Chapter Seven React Questions Answers
React 1
This provides a concrete example showing energy and frequency are directly related
and energy and wavelength are inversely related.
Some students have the misconception that wavelength and amplitude mean the same
thing or are somehow related. This should bring the misconception to light (no pun
intended).
React 2
Give the students time to think about this, and to help them, ask them about the
significance of white light. A light bulb lights up because a thin filament in the bulb is
heated (electric current is sent through the thin filament) and this heat (energy) is
released as light. Energy is related to the wavelength of electromagnetic radiation, and
the wavelength (if visible) is related to color. Because the wire in the light bulb is
heated, energy of all wavelengths is emitted, and therefore wavelengths of all colors are
emitted. When all of the colors of visible light are mixed, the result is white light.
It is quite significant that the light from the hydrogen emission spectrum is not white, but
has a characteristic color associated with each different salt. This must mean that not
all wavelengths of light are emitted. In other words, the electrons cannot go to any
excited state and then return to any other state. If this occurred, all wavelengths of light
would be emitted and the light would be white. It must mean that the electrons can only
go to certain energy levels. This would account for only certain colors being emitted.
Emphasize to the students that this is a surprising and non-intuitive result. Energy
levels in an atom are “quantized”; that is, only certain levels are possible.
React 3
The answers are: I. blue, II. green, III. red.
Split the class into three groups and have smaller “subgroups” (2-3 students) in each of
these larger groups solve only one of these. Then collect results from all groups.
The students will need to calculate the energy change and then use this to calculate the
wavelength. The students then need to look at the EMR spectrum (such as Figure 7.2)
to determine the color.
Let the students have time to figure out what they need to do to solve this problem
(don’t tell them which equations they have to use). When they are finished, let them
know that they do not have to memorize the EMR spectrum. If there is a problem like
this on the exam, we will tell them the relevant wavelengths. Also, make sure the
students realize that the energy change is negative, as expected (exothermic for an
electron becoming more stable), but that wavelength is always positive.
React 4
Only four of all the possible transitions occur in the visible region of the EMR spectrum.
React 5
The answer is yes.
This problem is similar to React 4. The students will have to know something about
ionization energy. The ionization energy is the energy required to completely remove an
electron from the ground state (n1 = 1 to n1 = ).
One way of answering this question is to calculate the energy of the gamma ray, and
compare it to the ionization energy (IE) of hydrogen. Using the Rydberg equation, we
get 2.178 x 10-18 J for the IE of hydrogen. The energy for the gamma ray is 1.99 x 10 -15
J. Thus, the answer is no. Give some time for the students to think about how to set up
this problem.
Point out that the Bohr model has limited use. It works great for all one-electron atoms
or ions. The Rydberg equation for one-electron ions must account for the increased
nuclear attraction for the one electron. Also, it just doesn't work for atoms/ions
containing more than one electron! This is the vast majority of all atoms/ions. The
reason for this is that it is a fundamentally incorrect physical model. It is too simple of a
model, and assumes we know how the electrons move and where they are, but we
don’t. So why consider it? It introduces us to the idea of quantized energy levels, which
we are going to keep. We also will still use our understanding of energy emission and
absorption and the sign convention.
React 6
Students should understand that all of the orbitals in a given energy level for the
hydrogen atom are degenerate. This is not true for other atoms. This is because of
complications due to having two electrons. Discussing this question should provide a
good opportunity to go over electron shielding and the penetration effect.
React 7
In order to place electrons in the lowest energy levels, we must know the energy order
of the atomic orbitals. The energy order of atomic orbitals has been determined from
experiment.
As it turns out, the order (lowest energy to highest energy) can be obtained from the
periodic table: ls < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p
< 7s < 5f< 6d < 7p < 8s. Students need not memorize this order (or use various
“shortcuts”). They should see how it comes from the periodic table. Go to the periodic
table and point out the s block, the p block, the d block and the f block of elements in
the periodic table. Then discuss how the periodic table can be used to give the energy
ordering of the atomic orbitals.
React 8
This should be straightforward with the exception of the transition metal ions. The rules
for TM ions are an exception and the students should know this.
S: [Ne] 3s23p4
Ba: [Xe]6s2
Ni2+: [Ar]3d8
Eu: [Xe]6s25d14f6
Ti+: [Ar]3d3
Explain to students that they should use the periodic table to obtain electron
configurations.
React 9
The hydrogen 1s orbital is larger, and the lithium 1s orbital is lower in energy.
Because the nucleus of lithium has a greater attractive charge, the electrons in the 1s
orbital for lithium are more attracted to the lithium nucleus than the electron in the 1s
orbital in hydrogen is attracted to the hydrogen nucleus.
Reacts 10-15 are good to use because they start with the extreme cases of atomic size
and ionization energy trends and have the students develop general trends from these.
Make sure the students can explain these trends.
React 10 and React 11
More energy is required to remove an electron from Cl than Na, and Li than Cs.
Students will try to memorize trends on the periodic table. Don’t let the students get
away with an answer such as “it wants to be like a Noble gas”. Have them think about
the energy level of the electron in question, and the attractive force of the nucleus. For
example, in comparing Na and Cl, the energy level is the same but the attractive forces
of the nuclei are different (it is greater for Cl, thus Cl requires more energy to remove an
electron).
React 12
Once the students have considered Reacts 10 and 11, have them speak of the general
trends (IE increases up and to the right). Again, the emphasis should be on the
understanding of these trends in terms of energy levels and nuclear charge and
attraction.
React 13 and React 14
Na is a larger atom than Cl and Cs is a larger atom than Li.
Like React 10 and React 11, have the students think over memorize. Have them think
about the energy level of the electron in question, and the attractive force of the
nucleus. For example, in comparing Na and Cl, the energy level is the same but the
attractive forces of the nuclei are different (it is greater for Cl, thus Cl is a smaller atom).
React 15
Once the students have considered Reacts 13 and 14, have them speak of the general
trends (atomic size increases down and to the left). Again, the emphasis should be on
the understanding of these trends in terms of energy levels and nuclear charge and
attraction.
React 16
Increasing ionization energy: S, O, F
Increasing atomic radius: F, O, S
Make sure the students can explain these answers without resorting to “because it
follows the trend” or “because the elements want to be like Noble gases”.
React 17
Looking at the electron configuration for N and O, the major difference is a paired set of
p electrons in O vs. all unpaired p electrons in N. The decrease in ionization energy in
going from nitrogen to oxygen reflects the extra electron-electron repulsions in the
doubly occupied oxygen 2p orbital. It takes less energy to remove the paired O electron
than the unpaired N electron. The reason for Be-B exception is the 2p vs the 2s
electron. The higher energy 2p electron in B is easier to completely remove than the
lower energy 2s electron in Be. Thus B has a lower ionization energy than Be.
React 18
The answer is III.
Have each student individually write down his/her answer to this question and take a
poll of answers (you can have them submit this anonymously if you wish). Look for all
responses to appear – they get at the typical misconceptions students have. This type
of question helps the students look for reasons as opposed to memorized phrases
(such as “wants to be like a Noble gas”).
React 19
Because the 3rd ionization energy is so much higher than the 2nd, these elements are in
Group 2 (the alkaline earth metals). Because the ionization energy of element Y is
higher than that of element X, Y must be higher on the periodic table than X. Any
answer that reflects these ideas is reasonable.
React 20
The second ionization energy for lithium will be much higher because that second
electron will be taken from the first energy level; the second electron taken from
beryllium will be taken from the second energy level.