Name_____[KEY]______________________
[KEY]
Date:______________
Period:______
REVIEW Unit 1 Test (Chp 6, 7): Atoms, Electrons, & Periodic Properties
1.
The frequency of electromagnetic radiation is inversely proportional to the wavelength and
directly proportional to the energy of the waves.
2.
Of the following, microwave radiation has the longest wavelength and X-ray radiation has the
most energy.
X-ray
infrared
microwave
3.
What color of visible light has the shortest wavelength? violet
4.
In which orbital does an electron in a bromine atom experience the least shielding?
(A) 3s
(B) 3p
(C) 3d
(D) 4s
(E) 4p
(closest to nucleus)
5.
[Kr] 5s2 4d10 5p2 is the electron configuration of a(n) ____ atom.
(A) Ge
6.
(B) Sr
(C) I
(D) Sn
(E) Te
List all of the elements in the 4th period that are paramagnetic.
K, Sc, Ti, V, Cr, Mn, Fe. Co, Ni, Cu, Ga, Ge, As, Se, Br
7.
The principle quantum number for the outermost electron in a scandium atom is 4.
–Transition metals have the outer s electrons (like 4s) as valence not the inner d electrons (like 3d).
8.
Write the electron configuration for the Fe+3 ion.
Fe atom: 1s2 2s2 2p6 3s2 3p6 4s2 3d6
or
[Ar] 4s2 3d6
Fe3+ ion: 1s2 2s2 2p6 3s2 3p6 3d5
or
[Ar] 3d5
or
[Ne] 3s2 3p6 3d5
(loses the two 4s electrons first, then a 3d electron as well)
9.
Electrons in the 2p subshell are much further from the nucleus in C than in Sn due to less effective
nuclear charge in C. (due to fewer protons in the nucleus of C)
10.
List one element that could replace calcium in a chemical reaction.
Mg and Sr are best. (same group, closest in size within that group)
1
11.
In general, as you go down a group and across a period in the periodic table:
(1) the atomic radius gets bigger and smaller.
(2) the electronegativity gets lower and higher.
(3) the first ionization energy gets lower and higher.
12.
Electronegativity gets ____larger_____ across a period toward the noble gases which all have
_zero__ electronegativities.
(larger across a period due to greater Zeff (+8) with more protons and same shielding)
(smaller down a group due to more shielding from more energy levels)
(noble gases have zero EN and cannot gain an e– b/c Zeff would be 0 on the new outer e– which is completely shielded)
13.
Name the group with the lowest first ionization energy and the group with the highest.
Lowest: alkali metals (Group 1)
14.
Highest: noble gases (Group 18)
Write an equation for the 1st ionization of magnesium and an equation for the 2nd ionization of Mg.
1st:
Mg(g) → Mg+(g) + e–
2nd:
Mg+(g) → Mg2+(g) + e–
Ionization Energies for Element X
Ionization Energy
(kJ mol-1)
1st
2nd
3rd
4th
5th
6th
7th
787
1,580
3,200
4,400
16,000
20,000
24,000
___15. The first seven ionization energies of element X are shown in the above. On the basis of these data,
element X is most likely a member of which of the following groups (families) of elements?
(A)
(B)
(C)
(D)
(E)
Alkaline earth metals
Boron group
Carbon group the large jump in IE from 4th to 5th is caused by the 5th e– removed from an inner shell
Nitrogen group
Halogen group
___16. If a metal M forms an ionic chloride with the formula MCl3, then which of the following formulas is
most likely to be that of a stable sulfide of M?
(A) MS2
(B) M2S3
(C) MS6
(D) M(SO3)3
(E) M2(SO3)3
chloride is Cl– so M is M3+ in MCl3, sulfide is S–2 so with M3+ it forms M2S3
17.
1s2 2s2 2p6 3s1
Atoms of an element, J, have the electronic configuration shown above. The compound element J
will most likely form with oxygen, O, is… J2O
J must form J+ when ionized (loses one e–) so it forms J2O with O2–
2
18.
List at 5 characteristics of metals
malleable (hammer to sheets)
ductile (drawn into wire)
luster (shiny)
good conductors
low ionization energies
19.
(remember this one especially as a property of more metallic elements)
Which group has the most reactive metals?
alkali metals (Group 1) loses e– easiest with lowest IE (ionization energy)
Which group has the most reactive nonmetals?
halogens (Group 17) gains e– easiest with most neg (exothermic) EA (electron affinity)
___20. In which of the following groups are the three species isoelectronic?
(i.e. have the same number of electrons?)
(A)
(B)
(C)
(D)
(E)
21.
S2– ,
Sc ,
O2– ,
Mg2+ ,
Cs ,
K+ ,
Ti ,
S2– ,
Ca2+ ,
Ba2+ ,
Ca2+
V2+
Cl–
Sr2+
La3+
all have elecrton confugurations of argon (1s2 2s2 2p6 3s2 3p6)
Place the answer chosen in #20 above in order of increasing radius.
Ca2+ < K+ < S2–
all have 18 electrons and equal shielding, but most effective nuclear charge in
Ca2+ with 20 protons pulls in outer level closer
Questions 22-23
Consider atoms of the following elements. Assume that the atoms are in the ground state.
(A) S
(B) Ca
(C) Ga
(D) Sb
(E) Br
A 22. The atom that contains exactly two unpaired electrons
1s22s22p63s23p4
↑↓ ↑ ↑
3p
.
C 23. The atom that contains only one electron in the highest occupied energy sublevel
Ga: 1s22s22p63s23p64s23d104p1
↑↓ ↑ ↑
4p
(4p is the highest occupied energy SUBLEVEL)
3
Questions 24-27 refer to neutral atoms for which the atomic orbitals are represented below.
(A) ↑↓
1s
(B) ↑↓ ↑
1s 2s
↑ ↑ ↑.
2p
(C) ↑↓ ↑↓
1s 2s
↑ ↑ ↑
2p
(D) ↑↓ ↑↓
1s 2s
↑↓ ↑↓ ↑↓
2p
(E) [Ar] ↑↓
4s
↑↓ ↑↓ ↑↓ ↑ ↑
3d
B 24. Is in an excited state aufbau principle not followed so must have an excited electron instead of 2s22p2
C 25. Has exactly five valence electrons
highest level is 2, so 2s2 and 2p3 electrons add up to 5
D 26. Has the highest first ionization energy
has the greatest Zeff experienced by outer level (most protons)
E 27. Forms an aqueous cation that is colored 3d shows it’s a transition metal which usually have colored solutions
Use these answers for questions 28-31.
(A) 1s2 2s2 2p6 3s2 3p5
(B) 1s2 2s2 2p6 3s2 3p6
(C) 1s2 2s2 2p6 2d10 3s2 3p6
(D) 1s2 2s2 2p6 3s2 3p6 3d5
(E) 1s2 2s2 2p6 3s2 3p6 3d3 4s2
C 28. An impossible electronic configuration
there is no 2d sublevel that can hold 10 electrons
E 29. The ground-state configuration for the atoms of a transition element (E) has d orbitals, but (D) is excited state
B 30. The ground-state configuration of a negative ion of a halogen Cl gained one e– for Ar configuration
B 31. The ground-state configuration of a common ion of an alkaline earth element
lost two e– for noble gas configuration (Group 2)
Use these answers for questions 32-34.
(A) O
(B) La
(C) Rb
(D) Mg
(E) N
A 32. What is the most electronegative element of the above? furthest to the top right in table (Zeff , Shielding)
E 33. Which element exhibits the greatest number of different oxidation states? you’ll get this later
D 34. Which of the elements above has the smallest ionic radius for its most commonly found ion?
Mg2+ has same # of electrons as N3– but has more protons so greater effective nuclear charge
4
___35. The elements in which of the following have most nearly the same atomic radius?
(A) Be, B, C, N
(B) Ne, Ar, Kr, Xe
(C) Mg, Ca, Sr, Ba
(D) C, P, Se, I
(E) Cr, Mn, Fe, Co all have 4s as outer most valence level (though 3d as last level filled with electrons)
___36. Which of the following properties generally decreases across the periodic table from sodium to
chlorine?
(A)
(B)
(C)
(D)
(E)
37.
First ionization energy
Atomic mass
Electronegativity
Maximum value of oxidation number
Atomic radius
Define the following terms:
(A) nuclear charge – total positive charge in an atom’s nucleus (# of protons) (atomic number)
(B) effective nuclear charge – net positive charge experienced electrons estimated by subtracting
# of inner core electrons from total nuclear charge (# of protons) (Zeff < Z due to shielding)
(C) shielding effect – the lowering of effective nuclear charge experienced by outer electrons by
shielding the nucleus with other inner electrons in an atom.
(D) ionization energy – the energy required to remove an electron from a gaseous atom
(E) electronegativity – the ability to attract electrons when bonded to another atom
___38.
The photoelectron spectra below show the energy required to remove a 1s electron from a nitrogen
atom and from an oxygen atom. Which of the following statements best accounts for the peak in the
upper spectrum being to the right of the peak in the lower spectrum?
(A) Nitrogen atoms have a half-filled p subshell.
(B) There are more electron-electron repulsions
in oxygen atoms than in nitrogen atoms.
(C) Electrons in the p subshell of oxygen atoms
provide more shielding than electrons in the
p subshell of nitrogen atoms.
(D) Nitrogen atoms have a smaller nuclear
charge than oxygen atoms.
5
Section II Free Response
Calculator Allowed
CLEARLY SHOW THE METHODS USED AND STEPS INVOLVED IN YOUR ANSWERS.
It is to your advantage to do this, because you may earn partial credit if you do and little or no credit if you
do not. Attention should be paid to significant figures.
1. Answer the following questions regarding light and its interactions with molecules, atoms, and ions.
(a) A certain line in the spectrum of atomic hydrogen is associated with the electronic transition in
the H atom from the second energy level (n=2) to the fifth energy level (n=5).
(i) Indicate whether the H atom emits energy or whether it absorbs energy during the transition.
Justify your answer.
Absorbs energy because energy is required to move an electron up to higher energy levels.
(ii) Calculate the wavelength, in nanometers, of light that corresponds to this line with an
energy of 4.63 x 10–19 J.
E = hν
c = λν
4.63 x 10–19 J = (6.63 x 10–34) ν
2.998 x 108 = λ (6.98 x 1014 s–1)
ν = 6.98 x 1014 s–1
λ = 4.30 x 10–7 m or 430 nm
O3(g) UV
 O2(g) + O(g)
(b) In the upper atmosphere, ozone molecules decompose as they absorb ultraviolet (UV) radiation,
as shown by the equation above. Ozone serves to block harmful ultraviolet radiation that comes
from the Sun. A molecule of O3(g) absorbs a photon with a frequency of 1.001015 s-1.
(i) How much energy, in joules, does the O3(g) molecule absorb per photon?
E = hν
E = (6.626 x 10–34) (1.00 x 1015)
E = 6.63 x 10–19 J
(ii) The minimum energy needed to break an oxygen-oxygen bond in ozone is 387 kJ mol–1.
Does a photon with a frequency of 1.001015 s–1 have enough energy to break this bond?
Support your answer with a calculation.
6.63 x 10–19 J x 6.022 x 1023 photons x 1 kJ = 399 kJ/mol
1 photon
1 mol
1000 J
399 kJ > 387 kJ so this is enough energy to break the bond.
6
Your responses to these questions will be graded on the basis of the accuracy and relevance of the
information cited. Explanations should be clear and well organized. Examples and equations may be
included in your responses where appropriate. Specific answers are preferable to broad, diffuse responses.
2. Use principles of atomic structure and/or chemical bonding to answer each of the following.
K
Ca
Ionization Energy
(kJ/mol)
First
Second
419
3,050
590
1,140
(a) Explain the difference between Ca and K in regard to
(i) their first ionization energies,
Ca and K both lose an electron from the 4s with equal shielding, but Ca has a greater effective nuclear charge
with 20 protons making its 4s electron more difficult to remove.
(ii) their second ionization energies
Ca loses its 2nd electron from the 4s which is further from the nucleus (more shielded) than the 2nd electron in K
from the 3p making the 2nd electron in Ca easier to remove.
(b) The first ionization energy of aluminum is lower than the first ionization energy of magnesium.
(First ionization energies: 12Mg = 7.6 eV; 13Al = 6.0 eV)
The 3p orbital of Al already represents more energy than the 3s orbital of Mg so not as much energy is required
to be added to remove an electron; and the 3p electron in Al is slightly better shielded from its nucleus than a 3s
electron in Mg.
(c) For magnesium, the difference between the second and third ionization energies is much larger
than the difference between the first and second ionization energies.
(Ionization energies for Mg: 1st = 7.6 eV; 2nd = 14 eV; 3rd = 80 eV)
In Mg, the first two electrons lost are removed from the 3s orbital whereas the 3rd electron comes from a 2p
orbital which is much lower in energy than the 3s; so more energy is needed to remove a 2p electron.
(d) A neutral atom of which of these four elements has the largest radius? K, Ca, Mg, Al
K is larger than Mg and Al because its valence electron shell is n = 4 and it is larger than Ca because it has a less
effective nuclear charge than Ca so its valence electrons are not attractive as tightly to the nucleus.
(e) Which of these four elements (K, Ca, Mg, Al) is the least metallic?
Al because it is furthest to the top right in the periodic table and it has the highest first ionization energy.
(f) Write the complete electron configuration for calcium.
1s2 2s2 2p6 3s2 3p6 4s2
(g) What is the most common charge for calcium ions? Explain.
+2 because Ca easily loses two valence electrons from its 4s and does not easily lose a 3rd electron from its 3p
noble gas core which is held more tightly with a greater effective nuclear charge and less shielding.
7
3. Consider an unknown element J that has the electron configuration [Rn]7s25f146d107p1.
(a) What is the number of the group to which J belongs in the periodic table?
Group 13
(b) Would J be a metal or a nonmetal? Explain in terms of electron configuration.
Metal because it has three valence electrons (7s2 7p1) and must have a lower first ionization energy than most
nonmetals due to having a lower effective nuclear charge than nonmetals in period 7 that have more protons in
the nucleus attracting the valence level (7) electrons.
(c) On the basis of periodic trends, would J have the largest atomic radius in its group or would it
have the smallest? Explain in terms of electronic structure.
It would have the largest b/c it is at the bottom of its group and it would have 7 energy levels whereas other
group members will have fewer.
(d) What would be the most likely charge of the J ion in stable ionic compounds?
+3
(e) Assume that J reacts to form a nitrate compound.
–
Write the formula for the compound formed between J and the nitrate ion, NO3 .
J+3 NO3–1
J(NO3)3
8
4. Using principles of atomic and molecular structure and the information in the table below, answer the
following questions about atomic fluorine, oxygen, and xenon, as well as some of their compounds.
Atom
First Ionization
Energy (kJ mol–1)
F
1,681.0
O
1,313.9
Xe
?
(a) Write the equation for the ionization of atomic fluorine that requires 1,681.0 kJ mol–1.
F(g) → F+(g) + e–
(b) Account for the fact that the first ionization energy of atomic fluorine is greater than that of
atomic oxygen. (You must discuss both atoms in your response.)
In both cases the electron removed is from the same energy level (2p) (same shielding), but fluorine has a greater
effective nuclear charge due to one more proton in its nucleus (the electrons are attracted more and thus take
more energy to remove).
(c) Predict whether the first ionization energy of atomic xenon is greater than, less than, or equal to
the first ionization energy of atomic fluorine. Justify your prediction.
The first ionization energy of Xe should be less than the first ionization energy of F. To ionize the F atom, an
electron is removed from a 2p orbital. To ionize the Xe atom, an electron must be removed from a 5p orbital.
The 5p is a higher energy level with more shielding and is further from the nucleus than 2p so it takes less energy
to remove the first electron from Xe.
(d) The radius of the fluorine atom is smaller than the radius of the fluoride ion, F–.
The F– ion is larger because there is an extra electron causing repulsion without an extra proton in the
nucleus to attract the valence electrons more tightly to the nucleus.
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Answer Key
1.
2007B #2(b)(c) (modified)
2.
1994D (c)(d) 1987D (c)(d)
3.
2006 #8
4.
2008 #5(a)(b)(c)
(ATOMIC THEORY) (modified)
(modified)
(d) added
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