Chapter 7 Atomic Structure and Periodicity. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 7–2 QUESTION Suppose that a microwave

Chapter 7

Atomic StructureAtomic Structureand Periodicityand Periodicity

Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 7–2

QUESTIONSuppose that a microwave oven uses photons with an energy of 1.42 10–23 joules to provide you with a cooked popcorn snack. You have less than two minutes before the corn pops to determine the wavelength of the microwaves. (Note: use Planck’s constant as 6.626 10–34 Js.)

1. 1.40 cm2. 2.14 1010 cm3. 0.714 cm4. This can’t be solved unless you are given the frequency.


ANSWERChoice 1 provides the correct response. The equation E = hcan also be written E = hc/ so that can be determined from the given data.

Section 7.1: Electromagnetic RadiationSection 7.2: The Nature of Matter


QUESTIONThe wavelength of the laser light that allows you to listen to your favorite tunes on a CD player lies in the red area of the visible spectra. If one mole of the photons delivers 1.54 105 J, what is the frequency of this useful energy?

1. 2.59 10–15 s–1

2. 3.86 1014 s–1

3. 1.56 1023 s–1

4. I don’t know (I only listen to vinyl).


ANSWERChoice 2 correctly reports the frequency. The total energy for a mole must be converted to the energy for just one photon (divide by Avogadro’s number) then E = h could be used to determine .

Section 7.2: The Nature of MatterSection 7.3: The Atomic Spectrum of Hydrogen


QUESTIONWhile you are sitting in the dentist’s chair you are being hit with several forms of electromagnetic radiation. The bright light is sending energy in the visible region, the body heat of the dentist is emanating energy in the infrared region; perhaps a radio is playing and some energy in the radio waves region approach your ears; and you are about to receive an x-ray. Which arrangement places the EM waves in proper order with respect to increasing energy per photon?

1. Visible; infrared; radio; x-rays2. X-rays; visible; radio; infrared3. Infrared; radio; visible; x-rays4. Radio; infrared; visible; x-rays


ANSWERChoice 4 sorts the EM waves in order of increasing energy per photon. Energy is directly proportional to frequency. The arrangement is also in order of increasing frequency.

Section 7.1: Electromagnetic RadiationSection 7.2: The Nature of Matter


QUESTIONDiffraction is known to be a characteristic of waves. Yet, when electrons, with mass, pass through the openings in crystals a diffraction pattern appears. Which of the following is consistent with this information?


QUESTION (continued)1. Diffraction patterns are caused as the waves exhibit

destructive interference only. Electrons annihilate each other.2. Electrons and EM radiation produce diffraction patterns

meaning mass must have wavelike properties and waves must have mass.

3. Electrons can exhibit diffraction patterns because as they pass through the regular patterns of openings in crystals they get lodged in the crystal causing light to be emitted.

4. Mass and wavelength are directly related. The small mass of an electron allows it to have a small enough wavelength to cause diffraction.


ANSWERChoice 2 is consistent with known observations. The work of DeBroglie, Davisson and Germer helped to develop the concept of wave-particle duality.

Section 7.2: The Nature of Matter


QUESTIONIf you dropped your textbook, you might claim that “it is difficult to hold on to waves” with some validity. After all, mass does possess wave-like properties. However, what would be the wavelength of a 855 g textbook moving 9.8 m/s?

1. 7.9 10–38 m2. 7.7 10–37 m3. 7.9 10–35 m4. I don’t see how something with mass could have a wavelength.


ANSWERChoice 3 provides the correct value. The wave-particle duality equation developed by DeBroglie should be used here after the mass has been converted to kg (855 1/1000).

Section 7.2: The Nature of Matter


QUESTIONOf the following, which provides an accurate statement about the importance of the discrete line spectrum of H2?

1. Some aspect of hydrogen’s structure must allow electrons tobe attracted to the nucleus in such a way to have energyavailable from the visible spectrum.

2. The discontinuous nature of the spectrum indicates thatelectrons somehow have destructive interference.

3. Since only certain wavelengths are observed, electrons mustonly have certain allowed energy levels around the nucleus.

4. The quanta of energy associated with an electron changingenergy levels must be indirectly related to its frequency.


ANSWERChoice 3 is the hypothesis scientists formed as a result of their observations. The difference in the energy levels was shown by the light emitted. Since the light was emitted only in discrete wavelengths it indicated that electron energy levels are quantized.

Section 7.3: The Atomic Spectrum of Hydrogen


QUESTIONAccording to Bohr’s calculation, the energy for an electron in a hydrogen atom’s first energy level is –2.178 10–18 J. What would be the energy of an electron, of the same electron, in the third level?

1. –6.534 10–18 J2. –2.420 10–19 J3. –1.960 10–17 J4. Without more information, I am unable to solve this.


ANSWERChoice 2 indicates that the energy of the electron in the third level is higher (less negative) by 9 times. The squared value of the energy level is inversely related to the energy of an electron in the atom.

Section 7.4: The Bohr Model


QUESTIONAccording to Bohr, electrons in hydrogen atoms made “leaps” from one level to another. Which transition depicted here would require the electron to absorb the most energy?

1. 4th level to the 7th level2. 5th level to the 2nd level3. 2nd level to the 4th level4. 1st level to the 2nd level


ANSWERChoice 4 relates the correct relationship between energy level number and gaps between levels. Note that the difference between energy levels decrease as the level number increases.

Section 7.4: The Bohr Model


QUESTIONSchrodinger and Heisenberg interpreted electron distribution in atoms with a quantum momentum and probability. This revolutionary view could be summarized accurately by which of the following?

1. The probability for finding an electron (e-) is greatest near thenucleus, but the radial probability increases to a maximumthen decreases.

2. The orbital of an electron must replace the term orbit due touncertainties in its position. The orbital is where the e– maybe found 50% of the time.

3. The volume of a particular sphere of e– movement around thenucleus increases with distance; therefore, the total e–

probability must decrease.4. Like Heisenberg, I am uncertain.


ANSWERChoice 1 contains the correct reasoning in this revolutionary treatment of e– and atoms. Note that as the distance from the nucleus increases the probability of finding an e– at a certain position does decrease. However, if we sum the probabilities over that volume, the total probability reaches a maximum (basically the atomic radius), then decreases. Note the view shown in Figure 7.12 on page 293.

Section 7.5: The Quantum Mechanical Model of the Atom


QUESTIONOne key result of the Schrodinger equation for general chemistry is the ability to use quantum numbers to discuss and predict the distribution of energy among an atom’s electrons. This is done through a series of quantum numbers. How many total orbitals should be present for electrons in the fourth energy level?

1. 42. 73. 164. 32


ANSWERChoice 3 correctly predicts the value, although that does not mean that all would be occupied in a given atom or ion.

Section 7.6: Quantum Numbers


QUESTIONTwo electrons in an atom of helium have the same value of n, l, and m. It follows then, that…

1. the two electrons can only be in the first level (1s).2. Heisenberg’s uncertainty principle works because we can never

simultaneously know both the exact position and momentum ofelectrons. So, with the small level of uncertainty both electronscan have those three values.

3. The two electrons must have an opposite spin.4. The electrons must now be in separate orbitals.


ANSWERChoice 3 is correct, based on Pauli’s Exclusion Principle. No two electrons in an atom may have the same two quantum numbers. The quality of spin (either assigned +½ or –½) allows the fourth quantum number to distinguish the two electrons.

Section 7.8: Electron Spin and the Pauli Principle


QUESTIONQuantum numbers, of course, are applied to atoms other than hydrogen. When examining electron energy levels and their distribution in polyelectronic atoms, what key precept must be kept in mind?

1. The penetration effect predicts that the value of n determineshow close an e– may be to the nucleus.

2. The penetration toward the nucleus for an e– is the same as itsmost probable distance from the nucleus.

3. The electron treatment should take into account the nuclearpull and the other electron repulsions.

4. The electron treatment is based on penetration toward thenucleus, which is based on its four quantum numbers.


ANSWERChoice 3 takes into account the two major factors for the energy of an electron and the electron’s attraction toward the nucleus. Penetration toward the + nucleus is a key factor, but other negative electrons will repel each electron. The diagram in Figure 7.21 may be helpful to review.

Section 7.9: Polyelectronic Atoms


QUESTIONUsing Hund’s rule and the Aufbau principle, compare the number of unpaired electrons in one atom each of C, O and Ne.

1. All are in the same row on the periodic table so they have the same number of unpaired electrons.

2. C has the fewest; O has the most, and Ne has none3. O has two; C and Ne have none4. C and O have two; Ne has none


ANSWERChoice 4 produces the correct comparison. Note that carbon’s two “p” electrons, based on Hund’s rule would be in separate orbitals thus they would both be unpaired. Oxygen also has two unpaired “p” electrons, since the three orbitals must accommodate six electrons before electrons enter a new level.

Section 7.11: The Aufbau Principle and the Periodic Table


QUESTIONArrange the following metals in order from most unpaired electrons to the least considering only their 3d sublevel:

Cu; Mn; Cr; Sc

1. Cu; Mn; Cr; Sc2. Cr and Mn are the same; then Sc, and then Cu3. Mn; Cr; then Cu and Sc which have fewer but are the same4. Cr and Mn are the same; then Sc and Cu, which have fewer

but are the same


ANSWERChoice 2 provides the proper sorting from most unpaired electrons in the 3d to least among those four metals. Remember that the question for both Cr and Cu must be examined carefully.

Section 7.11: The Aufbau Principle and the Periodic Table


QUESTIONSodium and aluminum both have one unpaired electron in their neutral ground state atoms. Why is the first ionization energy of Na lower than Al, but the second ionization of energy of Al lower than the second ionization energy of Na?

1. The smaller size of Al makes it difficult to ionize at first, butafter losing one e–, the other atoms can expand more and make it easier to ionize a second e–.

2. The 3s electrons of Na does a better job of shielding than the3p electron of Al.

3. The second e– taken from Na must be taken from a new lowerlevel; the second e– from Al is in same level as before.

4. The first e– ionized from both is single, but the second one inNa is paired, while the second in Al is still single.


ANSWERChoice 3 is correct. When removing the first e– from the larger Na atom, less energy is needed than the smaller Al atom. However, removing the next electron from Na means taking an e– from the second energy level. This is closer to the nucleus making it more difficult than Al’s second e–, which is still in the third energy level.

Section 7.12: Periodic Trends in Atomic Properties


QUESTIONWhich of the following statements correctly interprets the relationship between ionization energy (I1), atomic radius and electron affinity when comparing two atoms? (Answer this thinking about the general trends rather than some exceptions.)

1. When atomic radius increases, I1 decreases and electronaffinity becomes less negative.

2. When atomic radius increases, I1 increases and electronaffinity becomes more negative.

3. When atomic radius increases, I1 decreases and electronaffinity becomes more negative.

4. When atomic radius increases, I1 increases and electronaffinity becomes less negative.


ANSWERChoice 1 indicates the proper relationship when comparing two atoms. As atomic radius increases the negative electron is further from the positive nucleus, which decreases their attraction, thus lowering the I1. Since electron affinity also deals with the attraction of the nucleus for electrons (new ones), a large size would generally decrease the attraction, thus causing the affinity to be less exothermic (less negative.)

Section 7.12: Periodic Trends in Atomic Properties


QUESTIONChemical reducing agents tend to give up electrons easily. Based on their first ionization energy Na should give up its outer e– more easily than Li. In fact, this is true unless the two are being compared in aqueous solution. What is the best explanation for this apparent anomaly?

1. Lithium ions are larger than sodium ions.2. In water, Li+ ion behaves like He and evaporates away thus not

having the opportunity to regain its electron. Na+ stays in thewater and can regain their electron.

3. The larger charge density of the Li+ attracts water more strongly than Na+ thus enhancing the reaction that leads to forming the ion. 4. This still does not seem logical, am I missing something?


ANSWERChoice 3 provides the subtle reasoning behind this apparent anomaly. Both Li and Na become +1 ions, but the charge in a Li ion is in a much smaller volume. Thus, the greater charge density attracts water (hydration). Since this is greater for Li than Na, it increases the reaction going from atom to ion.

Section 7.13: The Properties of a Group: The Alkali Metals

Documents

Chapter 7 Atomic Structure and Periodicity. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 7–2 QUESTION Suppose that a microwave