chem1300 fall 2011 with solutions

CHEM 1300 final exam, Fall 2011 December 14, 2011 Version 1

3 hour exam

Part A

1. Which exam version are you writing (see the header above)? This question is not worth any

marks, but no concessions will be made if you answer it incorrectly.

a. Version 1

b. Version 2

2. The concentration of a solution prepared by dissolving 10.0 g of sodium chloride in sufficient

water to make 750.0 mL of solution is:

a. 0.133 M

b. 0.228 M

c. 0.0133 M

d. 0.456 M

e. 0.0750 M

3. Electrical (Coulomb) energy:

a. Increases in magnitude as the distance between charged particles decreases.

b. Is greater for two Ca2+

cations 10 pm apart than for two Na+ cations 10 pm apart.

c. Is negative for two oppositely charged particles.

d. Is the potential energy associated with the forces of attraction and repulsion between

electrically-charged particles.

e. All of the above.

4. Which radiation has the highest frequency?

a. green visible light

b. radio waves

c. infrared radiation

d. microwave radiation

e. cell phone radiation

5. CANCELLED. A police radar unit is operating at a frequency of 9.527 gigahertz (109 Hz).

What is the wavelength of the radiation being employed?

a. 631.3 nm

b. 6.313×10–24

m

c. 2.858 nm

d. 31.47 cm

e. 2.858×1018

m


3 hour exam

6. The figure below represents the energy level diagram for the hydrogen atom, and the arrows labeled

a – f are possible electron transitions. Rank the transitions that involve the emission of a photon

from lowest to highest frequency.

a. a < c < b < e < f < d

b. d < f < e < b < c < a

c. d < e < c

d. c < e < d

e. a < b < f

7. The notation for the subshell with n = 5 and l = 3 is:

a. 3d

b. 3h

c. 5d

d. 5p

e. 5f

8. A possible set of quantum numbers for an electron in the partially filled subshell in a potassium

atom in its ground state configuration would be

n l ml ms

a. 3 0 0 ½

b. 3 1 0 -½

c. 4 0 0 -½

d. 4 1 0 ½

e. 4 2 1 ½

9. Which one of the following pictures shows a 2p orbital?

a. b. c. d. e.

10. Which atom has the smaller 3s orbital?

a. An atom with more protons

b. An atom with fewer protons

c. An atom with more neutrons

d. An atom with fewer neutrons

e. The size of the 3s orbital is the same for all atoms.

n = 1

n = 2

n = 3

n = 4c

d

a b

e

f

Ener

gy


3 hour exam

11. In the ground state electron configuration of gold (Au), how many electrons are there in the 3p

orbitals?

a. 0

b. 1

c. 2

d. 3

e. 6

12. Place the following species in order of decreasing radius.

2Te F 2O

a. F > 2O

> 2Te

b. F > 2Te

> 2O

c. 2Te >

2O

> F

d. 2Te

> F >

2O

e. 2O

> F >

2Te

13. Which one of the following orbital diagrams represents the ground state electron configuration

of a Cr3+

cation?

4s 3d 4s 3d

a. [Ar]

d. [Ar]

b. [Ar]

e. [Ar]

c. [Ar]

14. Choose the diamagnetic species from below.

a. Sn2+

b. Br

c. P

d. Cr

e. None of the above are diamagnetic.

15. Which one of the following is a set of isoelectronic species?

a. C, N, O

b. F, Cl, Br

c. Na+, Mg

2+, Cl

–

d. S2–

, Ar, Ca2+

e. Rb+, Sr

2+, Zr

2+


3 hour exam

16. Place the following in order of increasing first ionization energy:

N F As

a. N < As < F

b. N < F < As

c. F < N < As

d. As < N < F

e. As < F < N

17. Which one of the following species has the highest third ionization energy?

a. Al

b. Ca

c. Mg

d. P

e. P3–

18. Which one of the following has 4 valence electrons?

a. Al

b. Si

c. P

d. As

e. Be

19. The primary reason that an ionic compound of formula LiBr2 does not form is:

a. the electron affinity of Br is endothermic.

b. the lattice energy for LiBr2 is smaller than that of LiBr.

c. the total ionization energy required to produce Li2+

is too high.

d. the second electron affinity of Br is positive.

e. the bond energy of Br2 is prohibitively large.

20. Which one of the following molecules is an alcohol?

a. b. c. d. e.

OH

O O

NH2

O OH

O

O

O

21. Which one of the following bonds is least polar?

a. C–B

b. C–C

c. C–N

d. C–O

e. C–F


3 hour exam

22. Draw the best Lewis structure for the BF‹ anion. What is the formal charge on the boron atom?

a. +2

b. 0

c. +1

d. –1

e. –2

23. C is the central atom in the thiocarbonate ion, CO2S2–

. Which of the following would best reflect

the actual charges on the various atoms based on its Lewis structure(s)? Note: Oa and Ob are the

two oxygen atoms in the thiocarbonate ion.

C Oa Ob S

a. –½ –½ –½ –½

b. –1 0 0 –1

c. 0 0 –1 –1

d. 0 –1 –1 0

e. 0 0 0 –2

24. Which one of the following molecules is nonpolar?

a. CH2Br2

b. SeCl6

c. SO2

d. PCl3

e. CH3OCH3

25. Place the following species in order of decreasing XO bond length, where X represents the central

atom in each species.

CO2 CO⊀‾ SiO⊀‾

a. CO2 > CO⊀‾ > SiO⊀‾

b. CO⊀‾ > SiO⊀‾ > CO2

c. CO2 > SiO⊀‾ > CO⊀‾

d. SiO⊀‾ > CO2 > CO⊀‾

e. SiO⊀‾ > CO⊀‾ > CO2

26. How many - and -bonds are there in the CO molecule?

a. 1 -bonds, 2 -bonds

b. 2 -bonds, 1 -bonds

c. 1 -bonds, 1 -bonds

d. 1 -bonds, 0 -bonds

e. 2 -bonds, 0 -bonds


3 hour exam

27. Based on MO theory, what is the bond order in the LiBe molecule?

a. 0

b. 0.5

c. 1

d. 1.5

e. 2

28. The picture below illustrates the delocalized π-bonding molecular orbital in ozone constructed

from the overlap of the three parallel p-orbitals on adjacent O atoms. How many electrons can

this molecular orbital contain?

a. Up to two electrons having opposite spin.

b. Up to four electrons, two in each lobe.

c. Up to six electrons, two for each of the three O p-orbitals

d. All of the electrons involved in the σ-bonding in ozone.

e. All of the lone pair electrons from the Lewis structure of ozone.

29. How many delocalized -electrons are there in the following molecule?

a. 5

b. 6

c. 8

d. 10

e. 12

30. Which one of the following figures is the best representation of the electronic band structure of

silicon doped with arsenic atoms?

a. b. c. d. e.

O


3 hour exam

31. Place the following molecules in order of increasing boiling point for the pure liquid.

a. < <

b. < <

c.

< <

d.

< <

e. <

<

32. Which one of the following liquids is expected to have the highest vapour pressure at 20ºC?

liquid Surface tension at 20ºC (J/m2)

a. hexane 0.0184

b. acetone 0.0237

c. cyclohexanol 0.0344

d. water 0.0728

e. mercury 0.425

33. Solid Kr and solid Cu both have a face-centred cubic unit cell. Why are their physical

properties so different?

a. Kr atoms are bigger than Cu atoms.

b. Cu is more electronegative than Kr.

c. Kr atoms in the solid interact via dispersion forces while Cu atoms in the solid interact via

metallic bonding.

d. Cu atoms have unpaired electrons while all electrons are paired for Kr atoms.

e. Cu is more expensive than Kr.


3 hour exam

34. Which of the following images represents a network covalent solid?

a.

b.

c.

d.

e. All of these images represent network covalent solids.

35. In the unit cell pictured to the right, the solid spheres

represent oxide anions (in all corner and face positions) and

the open spheres with a T inside represent the tetrahedral

holes filled by metal cations (all completely inside the unit

cell). Determine the empirical formula for this ionic

compound, where X represents the metal.

a. XO

b. X2O

c. X3O

d. XO2

e. XO3

36. Order the following crystal structures for increasing number of atoms in the unit cell.

a. simple cubic < face-centred cubic < body-centred cubic

b. face-centred cubic < simple cubic < body-centred cubic

c. face-centred cubic < body-centred cubic < simple cubic

d. simple cubic < body-centred cubic < face-centred cubic

e. body-centred cubic < simple cubic < face-centred cubic

37. For solids with a hexagonal close packed arrangement, what is the coordination number and the

layer packing order?

a. Coordination number = 6, layer packing order = ABABAB

b. Coordination number = 8, layer packing order = ABCABC

c. Coordination number = 8, layer packing order = ABABAB

d. Coordination number = 12, layer packing order = ABABAB

e. Coordination number = 12, layer packing order = ABCABC

http://www.google.ca/imgres?imgurl=http://intro.chem.okstate.edu/1515SP01/Lecture/Chapter12/Salt.GIF&imgrefurl=http://intro.chem.okstate.edu/1515SP01/Lecture/Chapter12/Lec2701.html&usg=__6N3ted5dqEyO4q7L0jwSTc6rwUs=&h=249&w=387&sz=31&hl=en&start=1&zoom=1&tbnid=H6Rk9bRb35YRnM:&tbnh=79&tbnw=123&ei=aM_KTomnDdTH0AHK2JkC&prev=/search?q=ionic+solid&um=1&hl=en&sa=N&rlz=1T4HPIA_enCA312CA312&tbm=isch&um=1&itbs=1

http://www.google.ca/imgres?imgurl=http://wps.prenhall.com/wps/media/objects/3311/3391416/imag1108/AAAUBAF0.JPG&imgrefurl=http://wps.prenhall.com/wps/media/objects/3311/3391416/blb1108.html&usg=__NsLEQ-1sBlqc5iIE_PU8lw4dQQI=&h=300&w=418&sz=22&hl=en&start=8&zoom=1&tbnid=Ylq7pnUrs7kebM:&tbnh=90&tbnw=125&ei=o8_KTrjUHqne0QHAxrkJ&prev=/search?q=metallic+solids&hl=en&gbv=2&tbm=isch&itbs=1


3 hour exam

Part B

Question 1: /1 point

Balance the combustion reaction for butane:

C4H10 (g) + 13/2 O2 (g) → 4 CO2 (g) + 5 H2O (l)

or

2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (l)

Question 2: /2 points

Write the chemical formula for each compound:

cobalt (III) chloride hexahydrate CoCl3·6H2O

sodium sulfate Na2SO4

dinitrogen pentoxide N2O5

benzene C6H6

Question 3: /1 point

For each set of quantum numbers below, indicate whether the set is valid. If a set is not valid,

provide a new set of quantum numbers that corrects the problem.

Set of quantum numbers Valid (circle) Only if not valid, correct the problem

n = 3, l = 3, ml = 2, ms = –½ No n = 3, l = 2, ml = 2, mS = –½

or

n = 4, l = 3, ml = 2, mS = –½

n = 5, l = 2, ml = –1, ms = ½ Yes


The metalloid antimony is able to form three different stable ions in its compounds.

Sb [Kr] 5s2 4d

10 5p

3 or 1s

2 2s

2 2p

6 3s

2 3p

6 4s

2 3d

10 4p

6 5s

2 4d

10 5p

3

Sb3–

[Kr] 5s2 4d

10 5p

6 or 1s

2 2s

2 2p

6 3s

2 3p

6 4s

2 3d

10 4p

6 5s

2 4d

10 5p

6 or [Xe]

Sb3+

[Kr] 5s2 4d

10 or 1s

2 2s

2 2p

6 3s

2 3p

6 4s

2 3d

10 4p

6 5s

2 4d

10

Sb5+

[Kr] 4d10

or 1s2 2s

2 2p

6 3s

2 3p

6 4s

2 3d

10 4p

6 4d

10

Rank the four antimony species listed above in

order of increasing size (radius):

Sb5+

< Sb3+

< Sb < Sb3–

Rank the four antimony species listed above in

order of increasing first ionization energy:

Sb3–

< Sb < Sb3+

< Sb5+


3 hour exam


Platinum is a highly inert

metal used in electrical

contacts and electrodes, in

dentistry equipment, and

in jewelry.

The graph to the right

plots the kinetic energy of

ejected electrons as a

function of frequency of

light shone on a platinum

surface.

What is the binding

energy of platinum in

kJ/mol?

From graph, minimum frequency required to eject electrons = 1.3 × 1015

Hz

E = nhν = (6.022×1023

mol–1

)(6.626×10–34

Js)( 1.3 × 1015

s–1

)(1 kJ / 1000 J) = 520 kJ/mol


One of the less prominent lines in the spectrum of mercury is at 1013.975 nm. How many photons

of this wavelength are required to achieve a total energy of 100.0 kJ?

photons 10104.5photonper J 10959.1

photonsmany for J101.000 photons ofnumber

photon 1for J 10959.1m10 1013.975

m/s 102.998s J 106.626hc nm) E(1013.975

23

19

5

19

9-

8-34


What mass of FeS is produced when 4.71 g of Fe reacts with 4.25 g S8?

8 Fe (s) + S8 (s) → 8 FeS (s)

mol Fe = 4.71 g / 55.85 g/mol = 0.0843 mol Fe

mol FeS if all Fe reacts = 0.0843 mol Fe × (8 mol FeS / 8 mol Fe) = 0.0843 mol FeS if all Fe reacts

mol S8 = 4.25 g / 256.48 g/mol = 0.0166 mol S8

mol FeS if all S8 reacts = 0.0166 mol S8 × (8 mol FeS / 1 mol S8) = 0.133 mol FeS if all S8 reacts

thus, limiting reagent = Fe

mass FeS = 0.0843 mol × 87.91 g/mol = 7.41 g

0

200

400

600

800

0 0.5 1 1.5 2 2.5 3

Kin

etic

En

erg

y (

kJ

/mo

l)

Frequency (1015 Hz)


3 hour exam


Draw two sketches that illustrate how orbitals overlap to form all of the bonds in propene, C3H6.

The first sketch should show the -bonding framework, while the second should show the -

bonding framework. You may want to first draw the Lewis structure. Label the orbitals involved in

the formation of bonds.

C C

CH

H H

H

HH -bond

CH3

p - p

s - sp2s - sp3

s - sp3

s - s

p3

s - sp2

s - sp2

sp2 - sp2

sp2 - sp3


Draw the best Lewis structure(s) for SO⊀‾, including all equivalent resonance structures, if any.

SO

O

OS

O

O

OS

O

O

O

-2


Draw the best Lewis structures for Br„ and GaCl3, then fill in the table below.

Br„ GaCl3

Br Br Br-

Ga

Cl

Cl

Cl

Molecule

What is the name

of the electron

group geometry?

What is the name

of the molecular

geometry?

Is the molecule

polar or nonpolar?

Circle one.

List the value(s) of

the bond angles.

Br„ trigonal

bipyramidal

linear nonpolar 180º

GaCl3 trigonal planar trigonal planar nonpolar 120º


3 hour exam


Quinidine, pictured on the

right, is a medication used to

suppress abnormal rhythms

of the heart.

For each of the labeled bond

angles, write the theoretical

(VSEPR) angle value and the

hybridization of the central

atom in the table to the right.

N

O

N

OH

a

bc

d

Hybridization of

central atom

Theoretical

angle

a. sp3 109.5º

b. sp2 120º

c. sp3 109.5º

d. sp3 109.5º

Functional group at position a: ether Functional group at position d: amine


a. Indicate with an X which intermolecular force(s) must be overcome when the following pure

liquids are vapourized.

b. Circle the compound that has the highest viscosity at room temperature.

Compound Ion-dipole

forces

Dispersion

forces

Dipolar

forces

Hydrogen

bonding

C4H9OH X X X

C4H9F X X

C5H12 X

CH3COCH2CH3 X X

Question 13: Major typo! Final decision pending

Fill in the MO diagram below for O2–

and answer the questions:

EN

ER

GY

σ2p* –––

Bond order of O2– _______

What is the bond order of this species

after it absorbs a photon of light with

sufficient energy to cause ionization? _______

Provide a chemical equation for this ionization:

___________________↓___________________

π2p* ––– –––

π2p ––– –––

σ2p –––

σ2s* –––

σ2s –––


3 hour exam


Draw a line structure (e.g. ) or condensed structure representation (e.g. CH3–CH2–CH3; in

this case you must correctly include all H atoms) of 6,6-dichloro-3-ethyl-4-nonyne (also called:

6,6-dichloro-3-ethyl-non-4-yne).

OR


Explain the trend in boiling points shown in the graph below (be concise!).

The boiling points of HCl, HBr, and HI increase because the number of electrons and

hence the attractive strength of dispersion interactions increases as: HCl < HBr < HI

The boiling point of HF is much higher because HF molecules can hydrogen bond while

HCl, HBr, and HI molecules cannot.

150

200

250

300

2 3 4 5

Boil

ing P

oin

t (K

)

Period of Halogen

HF

HCl HBr

HI

ClCl

C C CHC

CH2

CH2 CH3

CH3

CH2CH2

CH3

Cl Cl

chem1300 fall 2011 with solutions

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