Name____________________________________ Date:______________ Period:______
1.
2.
[KEY] REVIEW Unit 2 Test (Chp 8,9) : Bonding and Molecular Geometry [KEY]
How does oxygen demonstrate the “octet rule” when forming ionic bonds?
Oxygen atoms tend to gain 2 e
–
.
Ionic compounds of the greatest lattice energy form from cations that are small and more positive and from anions that are small and more negative
___3. The electron-dot structure (Lewis structure) for which of the following molecules would have two unshared pairs of electrons on the central atom?
(A) H
2
S (B) NH
3
(C) CH
4
(D) HCN
4. Describe the trend in electronegativity in the periodic table.
Increases across a period and decreases down a group.
___5. Of the bonds below, __________ is the least polar.
(E) CO
2
(A) Na-S (B) P-S (C) Na-Cl (D) C-F (E) Si-Cl
___6. Which of the following compounds has the least ionic character?
(A) PH
3
(B) Na
2
O (C) ZnS (D) BeF
2
(E) GaN least difference in electronegativity (closest together in the periodic table) least difference in electronegativity (closest together in the periodic table, recall that H is near equal to C)
___7. Which of the following molecules has the shortest bond length?
(A) N
2
(B) O
2
(C) Cl
2
(D) Br
2
(E) I
2 triple bond requires side-by-side p orbital overlap even more than a double (in O
2
) and singles which are longer head-to-head sigma overlap.
1

Questions 8-11 refer to the following molecules.
(A) CO
2
(B) CH
4
(C) HF (D) NH
3
A 8. Contains 2 π-bonds.
C 9. Has the highest dipole moment.
D 10. Has a molecular geometry that is trigonal pyramidal.
(E) F
2
A_11. Has the largest bond dissociation energy.
8) The C is sp hybridized (two domains) so the first bond to each O is a σ bond (head-to-head overlap); the double bonds are π bonds (closer side-to-side overlap) from the 2 unhybridized p orbitals of C and each O.
9) most polar, greatest diff. in electronegativity
10) 4 domains but one is e
–
pair (AB
3
E)
11) More energy is required to break (dissociate) stronger bonds, so the double bond in CO
2
has the highest BDE.
___12. Which of the following has a zero dipole moment?
(A) HCN (B) NH
3
(C) SO
2
(D) NO
2
(E) PF
5
CCl
4
, CO
2
, PCl
3
, PCl
5
, SF
6
___13. Which of the following does not describe any of the molecules above?
AB
5
gives a symmetrical nonpolar trigonal bipyramidal molecule
(A) Linear
(B) Octahedral
( CO
2
)
( SF
6
)
(C) Square planar requires 6 domains as AB
4
E
2
(D) Tetrahedral (
CCl
4
)
(E) Trigonal pyramidal ( PCl
3
)
B
B
B
B
___14. Molecules that have planar configurations include which of the following?
I. BCl
3
II. CHCl
3
III. NCl
3
(A) I only (B) III only (C) I and II only (D) II and III only (E) I, II, and III
2
Cl
Cl
Cl

___15. Types of hybridization exhibited by the C atoms in propene, CH
3
CHCH
2
, include which of the following?
I. sp
II. sp
2
III. sp
3 the leftmost C is sp 3 (4 domains) the middle C is sp 2 (3 domains) the rightmost C is sp 2 (3 domains)
(A) I only (B) III only (C) I and II only (D) II and III only (E) I, II, and III
___16. What is the molecular geometry of the SbCl
5
molecule?
(A) tetrahedral
(B) T–shaped
(C) seesaw
(D) trigonal bipyramidal
(E) octahedral
17. The hybridization of the carbon atom labeled x in the structure below is sp 2
3 domains
The H-C-H bond angle is 120 o (or slightly less due to extra repulsion of lone e
–
pairs) 3 domains is trigonal planar
The hybridization of the carbon atom labeled y in the structure below is sp 3
The C-O-H bond angle is 109.5
o (or 104.5
o due to extra repulsion of lone e
–
pairs)
4 domains
4 domains is tetrahedral x y
___18. There are __________ unhybridized p atomic orbitals in an sp
3
hybridized carbon atom.
(A) 4 (B) 3 (C) 2 (D) 1 (E) 0 the s and all 3 p ’s have been hybridized leaving none unhybridized
___19. In a molecule in which the central atom has six electron pairs around it, each of the six electron domains is are directed toward the corners of
(A) a tetrahedron
(B) a square-based pyramid
(C) a trigonal bipyramid
(D) a square
(E) an octahedron octahedron = 8 sided figure
3

___20. A typical triple bond…
(A) is longer than a single bond
(B) is shorter than a double bond triple bond requires more side-by-side p orbital overlap than a double bond
(C) is weaker than a single bond
(D) has 1 sigma (σ) and 3 pi (π) bonds
(E) consists of 6 electron pairs
21. Explain why the bond angle of SO
2
is slightly less than 120 o
.
OR
With 1 e
–
pair on S it should have 3 domains for a trigonal planar geometry. But the nonbonded pair of electrons repels more (takes up more space) than the bonded pairs so it pushes the O atoms closer together than 120 o .
___22. Which of the following molecules contains bonds that have a bond order of 1.5?
(A) N
2
(B) O
3
(C) NH
3
(D) CO
2
(E) CH
2
CH
2 one single and one double bond in resonance (or 1 + 2 = 3 divided by two bonded domains = 1.5)
23. The molecular geometry of the XeF
4
molecule is square planar.
AB
4
E
2
is octahedral but with 2 nonbonded e
–
pairs.
24. According to Valence Bond Theory, describe the hybridization of atomic orbitals and its results on the ability of an atom to form bonds.
Hybridization is the blending of s and p orbitals to form new orbitals of energy between s and p . These new orbitals are identical and their new shape provides more effective overlap for bonding than the original p orbitals could.
25. Differentiate between ionic and covalent bonding.
Ionic bonding occurs from electrostatic attraction of oppositely charged ions formed by the transfer of e
– ’s, but covalent bonding is the sharing of valence electrons between atoms.
(single, double, triple)
26. Describe resonance and the delocalization of electrons as related to the character of a molecule’s bonds.
Resonance is used to represent a molecule when no single Lewis structure can explain its observed structure or properties.
Delocalization of pi e
– ’s gives covalent bonds intermediate character with bond orders between single, double, and triple.
4

Section II Free Response
CALCULATOR ALLOWED
Your responses to these questions will be graded on the basis of the accuracy and relevance of the information cited. Explanations should be clear and well organized . Examples and equations may be included in your responses where appropriate. Specific answers are preferable to broad , diffuse responses.
1.
Answer the following questions about the structures of ions that contain only sulfur and fluorine.
(a) The compounds SF
4
and BF
3
are made to react to form an ionic compound according to the following equation.
SF
4
+ BF
3
→
SF
3
BF
4
(i) Draw a complete Lewis structure for the cation SF
3
+ in the compound SF
3
BF
4
. (1)
F – S – F +
F
(ii) Identify the type of hybridization exhibited by sulfur in the SF
3 sp 3 (4 domains requires 4 orbitals, 1 s and 3 p ’s)
+
cation. (1)
(iii)Identify the geometry of the SF
3
+ cation that is consistent with the Lewis structure drawn in part
(a)(i). (1) trigonal pyramidal (due to the 4 domains in AB
3
E electron domain arrangement rather than AB
4
)
(iv) Predict whether the F-S-F bond angle in the SF
3
+
cation is larger than, equal to, or smaller than
109.5
o
. Justify your answer. (1)
The F-S-F bond angle in the SF
3
+ cation is expected to be slightly smaller than 109.5
o because the repulsion between the nonbonding pair of electrons and the S-F bonding pairs of electrons “squeezes” the F-S-F bond angles together slightly.
(b) The compounds SF
4
and CsF react to form an ionic compound according to the following equation.
SF
4
+ CsF
→
CsSF
–
(i) Draw a complete Lewis structure for the SF
5
5
anion in CsSF
5
. (1) –1
(ii) Identify the electron domain geometry exhibited by the SF
4
molecule. (1)
Five electron domains in an AB
4
E arrangement cause an electron domain geometry of trigonal bipyramidal.
[the actual molecular geometry of an AB
4
E molecule would be seesaw or distorted tetrahedron]
(iii) Identify the geometry of the SF
5
–
anion that is consistent with the Lewis structure drawn in
part (b)(i). (1) square pyramidal (caused by an AB
5
E electron domain arrangement)
(iv) Identify the formal charge of sulfur in the SF
5
–
anion. (1) –1
5

GeCl
4
SeCl
4
ICl
4
–
ICl
4
+
2.
The species represented above all have the same number of chlorine atoms attached to the central atom.
(a) Draw the Lewis structure (electron-dot diagram) of each of the four species. Show all valence electrons in your structures. (4)
Cl Cl
Cl Ge Cl Cl Se Cl
Cl Cl
Cl
Cl
I Cl
–
Cl
Cl I Cl
+
Cl Cl
(b) On the basis of the Lewis structures drawn in part (a), answer the following questions about the particular species indicated.
(i) What is the Cl–Ge–Cl bond angle in GeCl
4
? (1)
109.5
o
(ii) Is SeCl
4
polar? Explain. (1)
AP answer:
Yes. The SeCl
4
molecule is polar because the lone pair of nonbonded electrons in the valence shell of the selenium atom interacts with the bonding pairs of electrons, causing a spatial asymmetry of the dipole moments of the polar Se–Cl bonds. The result is a SeCl
4
molecule with a net dipole moment. (AP answer)
OR
Simple answer:
Yes. The SeCl
4
molecule is polar because the lone pair of nonbonded electrons on the Se atom cause an asymmetrical distribution on the polar Se–Cl bonds creating an overall dipole for the molecule. (simple answer)
(iii) What is the geometric shape formed by the atoms in ICl
4
–
? (1) square planar (caused by an AB
4
E
2
electron domain arrangement)
(electron domain geometry with six domains would be octahedral, AB
6
)
(iv) What is the geometric shape formed by the atoms in ICl
4
+
? (1) seesaw (or distorted tetrahedron if that’s easier to remember for you from the four atoms)
6

3.
Using principles of chemical bonding and molecular geometry, explain each of the following observations.
Lewis electron-dot diagrams and sketches of molecules may be helpful as part of your explanations. For each observation, your answer must include references to both substances.
(a) The bonds in nitrite ion, NO
2
–
, are shorter than the bonds in nitrate ion, NO
3
–
. (2)
Two resonance structures are required to represent the bonding in the NO
2
–
ion. The effective number of bonds (bond order) between N and O is 1.5 shown by (1 + 2)/2 = 1.5 .
Three resonance structures are required to represent the bonding in the NO
3
–
ion. The effective number of bonds between
N and O is 1.3 shown by (1 + 1 + 2)/3 = 1.3 . The greater the effective number of bonds, the shorter N–O bond length.
(b) The CH
2
F
2
molecule is polar, whereas the CF
4
molecule is not. (2)
The molecular geometry in both CH
2
F
2
and CF
4
is tetrahedral (or the same). The C-F bond is polar. In CF
4
, the molecular geometry arranges the C-F dipoles so that they cancel out and the molecule is nonpolar. The C-H bond is less polar than the C-F bond. The two C-H dipoles do not cancel the two C-F dipoles in CH
2
F
2
.
(c) The atoms in a C
2
H
4
molecule are located in a single plane, whereas those in a C
2
H
6
molecule are not.
(2)
The carbon atoms in C
2
H
4
have a molecular geometry around each carbon atom that is trigonal planar (AB
3
), so all six atoms are in the same plane. The carbon atoms in C
2
H
6
have a molecular geometry that is tetrahedral (AB
4
), so the atoms are not all in the same plane.
OR
The carbon-carbon double bond in C
2
H
4
results in a planar molecule whereas the carbon-carbon single bond in C
2
H
6 results in a non-planar (tetrahedral) site at each carbon atom.
(d) The shape of a PF
5
molecule differs from that of an IF
5
molecule. (1)
In PF
5
, the molecular geometry is trigonal bipyramidal because the phosphorus atom has five bonding pairs of electrons and no lone pairs of electrons. IF
5
has square pyramidal molecular geometry. The central iodine atom has five bonding pairs of electrons and one lone pair of electrons. The presence of the additional lone pair of electrons on the central iodine atom means the molecular geometry is different.
(e) Draw two possible Lewis structures for the CH
2
O molecule and identify the most significant
–
+
most significant because formal charges are minimized (zero on all atoms)
7

Answer Key
1. 2006 #7
2.
3.
2006B #6
2002B #6(a)-(d)
8