CH 301 Unit 2 Exam-Solutions

Version 396 – Unit 2 Exam – vandenbout – (51335) 1

This print-out should have 31 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 2.6 points

Which of the following statements is true re-garding electromagnetic radiation?

1. Electromagnetic radiation with a wave-length of 400 nm has a frequency that is lowerthan that with a wavelength of 600 nm.

2. Electromagnetic radiation with a wave-length of 400 nm travels faster than that witha wavelength of 600 nm.

3. Electromagnetic radiation with a wave-length of 600 nm travels faster than that witha wavelength of 400 nm.

4. Electromagnetic radiation with a wave-length of 600 nm has a frequency that is lowerthan that with a wavelength of 400 nm. cor-rect

5. The frequency of electromagnetic radia-tion determines how fast it travels.

Explanation:

v = f λ, so λ =1

f. Higher wavelengths cor-

respond to lower frequencies, and vice versa.

002 2.6 points

What is the frequency of a 500 nm photon?

1. 9.0× 10−17 Hz

2. 15 Hz

3. 4.0× 10−7 Hz

4. 5.0× 10−6 Hz

5. 1.5× 10−16 Hz

6. 6.0× 1014 Hz correct

Explanation:

frequency = speed of light / wavelength =

3e8 / 500e-9 = 6e14

003 2.6 points

Which color in the rainbow has the shortestwavelength?

1. green

2. blue correct

3. orange

4. red

5. yellow

Explanation:

Ordered by wavelengths,blue < green < yellow < orange < red

004 2.6 points

An orbital with n = 4 and ℓ = 2 would bedescribed as

1. 4s.

2. 2p.

3. 4d. correct

4. 2s.

5. 4p.

Explanation:

ℓ = 2 refers to the d orbital, so the orbitalis 4d.

005 2.6 points

Rank the following species from smallest tolargest atomic radius: K, Mg, Rb, Ca.

1. Mg < Rb < K < Ca

2. Mg < Ca < Rb < K

3. Mg < K < Ca < Rb

4. Mg < Ca < K < Rb correct


5. Rb < K < Ca < Mg

Explanation:

Atomic radii increase down and to the left.

006 2.6 points

When two H atoms are brought together fromlarge distances they can make H2. As thebond begins to form, the force between theatoms becomes more (attractive, repulsive)and the potential energy (increases, stays thesame, decreases).

1. attractive; increases

2. attractive; decreases correct

3. attractive; stays the same

4. repulsive; increases

5. repulsive; decreases

6. repulsive; stays the same

Explanation:

007 2.6 points

In a pure covalent bond, electrons are

1. uncharged.

2. shared equally. correct

3. transferred.

4. shared unequally.

Explanation:

008 2.6 points

In which compound is the bond the LEASTpolar?

1. HCl

2. HBr

3. HF

4. HI correct

Explanation:

009 2.6 points

The number of single bonds, double bonds,triple bonds, and unshared (lone) pairs on thecentral atom in the sulfite ion (SO2−

3 ) is

1. 3,0,0,1 correct

2. 2,1,0,0

3. 1,1,0,2

4. 1,2,0,1

5. 2,1,0,1

Explanation:

To draw the dot structure for SO2−3 wemust

first calculate the number of valence electronsavailable from the atoms:

A = 6× 1 (S atom) + 6× 3 (O atoms)

+ 2 (from −2 charge)

= 26 e−

We choose the least electronegative elementas the central atom (sulfur in this case) andplace the other atoms symmetrically aroundit.The correct dot structure for the molecule

should show a complete octet (8 electrons)around each atom and a total of 26 valenceelectrons for the entire structure:

S·· ····

··O

······O

······O

2−

As can be seen above, there is one lone pairon the central atom.(If using the S = N − A rule to determine

the dot structure,N = (8×1)+(8×3) = 32 e−

and S = 32−26 = 6 e−. This would indicate 3bonds and correspond to the structure shownabove.)

010 2.6 points

The carbon and oxygen in carbon monoxide


(CO) are joined by what type of bond orbonds?

1. two double bonds

2. one triple bond correct

3. one single bond

4. one double bond

5. four single bonds

Explanation:

The total number of valence electrons is

C 1× 4 e− = 4 e−

O 1× 6 e− = 6 e−

10 e−

and the Lewis dot structure is C Ob

b b

b

The 3 lines represent a triple bond betweenthe C and O.

011 2.6 points

Which of the following is the correct Lewisstructure of ethynol (C2H2O)?

1. CH C O H

b b

bb

2. CH C O H

b b

3. CH C O H

b b

4. CH C O H

b b

bb

correct

5. CH C O H

Explanation:

All of the other choices have octet rule vio-lations, and because Carbon and Oxygen arePeriod 2 elements, they cannot form hyperva-lent compounds.

012 2.6 points

Draw the Lewis electron dot structure forSO3. What is the total number of lone pairsof electrons on the three oxygen atoms?

1. 5

2. 3

3. 7

4. 8 correct

5. 9

Explanation:

To draw the dot structure for SO3 we mustfirst calculate the number of valence electronsavailable from the atoms:

A = 6× 1(S atom) + 6× 3(O atoms)

= 24 e−

We choose the least electronegative elementas the central atom (sulfur in this case) andplace the other atoms symmetrically aroundit.The correct dot structure for the molecule

should show a complete octet (8 electrons)around each atom and a total of 24 valenceelectrons for the entire structure:

S

······O

···· ··O

····O

As can be seen above, there are 8 lone pairson the oxygen atoms.(If using the S = N − A rule to determine

the dot structure, N = 8× 1 + 8× 3 = 32 e−

and S = 32−24 = 8 e−. This would indicate 4bonds and correspond to the structure shownabove.)Alternate Solution:

SO3 has 24 valence electrons. 6 of thesecome from S, and the remaining 18 come fromO (3 atoms at 6 electrons each). In draw-ing the Lewis structure for SO3, three reso-nance forms must be depicted for S to obeythe octet rule. Ultimately, 8 of the avail-able 24 electrons are used to form the threeS O bonds, and the remaining 16 electrons(8 pairs) exist as lone pairs on oxygen.


013 2.6 points

Which one of the following compounds doesNOT obey the octet rule?

1. CH4

2. BF3 correct

3. NaCl

4. H2O

Explanation:

014 2.6 points

How many lone pairs are on the centralatom of AsF5?

1. 4

2. 0 correct

3. 1

4. 2

5. 3

Explanation:

The Lewis dot structure for AsF5 is

As

F

b b

b

b

b

b

Fb

bb

b

b

b

Fb

b

b

b

b

b F

bb

b

b

b

b

F b

b

bb

b b

,

015 2.6 points

The chemical formula for potassium sulfate is

1. K3(SO4)2

2. KSO4

3. K(SO4)2

4. K2SO4 correct

Explanation:

K+ and SO2−4 forms K2SO4:

2(+1) + 1(−2) = 0

016 2.6 points

In the following molecule

C

H

H

H

C

O

b

b

b

b

b

b

Ob

b

b

b

⇐⇒ C

H

H

H

C

O

b

b

b

b

b

b

O

b

b

b

b

1. the carbon/oxygen bonds have two dis-tinct bond strengths and all carbon/hydrogenbonds have two distinct bond strengths

2. all the carbon/hydrogen bonds have thesame strength but the carbon/oxygen bondsare found to have two distinct bond strengths

3. all the carbon/oxygen bonds have thesame strength but the carbon/hydrogenbonds are found to have two distinct bondstrengths

4. all the carbon/oxygen bonds have thesame strength and all the carbon/hydrogenbonds have the same strength correct

Explanation:

Resonance is when there are two equiva-lent ways of drawing a Lewis structure, soyou draw both, as the true structure is some-where in between. The true structure is notswitching back and forth but is essentially theaverage of these two structures. Thus thecarbon/oxygen bonds are both 1.5 bonds.The molecule does not have two types of car-bon/oxygen bonds. The difficult is with ourdrawings not the molecule. Because we insiston having Lewis structure with only singleand double bonds the only way we can cap-ture this idea of a 1.5 bond is to draw twoequivalent resonance structures.

017 2.6 points

For an ionic compound, the lattice energyis the energy associated with

1. generating the ionic compound from theneutral elements


2. dissolving the compound in water toform aqueous ions.3. moving a pair of ions to infinite separa-

tion.4. breaking up the solid crystal into gas

phase ions. CORRECT

Explanation:

The lattice energy is the energy associatedwith taking the ions from the solid crystallinestructure and separating them into ions in thegas phase. The energy associated with simplyseparating two isolated ions is the Coulombenergy. The energy for dissolving the com-pound to for aqueous ions is the enthalpy ofsolution. And the energy for forming theionic compound from neutral elements is theenthalpy of formation.

018 2.6 points

Consider the first three energy levels for anelectron in a helium ion.

1st energy level : −6.41× 10−19 J

2nd energy level : −1.60× 10−19 J

3rd energy level : −0.72× 10−19 J

A 413 nm photon interacts with the ion.What might happen as a result of this in-teraction?

1. An electron could be excited from the 2nd

to 3rd energy level.

2. An electron could be excited from the 1st

to 3rd energy level.

3. None of the transitions listed would oc-cur.

4. An electron could be excited from the 1st

to 2nd energy level. correct

Explanation:

019 2.6 points

Which of the following is NOT correctlymatched name to formula

1. diselenium oxide SeO2 CORRECT

2. phosphorus pentachloride PCl53. sulfur trioxide SO34. carbon tetrachloride CCl45. dinitrogen pentoxide N2O5

Explanation:

SeO2 is selenium dioxide.

The prefix lable ”di” is misplaced in theanswer indicating a ratio of 2:1 Se to O.

020 2.6 points

The oxygen/oxygen bond in O2 is strongerthan in HOOH because

1. the oxygen/oxygen bond in O2 is morepolar than in HOOH.2. O2 has a oxygen/oxygen double bond

and HOOH has a oxygen/oxygen single bond.CORRECT3. O2 has only one structure, while HOOH

has a resonance structure that weakens thebond.4. O2 has an oxygen/oxgyen triple bond

where HOOH has an oxygen/oxygen doublebond.5. O2 has a resonance structure that pro-

vides extra stability compared to the HOOHwhich has only one contributing structure.6. the oxygen/oxygen bond in O2 is less

polar than in HOOH.

Explanation:

The Lewis dot structure of O2 shows it hasa oxygen/oxygen double bond. HOOH onlyhas a oxygen/oxygen single bond.

021 2.6 points

Which of the following species is isoelectronicwith S2−?

1. Br−

2. As3−

3. Rb+

4. Mg2+


5. Ar correct

Explanation:

Isoelectronic means ‘has the same numberof electrons as’. Locate each element in thePeriodic Table. The atomic number will givethe number of protons, which equals the num-ber of electrons for the neutral atom; for anion, you will have to add or subtract from thisvalue.

022 2.6 points

Let X be a hypothetical element.Which of the following would be largest?

1. X2+

2. X−

3. X2− correct

4. X

5. X+

Explanation:

X2− would be largest as it has 2 moreelectrons than protons, and the protons herewould be at the greatest disadvantage whentrying to draw the electrons towards the nu-cleus.

023 2.6 points

Cobalt-60 (6027Co) is an artificial radioisotopethat is produced in a nuclear reactor for useas a gamma-ray source in the treatment ofcertain types of cancer. If the wavelength ofthe gamma radiation from a cobalt-60 sourceis 1.00 × 10−3 nm, calculate the energy of aphoton of this radiation.

1. 2.0× 10−13 J correct

2. 2.0× 10−16 J

3. 2.0× 10−19 J

4. 6.4× 10−14 J

5. None of these

6. 6.63× 10−34 J

7. 6.4× 10−10 J

Explanation:

λ = 1.00 × 10−3 nm Ephoton = ?

Ephoton = h ν =h c

λ

=(6.626× 10−34 J · s)(3.0× 108 m/s)

(1.00× 10−3 nm)(10−9 m/nm)

= 2.0× 10−13 J

024 2.6 points

An electron in a hydrogen atom moves fromthe n = 2 to n = 5 level. What is the wave-length of the photon that corresponds to thistransition and is the photon emitted or ab-sorbed during this process?

1. 276 nm; absorbed

2. 1875 nm; absorbed

3. 1875 nm; emitted

4. 434 nm; emitted

5. 434 nm; absorbed correct

6. 276 nm; emitted

Explanation:

025 2.6 points

Which of the following is correct with respectto the photoelectric effect?

1. A plot of the kinetic energy of the ejectedelectrons versus the frequency of the incidentradiation has a slope that is equal to the valueof the work function.

2. The kinetic energy of the ejected electronsincreases with the intensity of the incidentradiation.

3. A plot of the kinetic energy of the ejected


electrons versus the frequency of the incidentradiation is linear. correct

4. All metals have the same work function.

Explanation:

The photoelectric effect is summarized by

KE = h ν − φ, which fits the standard

equation for a line :

y = mx− c ,

where φ is the work function of the metal,and varies for different metals. The intensityof the incident radiation does not show hereas it is just the number of photons hitting asurface in a given time.

026 2.6 points

You shine white light through a sample ofneon atoms and then through a holographicdiffraction grating. From this experiment,you would expect to observe

1. mostly black space with thin lines ofcolor.

2. a continuous spectrum broken by thinblack lines. correct

3. a continuous spectrum.

Explanation:

027 2.6 points

Phosphorus has a higher ionization energythan silicon because

1. False; phosphorus has a lower ionizationenergy than silicon.

2. the outer orbitals of phosphorus are lo-cated closer to the nucleus than the outerorbitals of silicon.

3. the effective nuclear charge of phosphorusis less than that of silicon.

4. the outer orbitals of phosphorus are lo-cated farther from the nucleus than the outerorbitals of silicon.

5. the effective nuclear charge of silicon isless than that of phosphorus. correct

Explanation:

15P : [Ne] ↑↓ ↑ ↑ ↑

3s 3p

14Si : [Ne] ↑↓ ↑ ↑

3s 3p

In both cases, the valence electrons are lo-cated in 3s and 3p orbitals. However, P has agreater nuclear charge due to its greater num-ber of protons, so more energy is required toremove an electron from P relative to Si.

028 2.6 points

Which of the following will not have ionicbonds?

1. NaCl

2. methane (CH4) correct

3. calcium oxide

4. KBr

5. magnesium chloride

Explanation:

All of the options except CH4 have ionicbonds, which are formed between metals andnonmetals. CH4 has only covalent (non-metal-nonmetal) bonds.

029 2.6 points

The graph shows the radial distribution plotsfor the 1s wavefunctions for H, He, and He+.


A

B

C

radius

4πr2Ψ

2

Which plot is the 1s wavefunction for the He+

ion?

1. A correct

2. C

3. There is no way to know

4. B

Explanation:

H has one electron and one proton. He hastwo electrons and two protons. He+ has oneelectron and two protons. Therefore the elec-tron will have the greatest attraction to thethe nuclei with two protons. In He there willbe a slight reduction in the effective nuclearcharge due to the electron-electron repulsion.In He+, there is only one electron so the radiuswill be the smallest. Plot A is peaked closestto the nucleus so it will be the smallest.

030 2.6 points

Any neutral atom having an odd (i.e., un-even) atomic number must be

1. metallic.

2. non-magnetic.

3. paramagnetic. correct

4. inert.

5. a transition element.

Explanation:


031 22.0 points

This question is merely a placeholder forthe points in the hand-graded portion of theexam.

Explanation:

This question is merely a placeholder forthe points in the hand-graded portion of theexam.

CH 301 Unit 2 Exam-Solutions

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