Atomic Structure Packet Answers
MC: 1-B 2-D 3-A 4-C 5-E 6-D 7-C 8-E 9-B 10-A 11-E 12-D 13-A 14-A 15-E
16-D 17-C 18-C 19-A 20-A 21-E 22-E 23-A 24-E
FRQ:
1) a) Mg 1s2 2s22p6 3s2
Ar 1s2 2s22p6 3s23p6 (both must be correct)
b) Valence electrons for Mg and Ar are in the same principal energy level, but Ar is the
smaller and it has a greater nuclear charge. Therefore, the first and second ionization
energies of Mg are less than those of Ar. The removal of the third electron from Mg
involves the n = 2 energy level, and this electron experiences a very large nuclear charge.
(A maximum of 2 points was awarded for explanations based only on the stability of an
octet.)
c) MgCl2 forms, since Mg readily loses its two valence electrons to form a stable
configuration.
d) The formula is QCl
A very high second ionization energy indicates that there is only one valence electron.
2) a) It is impossible to determine (or measure) both the position and momentum of any
particle (or object, or body) simultaneously.
OR
The more exactly the position of a particle is known, the less exactly the position or
velocity of the particle is known.
OR
([delta] x) ([delta] p) [greater than or equal to] h-bar (or h / 4pi)
h = Planck's constant
[delta] x = uncertainty in position
[delta] p = uncertainty in momentum
Notes:
1 point is given for the notion of simultaneous determination. (A number of
students give the first sentence but omit the word simultaneously. (They got 2 out
of 3.) If the second or third versions of the answer are given, simultaneity is
understood.
If they give the equation, they must have a > sign, not just an = sign, or they lose
1 pt.
A student who gives a correct answer but adds erroneous material gets one point
deducted.
b) Bohr postulated that the electron in a H atom travels about the nucleus in a circular
orbitand has a fixed angular momentum. With a fixed radius of orbit and a fixed
momentum (or energy),
([delta] x) ([delta] p) < h/4pi
and violates the uncertainty principle.
Students receive 2 pts total for the above. If they say only "the electron travels in circular
orbits", they get 1 point. They also get credit for saying elliptical orbits. To receive full
credit, they must describe an aspect of the Bohr theory.
The following portion of the answer is worth three points.
The wavelength of a particle is given by the deBroglie relation:
[gamma] = h/mv
For masses of macroscopic objects, h/m is so small for any v that [gamma] is so small as
to be undetectable. For an electron, m is so small that h/mv yields a detectable [gamma].
OR
They may say the product of the uncertainties in position and velocity depends on h/m
and since h is so small (h = 6.63 x 10¯34 J s) unless m is very small, as for an electron, the
product of the uncertainties is too small to be detected.
Students may discuss the fact that measuring the position and momentum requires having
a photon strike the particle. The photon has an energy comparable to that of an electron
but small compared to that of a macroscopic object. They must stress mass rather than
size as the important distinction to get full credit.
A student who says [gamma] or ([delta] x) ([delta] p) depends on "size" and not mass
gets 2 points if discussion is otherwise okay.
:
:F:
: : :
: F :C: F :
: : :
:F:
3) (a)
:
tetrahedral
: :
: :
F : F
: : : :
: : Xe : :
F : F
: : : :
square planar
:
:F:
: Cl: : :F :
: : :
:F:
:
T-shaped
(b) CF4 = 4 bonding pairs around C at corners of regular tetrahedron to minimize
repulsion (maximize bond angles).
XeF4 = 4 bonding pairs and 2 lone pairs give octahedral shape with lone pairs on
opposite sides of Xe atom
ClF3 = 3 bonding pairs and 2 lone pairs give trigonal bipyramid with one pairs in
equatorial positions 120º apart.
4) a) Across the period from Li to Ne the number of protons is increasing in the nucleus
hence the nuclear charge is increasing with a consequently stronger attraction for
electrons and an increase in I.E.
b) The electron ionized in the case of Be is a 2s electron wheras in the case of B it is a 2p
electron. 2p electrons are higher in energy than 2s electrons because 2p electrons
penetrate the core to a lesser degree.
c) The electron ionized in O is paired with another electron in the same orbital, whereas
in N the electron comes from a singly-occupied orbital. The ionization energy of the O
electron is less because of the repulsion between two electrons in the same orbital.
d) The ionization energy of Na will be less than those of both Li and Ne because the
electron removed comes from an orbital which is farther from the nucleus, therefore less
tightly held.
.
N
: : :: :
O
:O: :
: :
:O :
:
N
:: :
:O
:
:
: O
O :N
::
:
:
5) (a)
(b) NO2- < NO2 < NO2+
NO2- - 3 charge centers around N; lone pair of electrons on N
NO2 - 3 charge centers around N; single electron on N
NO2+ - 2 charge centers on N
(c) NO2+ is linear, has sp hybridization - or - NO2/NO2- have sp2 hybridization
(d) NO2 will dimerize, because it contains an odd electron that will pair readily with
another, forming N2O4.
6) a) Electron configuration of Na and Mg (1 pt)
Any one earns a point:
Octet / Noble gas stability comparison of Na and Mg
Energy difference explanation between Na and Mg
Size difference explanation between Na and Mg
Note: If only Na or Mg is used 1 point can be earned by showing the respective electron
configuration and using one of the other explanations
Shielding/effective nuclear charge discussion earns the third point.
b) Correct direction and explanation of any one of the following:
shielding differences
energy differences
# of proton/ # of electron differences
c) Any one set earns one point:
(i) Ni unpaired electrons. paramagnetic
(ii) Zn paired electrons/ diamagnetic
(iii) Ni unpaired electrons/ Zn paired electrons
(iiii) Ni paramagnetic/ Zn diamagnetic
Orbital discussion/ Hund's Rule/ Diagrams earns the second point.
d) Expanded octet or sp3d hybrid of phosphorous (1 pt)
Lack of d orbitals in nitrogen (1 pt)
OR
nitrogen is too small to accommodate (or bond) 5 Fluorines or 5 bonding sites (2 pts)
7) (a) The valence electrons in a calcium atom are the 4s2. In a calcium ion these
electrons are absent and the highest energy electrons are 3p, which has a much
smaller size because the (-)/(+) charge ratio is less than 1 causing a contraction of the
electron shell.
(b) Lattice energy can be represented by Coulomb’s law: lattice energy = k(Q1Q2),/r2,
where Q1 and Q2 are the charges on the ions, in CaO these are +2 and -2 respectively,
while in K2O they are +1 and -2. The r (the distance between ions) is slightly smaller
in CaO, combined with the larger charges, accounts for the larger lattice energy.
(c) Electron arrangements: K = [Ar] 4s1, Ca = [Ar] 4s2
(i) Potassium has a single 4s electron that is easily removed to produce an [Ar] core,
whereas, calcium has paired 4s electrons which require greater energy to remove
one.
(ii) a K+ ion has a stable [Ar] electron core and requires a large amount of energy to
destabilize it and create a K2+ ion. Ca+ has a remaining 4s1 electron that is more easily
removed than a core electron, but not as easily as its first 4s electron.
(d) Electron arrangements,
Mg = [Ne] 3s2, Al = [Ne] 3s2, 3p1
It is easier to remove a higher energy, single, unpaired 3p electron from the
aluminum than to remove one electron from a lower energy, paired 3s orbital in
magnesium.
8) a) Response must contain a cogent discussion of the forces between the nucleus and
the outermost (or "ionized") electron. For example, a discussion of "the outermost
electron on K..." should include one of the following:
i. it is farther from nucleus than the outermost electron on Li
ii. it is more shielded from the nucleus (or "experiences a lower effective nuclear
charge") than the outermost electron on Li
iii. it is in a higher energy orbital (4s) than tne outermost electron on Li (2s)."
2 points for any one
Notes:"K is larger than Li" earns 1 point.
No points earned for "K electron is easier to remove" (or some other restatement).
b) Nitrogen has one less proton than oxygen 1 point
Nitride and oxide ions are isoelectronic 1 point
or,
In nitride ion the electron/proton ratio is greater, causing more repulsion; thus, nitride is
the larger ion. 2 points
c) A Zn atom has more protons (10 more) than an atom of Ca 1 point
Electrons in d orbitals of Zn have a lower principal quantum number; thus, they are not in
orbitals that are farther from the nucleus. 1 point
d) Correct identification of the orbitals involved (2s versus 2p) 1 point
Clear statement that the two orbitals have different energies 1 point
Note: Arguments that "the 2p orbital is farther out than the 2s orbital", or that "the Be
atom has a filled subshell, which is a more stable configuration" earn no explanation
point.
General note:For all parts (a) through (d), discussions of position in the periodic table
earn no points.
9) a) i. there are two other similar resonance structures for the carbonate ion.
(ii) the pi O=C double bond in CO2 is shorter than a single O-C resonance sigmabond (all are identical and are about 1 1/3 bond) found in a carbonate ion.
(b) (i)
••
F
•• •• ••
F C F
•• •• ••
F
••
(ii) in the tetrahedral CF4, the polar C-F bonds are cancelled out by the equiangular
pull of the 4 bonds. With an expanded octet and trigonal bipyramidal structure, SF4
has a pair of unbonded electrons at the center of the bipyramid, this gives a “seasaw”
shape to the molecule and an uneven pull to the polar S-F bonds.
10) (a) all the isotopes have 34 protons but a different number of neutrons in the nucleus.
(b) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4
2 unpaired electrons. 4s [] 4p [][ ][ ] Hund’s rule indicates that each of the
orbitals will be filled with a single electron before it gets paired.
(c) (i) in Se, the single paired 4p electrons has 1 electron easily removed to create the 3
unpaired 4p orbitals which is energetically favorable; in bromine, the removal of 1
electron still leaves a paired 4p orbital.
(ii) the shielding effect is stronger in Te and makes it easier to remove a electron
(lower ionization energy).
F
(d)
··
·F·
· ·
· Se
·
·· F ··
··
··
·F ·
· ·
F
··
Se
F
·· F ··
··
F
see-saw shape
Because F is very electronegative and the molecule is asymmetric with respect to the
fluorines, this molecule is polar.
11) (a) Be has 1 more electron in the 2s orbital than Li, not in another larger orbital. Be
also has 1 more proton to “pull in” the 2s orbital, making it smaller than Li.
(b) the second ionization energy (IE2) removes the second 4s orbital in calcium, leaving
a noble gas kernel. The IE2 in potassium is removing a very stable 3p orbital electron
in its noble gas kernel, which requires a great deal more energy.
(c) the carbon-to-carbon bond in C2H4 is a double bond, which is stronger than the
carbon-to-carbon single bond in C2H6.
(d) Cl2 has 34 electrons and Br2 has 70 electrons. more electrons mean greater vander
Waals attractions in Br2, more energy to overcome them when it boils and, therefore
a higher boiling point than Cl2.
12) (b) (i) 109.5˚
(ii) sp3d
(iii) see-saw
(c) (i) 4 sigma, 1 pi
(ii) structure 1;
In structure 1, oxygen has a formal charge of 0 (6 valence electrons – 6 assigned
electrons), each fluorine is 0 (7 valence electrons – 7 assigned electrons), phosphorus
is 0 (5 valence electrons – 5 assigned electrons),
In structure 2, oxygen has a formal charge of –1 (6 valence electrons – 7 assigned
electrons), each fluorine is 0 (7 valence electrons – 7 assigned electrons), phosphorus
is +1 (5 valence electrons – 4 assigned electrons)
According to the electroneutrality rule, the better Lewis structure is the one with the
smallest separation of formal charge, i.e., structure 1.