4A Chem 1A 4

4A Chem 1A 4th midterm Spring 08 Chabot College Dr J. McOmber Name __________________________
ΔH(reaction) is about equal to Bond energy(bonds broken)‐Bond energy(bonds made) The formal charge on an atom = valence ‐ # lone electrons ‐ # bonds E pairs and atoms
connected to center
2
Bond
bonding
sp
Bond angle
180
3
sp2
120
4
sp3
109.5
5
6
sp3d
sp3d2
120, 90
90
Bond
length
pm
Bond
Electron/atom shape
Linear
Shape, no electron pairs
Linear
Shape, one electron pair
Linear
Shape, 2 electron
pairs
Trigonal planar
Trignoal planar
Bent
Tetrahedral
Tetrahedral
Trigonal pyramidal
Bent
Trigonal bipyramid
Octahedral
Trigonal bipyramid
Octahedral
see-saw
Square pyramid
T shape
square planar
H-O
97
Bond
energy,
kJ/mole
464
C-N
Bond
length
pm
147
Bond
energy,
kJ/mole
305
H-H
74
436
C=N
128
615
H-C
11
414
C=N
116
891
H-F
92
565
C-Cl
178
339
H-Cl
127
431
N-N
145
163
H-Br
141
364
N=N
123
418
H-N
100
389
N-O
136
222
C-C
154
347
N=O
120
590
C=C
134
611
O-O
145
142
C=C
120
834
O=O
121
498
C-O
143
360
F-F
143
159
C=O
120
736
Cl-Cl
199
243
N-F
144
____
_ σ2p* ____ ____ π*2p
____ σ2p
____ ____
π2p
____ ________ σ*2s
____ ________ σ 2s
____ ________ σ*1s
____ ________ σ 1s
Molecular Orbital Chart Paramagnetic = is magnetic (lone electrons) Diamagnetic = is not magnetic (no lone electrons 4A Chem 1A 4th midterm Spring 08 Chabot College Dr J. McOmber Name __________________________
__d___1 .
Atoms having greatly differing electronegativities are expected to form:
a) no bonds
b)
polar covalent bonds c)
nonpolar covalent bonds
d) ionic bonds
e)
covalent bonds
__e (d ok)___2.
Which of the following molecules has a dipole moment (is a polar compound. Hint,
make Lewis dot structures and see)?
b)
CCl4 c)
CO2 d)
SO3
e)
none of these
a) CH4
__b___3.
Which of the following pairs is isoelectronic (has same number of electrons)?
a)
__d___4.
Li+ and K+
b) Na+ and Ne
c) I– and Cl–
d) S2– and Ne e) Al3+ and B3+
Which of the following has the smallest radius?
a)
__e___5.
F-
b)
Ne
O2-
c)
d)
Mg2+ e)
Na+
Using the following data reactions calculate the energy of an H–Cl bond.
H (kJ)
H2(g) + Cl2(g)  2HCl(g)
a)
–184
H2(g)  2H(g)
432
Cl2(g)  2Cl(g)
239
770 kJ
b)
856 kJ c)
518 kJ d)
326 kJ e)
428 kJ
__c___6.
As indicated by Lewis structures, which of the following would probably not exist as a stable
molecule?
a)
CH3OH b)
CH2O c)
CH3O d)
C2H2
e)
C3H4
In the cyanide ion (CN–), the nitrogen has a formal charge of
__c___7.
a)
-2
b)
-1
c)
0
d)
2
e)
2
__a___8.
Choose the electron dot formula that most accurately describes the bonding in CS2. (Hint:
Consider formal charges.)
a)
__d___9.
S C
S
b) C
S
S
c) S
C S
d) S
C S
e) S
C S
Complete the Lewis structure for the molecule(missing some double bonds, do Lewis dots):
CH3 O
CH3
CH
C
C
N
This molecule has __________ single bonds and __________ multiple bonds.
a)
__d___10.
a)
__e___11.
a)
4, 2
b)
6, 3
c)
11, 5
d)
11, 2
e)
13, 0
In the Lewis structure for SF6, the central sulfur atom shares __________ electrons.
4 b) 8
c) 10
d) 12
e)none of the above, because SF6 is an ionic compound
Of the following elements, which is most likely to form a negative ion with charge 1–?
Ba
b)
Ca
c)
Si
d)
P
e)
Cl
4A Chem 1A 4th midterm Spring 08 Chabot College Dr J. McOmber Name __________________________
The hybridization of the central atom in ClF2+ is:
__c___12.
a)
sp
__d___13.
a)
sp2
b)
c)
sp3
d)
dsp3
e)
d2sp3
Which of the following does not contain at least one pi bond?
H2CO
___a__14.
b) CO2 c) C2H4 d) C3H8 e) All of the above (a-d) contain at least one pi bond.
Consider the molecule
O
H2C CH C
1
a)
b)
c)
d)
e)
2
C–1
sp2
sp2
sp2
sp2
sp
CH2 C
3
4
C–2
sp2
sp2
sp2
sp2
sp
N
Specify the hybridization of each carbon atom.
5
C–3
sp2
sp2
sp3
sp3
sp
C–4
sp3
sp3
sp3
sp3
sp2
C–5
sp
sp3
sp
sp3
sp
__c___15. Which of the following statements is false?
a) A sodium atom has a smaller radius than a potassium atom.
b) A neon atom has a smaller radius than an oxygen atom.
c) A fluorine atom has a smaller first ionization energy than an oxygen atom.
d) A cesium atom has a smaller first ionization energy than a lithium atom.
e) All are true.
Note: First ionization energy is the energy required to pull off one electron.
For these molecular orbital questions, see the chart on the first page.
__a___16.
a)
___a__17.
What is the (molecular orbital) bond order of Ne2?
0
b) 1/2
c) 1
d)
1 1/2 e)
2
Which of the following (molecular orbital) statements is false?
a)
C2 is paramagnetic.
b)
C2 is diamagnetic.
c)
The carbon-carbon bond in C22– is stronger than the one in CH3CH3.
d)
The carbon-carbon bond in C22– is shorter than the one in CH3CH3.
e)
Two of the above.
___c__18.
a)
Which of the following substances contains two pi bonds?
C2H4
b)
C3H8
c)
C2H2
d)
C2H6
e)
CH4
___c__19. Consider the skeletal structure shown N—C—C—N Draw the Lewis structure and answer the
following: How many pi bonds does the molecule contain? (double bond = 1, triple bond = 2)
a)
0 b)
2
c)
4
d)
6
e)
7
4A Chem 1A 4th midterm Spring 08 Chabot College Dr J. McOmber Name __________________________
20. (20 pts) Draw Lewis Dot structures for the following and fill in the table (bond angle around center atom, bonding to center atom) Molecule NO2–
Lewis Dot Structure Bonding type bond angle (about center atom) Sp2 120 sp 180 Sp2 120 H:C:::N: sp 180 O::C::O sp 180 H:O:O:H Sp3 109.5 C2H2
HNO3
HCN
CO2
H2O2
21. (10 pts) Calculate the lattice energy for KI. That’s K+(g) + I-(g)  KI(s) given the following
I2(s)  I2(g) Hsublimation = 62.438
½ I2(g)  I(g) H = 75.619
I(g)  I-(g) Electron affinity = -295.16
K(s)  K(g) Hsublimation 163.60
K(g)  K+(g) First Ionization energy = 419
K(s) + ½ I2(s)  KI(s) Hformation = -327.90
I-(g) I(g)
295.16
I(g)  ½ I2(g) H = -75.619
1/2I2(g)  1/2I2(s) H = -62.438/2
K+(g) K(g) -419
K(g)  K(s) H -163.60
K(s) + ½ I2(s)  KI(s) Hformation = ‐327.90 K+(g) + I‐(g)  KI(s) H = 295.16-31.219-753619-419-163.60-327.90 =-722.2 kJ
4A Chem 1A 4th midterm Spring 08 Chabot College Dr J. McOmber Name __________________________
22. (10 pts) Use the bond breaking/bond forming method, calculate Hrxn for: CH3OH + 3/2O2  CO2 + 2H2O (see table in
front, hint; balance first)
3 C-H + C-O + O-H + 3/2O=O – 2 C=O-4 O-H = 3x414+360+1.5x498-2x736-3x464 = -515 kJ
23 (10 pts) Dinitrogen difluoride is known to have a nitrogen-to nitrogen bond length of 123 pm and nitrogen-to fluorine bond length
of 141 pm. Write a Lewis structure consistent with these data.
F:N::N:F It’s a single bond between F and N and a double bond between N and N