Download solutions

Chem 6 sample Exam 3 solutions
1. (a) CO, (c) BrF3 ,
BBr3 , like BF3 , is trigonal planar and the B-Br bond dipoles cancel out. The
same is true for trigonal bipyramidal PF5 .
Linear CO has a dipole moment since C and O have different electronegativities.
BrF3 is a distorted T-shape, and the bond dipoles don't cancel out.
XeF4 is square planar, and the bond dipoles cancel out.
2. (a) O3 has a structure based on the trigonal plane, since there are 2 oxygens
and 1 lone pair on the central O.
:
:
O+
+O
:O
–
O:
O:
:
: :
:
: :
:O
Therefore, its idealized geometry is bent with an O-O-O angle of 120°. Again a
distortion is expected because of the lone pair on O, to make the angle <120°.
(b) BrF5 has a structure based on an octahedron, with 5 Br-F bonds and one lone
pair on Br.
lone pairs shown only
on one F
F
:
F
90°
:
:
F
Br
90°
F
:
F
Its idealized structure is a square pyramid, but to avoid repulsion from the lone
pair, it will distort such that the bond angle F square-plane-Br-Faxial is less than the
ideal 90°.
3. (a) GaCl 4 – : this has 32 valence electrons, and its structure is based on
tetrahedral geometry at Ga (4 electron pairs around the central atom). It is
therefore tetrahedral (like CH4 ).
SbCl4– has 34 valence electrons, and its
structure is based on trigonal bipyramidal geometry at Sb (5 electron pairs
around the central atom). The lone pair goes in an equatorial position, so the
structure is a distorted see-saw like SF4 .
(b) SbCl 2+ (18 valence electrons) has a structure based on trigonal planar
coordination at Sb (3 electron pairs at Sb). One is a lone pair, so the structure is
bent.
GaCl2 + (16 valence electrons at Ga) has a structure based on linear coordination
at Ga (2 electron pairs around the central atom). It is linear.
Conclusion: the molecule is really [SbCl2 +][GaCl4 ]–
4. Use the MO scheme provided to develop the valence electron configurations
for these molecules. Note: I used the "Li-N" scheme since N is involved; if you
used the "O-Ne" scheme (since O is involved) the electron configurations will be
slightly different but you will reach the same conclusions.
NO+ has 10 valence electrons: (σ2s)2 (σ∗2s)2 (π2px , π2py )4 (σ2pz )2
NO has 11 valence electrons: (σ2s)2 (σ∗2s)2 (π 2px , π 2py )4 (σ2pz )2 (π∗2px , π∗ 2py
)1
NO– has 12 valence electrons: (σ2s)2 (σ∗2s)2(π 2px ,π 2py )4 (σ2pz )2 (π∗2px , π∗2py )2
The bond orders are:
for NO+, [8 bonding electrons – 2 antibonding electrons]/2 = 3
for NO, [8 bonding electrons – 3 antibonding electrons]/2 = 2.5
for NO–, [8 bonding electrons – 4 antibonding electrons]/2 = 2
Conclusion: in increasing bond order: NO– <NO<NO+
5. Consider the MO's in O2 and N2 from which an electron is removed when
these molecules are ionized. Then compare the energy of these MO's to that of
the AO's on O and N, from which the electron is removed when these atoms are
ionized.
For O2 , with 12 valence electrons, using the appropriate scheme
(σ2s)2 (σ∗2s)2 (σ2pz )2(π 2px ,π 2py )4 (π∗ 2px , π∗2py )2
On ionization, the electron will be removed from the highest-energy occupied MO
(the HOMO), which is one of the degenerate π∗ 2px , π ∗2py set. This is an
antibonding MO formed from a combination of 2p AO's on oxygen, so it is
higher in energy than those atomic orbitals.
The valence electron configuration of the O atom is [He]2s2 2p4. On ionization,
the electron will be removed from the highest-energy occupied AO, which is one
of the degenerate 2p set.
Therefore, ionizing O requires removing an electron from an orbital whose
energy is lower than that of the orbital involved in ionization of O2. This should
require more input of energy, so the IE of O is greater than that of O2.
For N2, with 10 valence electrons, the electron configuration is the same as NO+
above:
(σ2s)2 (σ∗2s)2 (π 2px ,π 2py )4 (σ2pz )2
On ionization, the electron will be removed from the highest-energy occupied MO
(the HOMO), which is the σ 2 p z . This is a bonding MO formed from a
combination of 2pz AO's on nitrogen, so it is lower in energy than those atomic
orbitals.
The valence electron configuration of the N atom is [He]2s 2 2p3. On ionization,
the electron will be removed from the highest-energy occupied AO, which is one
of the degenerate 2p set.
Therefore, ionizing N requires removing an electron from an orbital whose energy
is higher than that of the orbital involved in ionization of N2 . This should require
less input of energy, so the IE of N is less than that of N2.
6. (i) C2 has 8 valence electrons, so its electron configuration is
(σ2s)2 (σ∗2s)2 (π 2px ,π 2py )4 . The HOMO is one of the degenerate π2px ,π2py set
(a π-bonding MO formed from a combination of 2px or 2py orbitals on C) Since
this is a homonuclear molecule, there should be equal contributions from the
atomic orbitals to the MO. Sketch:
π bonding MO for C2
+
(ii) CN has 9 valence electrons, so its electron configuration is
(σ2s)2 (σ∗2s)2 (π 2px ,π 2py )4 (σ2pz)1
The LUMO will be one of the degenerate set of π*2px, 2py antibonding MO's.
Because of the relative energies of the C and N AO's, this MO will be polarized
on C (the relative contribution of the C AO to the MO will be greater than that of
the N) and look like this:
π* antibonding MO for CN,
polarized on C
–
C
N
CN
7. (a) Methylacrylonitrile has a C=C double bond and a C≡N triple bond.
H
H
bC
H
C c
a
C
N
:
C
H
H
d
(b) The hybridization is as follows:
Ca is sp2 with 120° bond angles
Cb is sp3 with 109.5° bond angles
Cc is sp2 with 120° bond angles
Cd is sp with 180° bond angles
8. (a)
+
M pz
σ-bonding MO
Ns
(b)
y
+
z
M dxy
π-antibonding MO
N py
9. ClO4 – is sp3 at Cl and tetrahedral.
ClO3 – is also sp3 at Cl and the structure
is based on a tetrahedron-since there is a lone pair it's pyramidal.
O–
O
O
O–
Cl
O
only 1 resonance structure is shown
10. (a) (iii)
(b) (i) and (iv)
(c) (i)
(d) (i)
O
O
Cl
:
only 1 resonance structure is shown
(e) (i) and (iv)