Name______Date:______Period:______
[KEY] REVIEW Unit 2 Test (Chp 8,9): Bonding and Molecular Geometry[KEY]
1.How does oxygen demonstrate the “octet rule” when forming ionic bonds?
Oxygen atoms tend to gain 2 e–.
2.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) H2S(B) NH3(C) CH4(D) HCN(E) CO2
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.
(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) PH3(B) Na2O(C) ZnS(D) BeF2(E) GaN
___7.Which of the following molecules has the shortest bond length?
(A) N2(B) O2(C) Cl2(D) Br2(E) I2
Questions 8-11 refer to the following molecules.
(A) CO2(B) CH4(C) HF(D) NH3(E) F2
A 8.Contains 2 π-bonds.
C 9.Has the highest dipole moment.
D 10.Has a molecular geometry that is trigonal pyramidal.
A_11.Has the largest bond dissociation energy.
___12.Which of the following has a zero dipole moment?
(A) HCN(B) NH3(C) SO2(D) NO2(E) PF5
CCl4, CO2, PCl3, PCl5, SF6
___13.Which of the following does not describe any of the molecules above?
(A) Linear(CO2)
(B) Octahedral(SF6)
(C) Square planar
(D) Tetrahedral(CCl4)
(E) Trigonal pyramidal(PCl3)
___14.Molecules that have planar configurations include which of the following?
I. BCl3
II. CHCl3
III. NCl3
(A) I only(B) III only(C) I and II only(D) II and III only(E) I, II, and III
___15.Types of hybridization exhibited by the C atoms in propene, CH3CHCH2, include which of the following?
I. sp
II. sp2
III. sp3
(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 SbCl5 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 sp2
The H-C-H bond angle is 120o(or slightly less due to extra repulsion of lone e– pairs)
The hybridization of the carbon atom labeled y in the structure below is sp3
The C-O-H bond angle is 109.5o (or 104.5o due to extra repulsion of lone e– pairs)
___18.There are ______unhybridized p atomic orbitals in an sp3 hybridized carbon atom.
(A) 4(B) 3(C) 2(D) 1(E) 0
___19.In a molecule in which the central atom has six electron pairs around it, each of the six electron domains is are directedtoward the corners of
(A) a tetrahedron
(B) a square-based pyramid
(C) a trigonal bipyramid
(D) a square
(E) an octahedron
___20.A typical triple bond…
(A) is longer than a single bond
(B) is shorter 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 SO2 is slightly less than 120o.
___22.Which of the following molecules contains bonds that have a bond order of 1.5?
(A) N2(B) O3(C) NH3(D) CO2(E) CH2CH2
23.The molecular geometryof the XeF4 molecule issquare planar.
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 porbitals 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 porbitals could.
25.Differentiatebetween 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.
26.Describe resonance and the delocalization of electrons as related to the character of a molecule’s bonds.
Resonance is used torepresenta molecule when no single Lewis structure can explainitsobserved structure or properties. Delocalization of pi e–’s gives covalent bonds intermediate character with bond orders between single, double, and triple.
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 preferabletobroad, diffuse responses.
- Answer the following questions about the structures of ions that contain only sulfur and fluorine.
(a)The compounds SF4and BF3are made to react to form an ionic compound according to the following equation.
SF4 + BF3 → SF3BF4
(i)Draw a complete Lewis structure for the cation SF3+in the compound SF3BF4. (1)
F – S – F +
F
(ii)Identify the type of hybridization exhibited by sulfur in the SF3+cation. (1)
sp3 (4 domains requires 4 orbitals, 1 s and 3 p’s)
(iii)Identify the geometry of the SF3+cation that is consistent with the Lewis structure drawn in part (a)(i). (1)
trigonal pyramidal (due to the 4 domains in AB3E electron domain arrangement rather than AB4)
(iv)Predict whether the F-S-F bond angle in the SF3+cation is larger than, equal to, or smaller than 109.5o. Justify your answer. (1)
The F-S-F bond angle in the SF3+ cation is expected to be slightly smaller than 109.5obecause 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 SF4and CsF react to form an ionic compound according to the following equation.
SF4 + CsF → CsSF5
(i)Draw a complete Lewis structure for the SF5–anion in CsSF5. (1)
(ii)Identify the electron domain geometry exhibited by the SF4molecule. (1)
Five electron domains in an AB4E arrangement cause an electron domain geometry of trigonal bipyramidal. [the actual molecular geometry of an AB4E molecule would be seesawor distorted tetrahedron]
(iii)Identify the geometry of the SF5–anion that is consistent with the Lewis structure drawn in
part (b)(i). (1)
square pyramidal (caused by an AB5E electron domain arrangement)
(iv)Identify the formal charge of sulfur in the SF5–anion. (1) –1
GeCl4SeCl4ICl4–ICl4+
- 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
ClGe Cl Cl Se Cl
ClCl
– +
ClCl
ClI Cl ClI Cl
ClCl
(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 GeCl4 ? (1)
109.5o
(ii)Is SeCl4 polar? Explain. (1)
AP answer:
Yes. The SeCl4 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 SeCl4 molecule with a net dipole moment. (AP answer)
OR
Simple answer:
Yes. The SeCl4 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 ICl4–?(1)
square planar (caused by an AB4E2 electron domain arrangement)
(electron domain geometry with six domains would be octahedral, AB6)
(iv)What is the geometric shape formed by the atoms in ICl4+? (1)
seesaw (ordistorted tetrahedron if that’s easier to remember for you from the four atoms)
- 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, NO2–, are shorter than the bonds in nitrate ion, NO3–. (2)
Two resonance structures are required to represent the bonding in the NO2– ion. The effective number of bonds (bond order) between N and O is 1.5shown by (1 + 2)/2 = 1.5 .
Three resonance structures are required to represent the bonding in the NO3–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 CH2F2 molecule is polar, whereas the CF4 molecule is not. (2)
The molecular geometry in both CH2F2 and CF4 is tetrahedral (or the same). The C-F bond is polar. In CF4, 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 CH2F2.
(c)The atoms in a C2H4 molecule are located in a single plane, whereas those in a C2H6 molecule are not.
(2)
The carbon atoms in C2H4 have a molecular geometry around each carbon atom that is trigonal planar (AB3), so all six atoms are in the same plane. The carbon atoms in C2H6 have a molecular geometry that is tetrahedral (AB4), so the atoms are not all in the same plane.
OR
The carbon-carbon double bond in C2H4 results in a planar molecule whereas the carbon-carbon single bond in C2H6 results in a non-planar (tetrahedral) site at each carbon atom.
(d)The shape of a PF5 molecule differs from that of an IF5 molecule. (1)
In PF5, the molecular geometry is trigonal bipyramidal because the phosphorus atom has five bonding pairs of electrons and no lone pairs of electrons. IF5has 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 CH2O molecule and identify the most significant structure. Justify your choice in terms of formal charge. (2)
most significant because formal charges are minimized (zero on all atoms)
Answer Key
1.2006#7
2.2006B#6
3.2002B#6(a)-(d)
1