lone pairs on carbon
linear electron geometry Þ sp hybrid orbitals
(b) CH2O
3 bonding groups, 0 lone pairs on carbon
trigonal planar electron geometry Þ sp2 hybrid orbitals
(c) SeF4
4 bonding groups, 1 lone pair on selenium
trigonal bipyramidal electron geometry Þ sp3d hybrid orbitals
2. Do the following for each molecule below.
(a) Determine the number of s bonds and p bonds.
(b) Determine the hybridization scheme used by each underlined atom.
CH3CHO
6 s bonds and 1 p bond
C1 Þ tetrahedral electron geometry Þ sp3 hybrid orbitals
C2 Þ trigonal planar electron geometry Þ sp2 hybrid orbitals
CH3CH2CCH
H H
½ ½
H – C1 – C2 – C3 º C4 – H
½ ½
H H
9 s bonds and 2 p bonds
C1 Þ tetrahedral electron geometry Þ sp3 hybrid orbitals
C2 Þ tetrahedral electron geometry Þ sp3 hybrid orbitals
C3 Þ linear electron geometry Þ sp hybrid orbitals
C4 Þ linear electron geometry Þ sp hybrid orbitals
3. Use molecular orbital theory to predict which of the following ions are expected to be paramagnetic: N22-, O2+, Be22+, and
Li2+. If a particular ion is paramagnetic, how many unpaired electrons does it possess?
N22- Þ 12 valence electrons
_____
s*2p
__ _ _
p*2p
2 unpaired electrons Þ paramagnetic
_¯_
s2p
_¯_ _¯_
p2p
_¯_
s*2s
_¯_
s2s
O2+ Þ 11 valence electrons
_____
s*2p
___ ____
p*2p
1 unpaired electron Þ paramagnetic
_¯_ _¯_
p2p
_¯_
s2p
_¯_
s*2s
_¯_
s2s
Be22+ Þ 2 valence electrons
____
s*2s
_¯_
s2s
Li2+ Þ 1 valence electron
____
s*2s
1 unpaired electron Þ paramagnetic
___
s2s
According to molecular orbital theory N22-, O2+, and Li2+ are expected to be paramagnetic.
4. Use molecular orbital theory to compare the superoxide ion (O2 –) and the peroxide ion (O2 2–) with regard to their:
(a) magnetic character, (b) bond order, (c) bond length, and (d) bond strength.
O2– Þ 13 valence electrons
_____
s*2p
_¯_ __
p*2p
1 unpaired electron Þ paramagnetic
_¯_ _¯_
p2p
Bond order = (8 – 5)/2 = 1.5
_¯_
s2p
_¯_
s*2s
_¯_
s2s
O22- Þ 14 valence electrons
_____
s*2p
_¯_ _¯_
p*2p
0 unpaired electrons Þ diamagnetic
_¯_ _¯_
p2p
Bond order = (8 – 6)/2 = 1
_¯_
s2p
_¯_
s*2s
_¯_
s2s
The superoxide ion is paramagnetic whereas the peroxide ion is diamagnetic. The higher bond order for the superoxide ion means that its bond is shorter and stronger than the bond in the peroxide ion.
