Chem 130 – First Exam
Name_______________________________
On the following pages you will find eight questions covering varies topics ranging from
nomenclature to periodic properties, and electromagnetic radiation to Lewis structures.
Read each question carefully and consider how you will approach it before you put pen or
pencil to paper. If you are unsure how to answer a question, move to another; working on a
new question may suggest an approach to the one that is more troublesome. If a question
requires a written response, be sure that you answer in complete sentences and that you directly and clearly address the question. Of particular importance for this exam: if a question asks you
to explain a periodic trend, it is insufficient to write that “the <insert your property> of atoms increases to
the right and to the top of the periodic table.” Instead, your answer must explain why this trend exists.
Partial credit is willingly given on all problems so be sure to answer all questions!
Question 1 _____/15
Question 5 _____/11
Question 2 _____/6
Question 6 _____/11
Question 3 _____/20
Question 7 _____/11
Question 4 _____/11
Question 8 _____/15
Total _____/100
Potentially useful equations and constants:
c = λν
E = hν = hc/λ
KE = hν – W
⎛ 1
1
1 ⎞
= 1.09737 × 10 − 2 nm⎜⎜ 2 − 2 ⎟⎟
λ
⎝ n1 n2 ⎠
𝑉∝
c = 2.998×108 m/s
𝑄! 𝑄–
𝑑
h = 6.626×10–34 Js
NA = 6.022×1023 mol–1
Problem 1. For each of the following, provide one example of an element satisfying the
stated condition. Do not include lanthanides and actinides in your answers, and do not use any element
more than once! All possible answers shown, although just one is needed for each.
(a) forms a monatomic ion with a charge of +1: H, Li, Na, K, Rb, Cs, Fr, Cu, Ag, Au,
Ga, In, Tl
(b) is an alkaline earth: Be, Mg, Ca, Sr, Ba, Ra
(c) is in the third period: Na, Mg, Al, Si, P, S, Cl, Ar
(d) has a valence shell electron configuration of ns2np2: C, Si, Ge, Sn, Pb
(e) forms a +1 ion with a noble gas electron configuration: H, Li, Na, K, Rb, Cs, Fr
(f) is in the d-block: Sc → Zn, Y → Cd, La → Hg, Ac, Rf, Db, Sg, Hs, Mt
(g) forms monoatomic ions with charges of +2 and +3: Sc, Ti, V, Cr, Mn, Fe, Co
(h) has exactly seven electrons in a d-orbital: Co, Rh, Ir, Mt
(i)
is not deflected by a magnetic field: He, Be, Ne, Mg, Ar, Ca, Zn, Kr, Sr, Cd, Xe, Ba,
Hg, Rn, Ra
(j)
has an electron with the quantum numbers n = 3, l =2, ml = 0, and ms = +½: any
element from Sc on (that is, it has a 3d electron)
(k) has a covalent radius smaller than that for silicon: H → Ne (but not Li), P, S, Cl, Ar
(l)
has exactly three valence electrons: B, Al, Sc, Ga, In, Tl, Y, La, Ac
(m) is a non-metal: H, He, C, N, O, F, P, S, Cl, Se, Br, Kr, I, Xe, Rn
(n) cannot expand beyond an octet: H, He, Li, Be, B, C, N, O, F, Ne
(o) has only two peaks in its photoelectron spectrum: Li, Be
Problem 2. Fill in the missing spaces in the following table:
Formula
Name
Covalent or Ionic?
NH4Cl
ammonium chloride
ionic
Cu3(PO4)2 copper(II) phosphate
ionic
N2O4
dinitrogen tetraoxide
covalent
Problem 3. The following data is available for element X. Use this information to answer
the questions that follow.
peaks in photoelectron spectrum: 5
covalent radius: 0.104 nm
first ionization energy: 999.6 kJ/mol
number of valence electrons: 6
(a) Identify element X by name. Explain the reason for your choice in no more than two
sentences.
Five peaks in the PES spectrum means that the element has 1s, 2s, 2p, 3s, and 3p electrons. Six valence electrons means that it must have two in the 3s orbital and four in the
3p orbital, which makes the element sulfur (S).
(b) What is the electron configuration for X? Underline the orbital that contains the electron
responsible for the first ionization energy.
1s2 2s2 2p6 3s2 3p4 or [Ne] 3s2 3p4
(c) Is the radius for X2– larger than, smaller than, or the same size as the radius for X. Explain the reason for your choice in no more than two sentences.
The radius of an anion is always larger than the radius for the element. Adding electrons
to the 3p orbital increases repulsion between electrons, causing the radius to increase.
(d) An element in the same group as X has a first ionization energy of 1313 kJ/mol. Does
this element lie above or below X in the group? Explain the reason for your choice in no
more than two sentences.
Ionization energies increase as we move up a group so the element must lie above X in
the periodic table. The reason for this trend is that the valence electrons for elements in
the same group feel the same effective nuclear charge, but valence electrons further up in
a group are closer to the nucleus and experience a greater force of attraction.
(e) A laser pointer has a wavelength of 655 nm. What is the energy of this photon in Joules?
Is the energy of this photon sufficient to eject an electron from X?
The energy of a single photon is given by the relationship E = hc/λ. Substituting in
Planck’s constant (h = 6.626!10–34 Js), the speed of light (c = 2.998!108 m/s), and the
wavelength (λ = 655!10–9 m) gives the energy (E) as 3.03!10–19 J. The ionization energy is given in kJ for a mole of X, so we need to convert it to J per photon; thus
(999.6 kJ/mol) ! (1000 J/kJ) ! (1 mol/6.022!1023) = 1.66!10–18 J
From this we see that it takes more energy to ionize an electron from X (1.66!10–18 J)
than provided by a photon from the laser pointer (3.03!10–19 J); thus, the laser pointer
is not capable of ejecting an electron from X.
Problem 4. Shown below is the Lewis structure for the compound XCl2+. Based on this
Lewis structure, identify a main group element that could be X (there are several possibilities). Explain your reasoning in no more than three sentences.
The Lewis structure has 20 electrons (four in the two bonds and 16 in
the eight lone pairs. Each of the two chlorines contributes seven electrons, which leaves
six electrons (20 – 14) unaccounted for. The structure is an ion with a +1 charge, so X
must bring in seven electrons, so that we have 7 + 7 + 7 – 1 = 20 electrons. This means
that X is F, Cl, Br, I, or At (although, for reasons we have not discussed, fluorine is an unlikely option).
Problem 5. Arrange the following ions
Li+
F–
Na+
Cl–
Br–
in order of increasing size, placing the smallest ion on the left and the largest ion on the right.
Explain your reason for this arrangement in no more than three sentences.
Li+ < Na+ < F– < Cl– < Br–
Removing an electron from Li or Na reduces its size as its valence electrons are now in a
shell closer to the nucleus; however, adding an electron to F, Cl, or Br increases its size due
to the greater repulsion of eight electrons relative to seven electrons in the valence shell. Although they are isoelectronic, Na+ is smaller than F– because Na+ has the greater nuclear
charge. Relative size always increases down a group as the valence electrons are at an increasingly greater distance from the nucleus, which explains the relative order for Li+and Na+,
and for F–, Cl–, and Br–.
Problem 6. Arrange these elements
C
N
O
Si
P
S
in order of increasing covalent radius, placing the smallest element on the left and the largest
element on the right. Explain your reason for this arrangement in no more than three sentences.
O < N < C < S < P < Si
Covalent radii decrease across a group because the increasing effective nuclear charge draws
the valence electrons in closer to the nucleus; thus, O is smaller than N, which is smaller
than C, and S is smaller than P, which is smaller than Si. Covalent radii increase down a
group because the valence electrons are in shells further from the nucleus; thus S, P, and Si
are all larger than O, N, and C.
Problem 7. Arrange the following atoms and ions, which have the same electron configuration
Ar
S2–
K+
in order of increasing ionization energy, placing the species with the smallest ionization energy on the left and the species with the largest ionization energy on the right. Explain your
reason for this arrangement in no more than three sentences.
S2– < Ar < K+
Although isoelectronic in their electron configuration, S2– has fewer protons and a smaller
Zeff than Ar, which has fewer protons and a smaller Zeff than K+. For this reason alone, we
expect it to be easier to remove an electron from S2– than from Ar, and easier to remove an
electron from Ar than from K+. The trend is enhanced because it is even easier to remove
an electron from anion as it decreases repulsion between electrons, and it is even harder to
remove an electron from a cation as we are trying to remove a negatively charged particle
from a species that has a positive charge.
Problem 8. The fulminate ion, CNO–, forms explosive metal salts. Draw all possible resonance structures for the fulminate ion and, for each structure, identify the formal charge on
all three atom. Circle the resonance structure that is the most important in determining fulminate’s properties? Explain your reason for selecting this structure in no more than three
sentences. Please work neatly!
The total number of valence electrons for which we need to account is 4 + 5 + 6 + 1 = 16,
leading to three possible resonance structures
The resonance structure on the left is the most important in determining fulminate’s properties because it has the smallest formal charges (all ±1), places a negative formal charge on
oxygen (instead of zero or +2), and avoids placing a negative formal charge on carbon,
which has a smaller electron affinity than oxygen.