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CH_223, section 651, Summer 1 2011
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Crystal Badger
CH_223, section 651, Summer 1 2011
Instructor: Kay Sandberg
Current Score : 9.25 / 10
Due : Monday, May 23 2011 09:00 PM EDT
Ask Your Teacher Extension Requests Print Assignment
Question
Points
1
2
3
4
5
6
7
8
9
1/1 0.25/1 0.5/0.5 0.5/0.5 0.6/0.6 3.2/3.2 1.4/1.4 0.2/0.2 1.6/1.6
Total
9.25/10 (92.5%)
Description
This assignment is to help you see if you have retained important knowledge from your general
chemistry course.
Instructions
If you did not score 100% on this assignment (or if you scored 100% but you were guessing) you
should view the following review presentations:
electronic configuration review presentation (click here)
formal charge review presentation (click here)
hybridization review presentation (click here)
Make sure you check the number of submissions for each question type. Click "points" link to see
the number of submissions allowed for each question part.
If the presentation does not work properly, see the "Troubleshooting" forum for help.
Assignment Submission
For this assignment, you submit answers by question parts. The number of submissions remaining for
each question part only changes if you submit or change the answer.
Assignment Scoring
Your last submission is used for your score.
The due date for this assignment is past. Your work can be viewed below, but no changes can be
made.
Important! Before you view the answer key, decide whether or not you plan to request an extension.
Your Instructor may not grant you an extension if you have viewed the answer key. Automatic
extensions are not granted if you have viewed the answer key.
Request Extension
1. 1/1 points | Previous Answers My Notes
Question Part
Points
Submissions Used
1
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4
0.25/0.25 0.25/0.25 0.25/0.25 0.25/0.25
1/2
1/2
1/2
1/2
1/1
Click here for the review presentation and here for a periodic table.
Give the number of:
a) protons in an O atom
18
8
b) electrons in a Li atom
23
3
c) electrons in a Cl1- ion
318
18
d) electrons in a Ca2+ ion
418
18
Solution or Explanation
Proton number is the atomic number. When the atom is neutral, the electron number equals the
proton number. For the anion electron number, you add the charge to the atomic number to determine
the electron number. For the cation electron number, you subtract the charge from the atomic number
to determine the electron number.
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2. 0.25/1 points | Previous Answers My Notes
Question Part
Points
Submissions Used
1
2
3
4
0/0.25 0.25/0.25 0/0.25 0/0.25
1/1
1/1
1/1
1/1
Total
0.25/1
Select the word(s) that makes the statement true.
Click here for review presentation.
a) An electron in a 3s-orbital is lower
higher
in energy than an electron in a 2s-orbital.
b) An electron in a 2s-orbital is lower
lower
in energy than an electron in a 2p-orbital.
c) A metal typically gains
loses
its valence electron(s) becoming a(n) anion
cation
.
Solution or Explanation
The lower the energy level the lower the energy of the electron.
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3. 0.5/0.5 points | Previous Answers My Notes
Question Part
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1
2
0.25/0.25 0.25/0.25
1/2
1/2
Give the symbol for:
Click here for review presentation.
Periodic Table
a) the cation possessing a positive one charge that is isoelectronic with neon.
(NOTE: input format for a charged species: symbol^#sign. Example, H^1+ represents H1+.)
1Na^1+
Na^1+
b) the anion possessing a negative two charge that is isoelectronic with argon.
(NOTE: input format for a charged species: symbol^#sign. Example, H^1- represents H1-.)
2S^2S^2Viewing Saved Work Revert to Last Response
4. 0.5/0.5 points | Previous Answers My Notes
Question Part
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1
2
0.25/0.25 0.25/0.25
1/3
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1/3
0.5/0.5
Following the energy ordering dictated by the periodic table, write the indicated configuration.
Input format: [Ne]3s23p1 would be typed in as [Ne]3s^2,3p^1
Note: a comma is used to separate sublevels in the valence shell as shown.
Periodic Table
Click here for review presentation.
a) Using a noble gas core write the electron configuration for S.
1[Ne]3s^2,3p^4
[Ne]3s^2,3p^4
b) Using a noble gas core write the electron configuration for Li.
2[He]2s^1
[He]2s^1
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5. 0.6/0.6 points | Previous Answers My Notes
Question Part
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Submissions Used
1
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3
0.2/0.2 0.2/0.2 0.2/0.2
1/2
1/2
1/2
Total
0.6/0.6
Give the formal charge of the indicated atom.
All hydrogens and lone pairs are shown.
Click here for review presentation.
Input format for non-zero formal charge: sign and then number, something like: -1
10
0 -or- 0 -or- 0
atom 1
2+1
1+ -or- +1 -or- +
atom 2
3-1
1- -or- -1 -or- atom 3
Solution or Explanation
FC = family # - possession #
Family # is off the periodic table
Possession # is determined by drawing a circle around the atom (making sure to all cut bonds in half)
and counting the # of electrons in the circle.
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6. 3.2/3.2 points | Previous Answers My Notes
Question Part
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Submissions Used
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16
0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0. 0.2/0.
2
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2
2
2
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2
2
2
2
2
2
2
2
2
2
2/2
2/2
1/2
1/2
1/2
1/2
1/2
2/2
1/2
2/2
1/2
1/2
2/2
1/2
2/2
2/2
Total
3.2/3.2
Each atom below has access to an octet of electrons (except for hydrogen which has access to a duet)
and all atoms are shown explicitly. Not shown are the lone pairs of electrons and the formal charges.
You will need to determine how many lone pairs, if any, are needed to give the indicated atom access to
8 valence electrons and the formal charge of the indicated "octeted" atom
Click here for review presentation.
(a) Select the number that corresponds to the number of lone pairs possessed by the
indicated atom.
(b) Select the formal charge of the indicated atom.
atom a lone pair(s) atom a formal charge atom b lone pair(s) atom b formal charge
2
-1
3
-1
2
-1
3
-1
atom c lone pair(s) atom c formal charge atom d lone pair(s) atom d formal charge
0
+1
1
-1
0
+1
1
-1
atom e lone pair(s) atom e formal charge atom f lone pair(s) atom f formal charge
1
+1
4
-1
1
+1
4
-1
atom g lone pair(s) atom g formal charge atom h lone pair(s) atom h formal charge
2
0
1
0
2
0
1
0
Solution or Explanation
For access to octet determination, count all of the electrons in the bond and then add as many lone
pairs as needed to equal 8 electrons. The formal charge is determined by the formula: Family # minus
possession #. The family # is from the periodic table. For example, oxygen is in the 6th family. The
possession number is determined by circling the atom for which you are determining the formal
charge making sure you cut all of the bonds in half. The electrons in the circle will include all of the
lone pairs and half of the bonding electrons.
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7. 1.4/1.4 points | Previous Answers My Notes
Question Part
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Submissions Used
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7
0.2/0.2 0.2/0.2 0.2/0.2 0.2/0.2 0.2/0.2 0.2/0.2 0.2/0.2
1/2
1/2
2/2
1/2
2/2
2/2
1/2
You may want to draw the Lewis structure for the HCN molecule before attempting this question.
Click here for review presentation.
b
a) Select the correct description for each region.
Region V pi bonding region
pi bonding region
Region U sigma bonding region
sigma bonding region
Region X lone pair region
lone pair region
Region Y sigma bonding region
sigma bonding region
b) Write the capital letter (U, V, W, X or Y) of the region that results from the overlap of 2
hybrid orbitals.
5Y
Y
Only input a single capital letter
c) Write the capital letter (U, V, W, X or Y) of one of the regions that results from the
overlap of 2 pure p orbitals.
6V
V -or- W
Only input a single capital letter
d) Type in the total number of electrons contained in all of the electron regions between the
C and N.
76
6
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8. 0.2/0.2 points | Previous Answers My Notes
Question Part
Points
Submissions Used
1
0.2/0.2
1/2
Total
0.2/0.2
Imagine the following orbitals gradually approaching each other as indicated by the arrows so that
they begin to overlap.
Click here for review presentation.
If two electrons with opposite spins reside in the overlap region, then which of the following is the true
statement?
1
A & C result in a pi bond and B results in a sigma bond A & B result in a sigma bond and C results
in a pi bond
B & C result in a sigma bond and A results in a pi bond
A & C result in
a sigma bond and B results in a pi bond
A & B result in a pi bond and C results in a sigma bond
& C result in a pi bond and A results in a sigma bond
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9. 1.6/1.6 points | Previous Answers My Notes
Question Part
Points
Submissions Used
1
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0.2/0.2 0.2/0.2 0.2/0.2 0.2/0.2 0.2/0.2 0.2/0.2 0.2/0.2 0.2/0.2
2/2
2/2
1/2
1/2
1/2
1/2
1/2
2/2
Total
1.6/1.6
Click here for review presentation.
The red labels a, b and c are to be used in part c only.
Use the structure below to answer the following questions.
Note: the structure below does not reflect its true 3-D shape. You should add lone pairs as needed so
that each non-hydrogen atom obeys the octet rule.
a) number of pi bonds in the entire molecule 13
3
b) number of sigma bonds in the entire molecule 211
11
B
c) Select the hybridization of the indicated atom.
a sp^2
b sp
c sp^3
sp^2
sp
sp^3
d) Give the approximate values of the indicated angles.
Choose from the following values: 60, 90, 109, 120, 180.
Do not include the degree symbol - I have provided that for you.
6120
7180
8109
120
180
109
o
o
o
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