CH101 SPRING 2014 MID TERM REVIEW QUESTIONS (CHAPTERS 3-6) SOLUTIONS
1. A gold nucleus is located at the origin of coordinates, and an electron is brought to a position
exactly 2 Å from the origin.
(a) Calculate the force between the gold nucleus and the electron.
(b) Calculate the potential energy of the gold nucleus and the electron.
The unit of electronic charge e is 1.602 x 10-19 C and the permitivity of vacuum constant 0 is
8.854 x10-12 C2J-1m-1.
Use Coulomb's Law, with r = 2 A (2 x 10-10 m) and Z = 79.
qe qn
(a) F(r) =
40
=
r2
_
( e)(+79e)
40r2
(-79)(1.602 x 10-19 C)2
=
= -4.56 x 10-7 N
(4) (3.142) (8.854 x 10-12 C2/J m) (2 x 10-10 m)2
qe qn
(b) V(r) =
=
40r
_
( e)(+79e)
= (-4.56 x 10-7 N)(2 x 10-10 m) = -9.11 x 10-17 J
40r
2. Write the best Lewis diagram for the sulfite anion SO 32-.
S is central atom (group VIA)
_ ..
:O :
0
:O :
_ ..
: ..O
_
..
O:
..
S
..
0
_ ..
: ..O
Rather than
+
..
S
_
..
with more
O:
..
formal charges
3. Write down Lewis diagrams, with formal charges on all atoms, for the molecules cyanic acid
(HOCN) and fulminic acid (HCNO). State which molecule is likely to be the more stable.
Lewis diagrams (with all formal charges):
0
H
0..
O
..
0
C
0
H
0
N:
-1
..
C
+1
N
0
0
..
O : or H
0
+1
C
N
-1
..
O:
..
HOCN can be written with no formal charges and hence is more stable
4. Write resonance structures for the carbonate ion (CO32-) and comment on the C-O bond length
in real carbonate salts.
..
.. _
.... _
_
O:
_
O:
O
..
.. :
..
..
..
:O
C
:O
C
:O
C
..
.. _
..
.. _
O
:
O:
O
..
.. :
..
O
O
C
O
2-
Resonance implies
equal C-O bond
lengths, between
true C=O and C-O
bond lengths (agrees with
experiment)
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5. Arrange the following in order of bond strength.
(a) HCl; HF; HBr; HI
(b) O2; F2; N2
(a) The order of atomic size is F<Cl<Br<I, hence order of bond strength is HF>HCl>HBr>HI.
(b) Bond orders are O2 = 2; F2 = 1; N2 = 3, hence order of bond strength is F2<O2<N2.
6. Use the VSEPR theory to determine the shape of the xenon tetrachloride (XeCl 4) molecule.
State whether XeCl4 is polar or nonpolar.
Cl
xx
Cl
Xe
Cl
xx
Lewis diagram
electron pairs = 6: octahedral arrangement
Cl
Cl
Cl
Cl
Xe
Cl
Square planar shape, with lone pairs as far apart as possible: nonpolar
7. Determine the excited electronic state of Li2+ that has the same energy as the first excited state
of hydrogen. [RH (Rydberg constant) = 3.29 x 1015 /s; c (speed of light) = 3.00 x 108 m/s; h
(Planck’s constant) = 6.626 x 10-34 Js]
For H, Z = 1 and for Li2+, Z = 3.
The energy of electronic levels is given in rydbergs by
En = -Z2
n2
For the first excited state of H, n = 2.
Hence
2
_ Z
n22
or
2
_ Z
n?2
=
H
_ 1
4
n?2
=
=
_
Li2+
9
n?2
36
(6 points)
n = 6 ( 5th excited state)
8. Determine the maximum wavelength of light that is needed to eject electrons from the surface
of sodium, if the work function of sodium is 4.41 x 10-19 J.
[Planck’s constant h = 6.626 x 10-34 Js; speed of light c = 2.998 x 108 m/s]
The maximum wavelength of light here is the one that supplies just enough
energy to overcome the work function ( ) at the surface of Na. The electron
kinetic energy here is zero and hence, from the Einstein equation, E = 
6.626 x 10-34 (Js) x 2.998 x 108 (m/s)
E =  = hc Hence,  =

4.41 x 10-19 (J)
= 4.50 x 10-7 m or 450 nm (visible region)
9. Sketch the plot of radial distribution function (r 2[R(r)]2) versus r for 1s, 2p, 3d and 4f orbitals,
relating the value of r for the maximum value of the function with a0, the Bohr radius.
2
10. Give n, l, number of orbitals, number of radial nodes and number of angular nodes for the
orbitals, 3s, 2p, 6d, 5f and 5p.
Orbital
3s
2p
6d
5f
5p
n
l
3
2
6
5
5
0
1
2
3
1
No. orbitals
1
3
5
7
3
No. radial nodes
2
0
3
1
3
No. angular nodes
0
1
2
3
1
11. (a) For each of the following pairs of atoms, state which you would expect to have the higher
first ionization energy.
(i) Bi or Xe (ii) Se or Te (iii) Rb or Y (iv) K or Ne.
(b) For each of the following pairs of atoms, state which you would expect to have the higher
electron affinity.
(i) Rb or Sr (ii) I or Rn (iii) Ba or Te (iv) Bi or Cl.
(a) (i) Xe (noble gas closed shell) (ii) Se (comes before Te in same group) (iii) Y (comes later in
same row) (iv) Ne (noble gas above Ar, which is nearest noble gas to K)
(b) (i) Rb (anion has outer configuration 5s2) (ii) I (EA decreases down group) (iii) Te (in
previous row to Ba) (iv) Cl (a halogen early in periodic table)
12. Photoelectron spectroscopy studies have determined the orbital energies of atomic chlorine to
be (in eV) –2835, –273, –205, –21, and –10. Assign these energies to the orbitals of Cl and
estimate the value of Zeff for Cl for each of these orbitals. Assume 1 rydberg = 13.61 eV.
The orbital identities are: –2835 (1s), –273 (2s), –205 (2p), –21(3s), and –10 (3p)
2
En ~ _ Zeff (17) or
n2
Zeff (1s) =
Zeff (2p) =
Zeff (3p) =
Zeff (17) ~
(+2835 eV)(12)
= 14.4
(13.61 eV/rydberg)
(+205 eV)(22)
= 7.76
(13.61 eV/rydberg)
(+10 eV)(32)
_
En n2 , where En is in rydbergs
Zeff (2s) =
Zeff (3s) =
(+273 eV)(22)
= 8.96
(13.61 eV/rydberg)
(+21 eV)(32)
= 3.7
(13.61 eV/rydberg)
= 2.6
(13.61 eV/rydberg)
13. Sketch MO energy diagrams (omit AOs: just give identities and relative energies of MOs) for
the following species, and arrange them in order of increasing bond order. BC, OF -, N2-, NO+, BN.
3
14. Sketch the LCAO correlation diagram for NH and predict its magnetic properties.
15. Use MO theory to predict which of the following has the strongest bond: NO +, N2+, O2+.
Identify and sketch the MO boundary surface of the NO+ highest occupied MO (HOMO).
4
16. (a) Write simple VB wave functions for the diatomic molecules B2 and O2. State the bond
order predicted by the simple VB model and compare with the LCAO bonding predictions.
(b) How do other predictions by the VB and LCAO models compare for these molecules?
(a) For B2 (valence electron configuration of B is 2s22p1), a single () bond is predicted, formed
by overlap of the 2p orbital on each atom.
bond (1,2, RAB) = C1(RAB) 2pA(1)2pB(2) + 2pA(2)2pB(1)
The LCAO model also predicts a single bond (bond order = 1).
For O2 (valence electron configuration of O is 2s22p4), overlap of the (half-filled) 2pz orbitals along
the bond axis gives a -bond, whereas overlap of orthogonal (half-filled) 2px or 2py orbitals gives
a -bond. Thus a double bond is predicted.
bond (1,2, RAB) = C1(RAB) 2pzA(1)2pzB(2) + 2pzA(2)2pzB(1)
bond (1,2, RAB) = C2(RAB) 2pxA(1)2pxB(2) + 2pxA(2)2pxB(1)
The LCAO model also predicts a double bond (bond order = 2)
(b) The two methods differ in the bonding and electron spin details. For B2, the VB model
does not specify the bond as a  or  type, but single bonds are normally of the  type. The VB
model predicts the B2 molecule to be diamagnetic. The LCAO model gives the ground state
configuration of B2 as (g2s)2(*u2s)2(u2px)1(u2py)1, thus predicting type bonding and a
paramagnetic molecule.
For O2, the VB model again predicts a diamagnetic molecule, whereas the LCAO model gives O 2
an electron configuration of (g2s)2(*u2s)2(g2pz)2(u2px)2(u2py)2(g2px)1(g2py)1, predicting it to be
paramagnetic.
17. Use the Valence Bond theory to sketch the molecule allene (CH2=C=CH2), showing
hybridization modes of C, types of sigma bonds, and pi bond formation.
(The central C atom forms one -bond on each side. These -bonds are orthogonal to -bond.
The central C atom must be sp-hybridized. The terminal C atoms are sp2 hybridized. These sp2
orbitals are -bonded with 1s orbitals of H atoms and the sp hybrid orbital of the central C atom.)
18. Formulate the VB/MO structure of NO2+ for localized  bonds and delocalized  bonds.
(a) Describe its shape.
(b) assign it as paramagnetic or diamagnetic.
(c) Compare with NO2 and NO2-.
(a) From the VSEPR/VB model, NO2+ is linear (compare with CO2) with sp hybridization on N, so
the VB/MO structure is
5
+
+
O
N
O
O
N
O
O
N
O
*x (antibonding)
or
+
+
or
O
N
nbx (nonbonding)
O
+
O
N
O
+
or
O
N
O
x (bonding)
sp-sp2 or sp-pz  framework
(b) It is diamagnetic.
(c) NO2 and NO2- are both bent molecules, whose  frameworks are based on sp2 hybridization of
the central N atom (see Fig. 6.42). Only NO2 is paramagnetic.
True/False type questions for chapters 3-6
Determine whether the following statements are true or false. Write T or F in the boxes adjacent
to the questions. (2 points for each correct answer, -1 point for each incorrect answer and 0 for no
answer)
(i) Xenon difluoride is a V-shape polar ( > 0) molecule.
F
(ii) 1,2,3-butatriene (C4H4) is a planar molecule with sp and sp2 hybridized carbon atoms. T
(iii) The diagrams below represent the VB boundary surfaces of the  bonds in CO2. F
O
C
O
O
C
O
(iv) The ion HHe- has a bond order of zero, according to the LCAO-MO theory and hence should
be unstable. T
(v) Addition of an electron to C2 leads to increased stability (higher bond energy), according to the
LCAO-MO theory. T
(vi) For the lithium atom, the principal quantum number (n) completely defines the energy of a
given electron. F
(vii) Elements with very high first ionization energies tend to have negative electron affinities
(often assigned <0). T
(viii) The ground state electronic configuration of Cu is [Ar]4s 23d9. F
(ix) The ground state electronic configuration of As3+ is paramagnetic. F
(x) The transition elements of the 6th period (Hf-Hg) have somewhat higher first ionization
energies than those of the 5th period, due to prior filling of the 4f orbitals, which are less effective
at screening the outer electrons from the nuclear charge than d orbitals. T
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