CHEM 1411
Practice problems (Chapters 10, 11)
Q1-9: Chapter 10; Q10-15: Chapter 11.
1. Which of the following has the arrangement of electron pairs (i.e. electron-pair geometry) identical to its
molecular geometry?
(a) H2S
(b) SF4
(c) XeF4
(d) SiCl4
(e) NH3
Hint: p.p. 400-409.For molecules in which the central atom does not have lone pairs, the arrangement of
electron pairs is identical to the molecular geometry.
2. Which of the following molecules have same molecular geometry?
(a) CO2 and CF4
(b) NH3 and BF3
(c) H2O and CS2
(d) NO3- and CO32(e) SF4 and CF4
Hint: p.p. 400-409.
3. Which of the following is a polar molecule?
(a) CF4
(b) CO2
(c) AlCl3
(d) NH3
(e) XeF4
Hint: p.p. 409-415. A molecule will be nonpolar if: (a) the bonds are nonpolar, or (b) there are no lone
pairs in the valence shell of the central atom and all the atoms attached to the central atom are the same
4. Which of the following underlined atoms have the same hybrid orbitals?
(a) CH3COOH
(b) CH3CH=O
(c) CH3-HC=CH-CH2OH
(d) H2C=C=CH2
(e) H3C-HC=CH-CH2OH
Hint: p.p. 421-429. For organic molecules, the hybridization type of central carbon atoms generally obey
the following rule: carnon single bond:sp3, carbon double bond sp2, carbon triple bond sp.
5. How many pi bonds and sigma bonds are there in the tetracyanoethylene molecule?
(a) 9, 9
(b) 4, 5
(c) 4, 9
(d) 9, 4
(e) 5, 4
Hint: p.p. 426-429.Single bond: one sigma bond, Double bond: One sigma and one pi bond, Triple bond:
One sigma and two pi bonds.
6. Which of the following pairs has approximately the same bond angles?
(a) BeCl2 and BCl3
(b) CCl4 and CH3Cl
(c) AlCl4- and AlCl3
(d) PF3 and PF5
(e) CO2 and H2O
Hint: p.p. 400-409.
7. What is the hybrid orbital for S in SF4?
(a) sp
(b) sp2
(c) sp3
(d) sp3d
(e) sp3d2
Hint: How to predict the hybridization of the central atom?
1. Hint: p.p. 417-426. Draw the Lewis structure of the molecule.
2. Count the number of lone pairs AND the number of atoms bonded to the central atom. Predict the
overall arrangement of electron pairs using VSEPR model ( Table 10.1)
3. Deduce hybridization of central atom by matching the arrangement of the electron pairs with those
of hybrid orbitals shown in Table 10.4.
8. Which of the following has the bond angle as 120 o?
(a) NH3
(b) AlCl3
(c) SF6
(d) PF3
(e) XeF4
Hint: p.p. 400-409. In order to get the bond angle, you must first predict the arrangement of electron pairs
on the central atom.
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9. Which of the following has the same bond order as H2+?
(a) H2(b) H2
(c) He2
(d) All of the above
(e) None of the above
Hint: p. 433. In order to get the bond order, you must first write the electron configuration using molecular
orbital theory. Remember the approximate relative energies of molecular orbitals in second period diatomic
molecules. (a) Li2 through N2, σ2sσ 2s* π2px π2py σ2pz π2px *π2py* σ 2pz*
(b) O2 through Ne2. σ2sσ 2s* σ2pz π2px π2py π2px *π2py* σ 2pz*
Bond order= ½( #electrons in bonding orbitals - # electrons in anti-bonding orbitals)
10. Which of the following order is correct concerning the boiling points?
(a) CH4 < SiH4 < GeH4 < SnH4
(b) NH3 < PH3 < AsH3 < SbH3
(c) H2S < H2Se < H2Te < H2O
(d) HF < HCl < HBr < HI
(e) CI4 < CBr4 < CCl4 < CF4
Hint: p.p. 453-458. Boiling points reflect the strength of intermolecular force. Dispersion forces usually
increase with molar mass (more electrons), or size of the atom. But there are exception in group 5A,6A and
7A, when hydrogen bonds are involved. The hydrogen bond is a dipole-dipole interaction between the
hydrogen atom in a polar N-H, O-H, or F-H bond and an electronegative O, N, or F atom. Hydrogen bond
is much stronger than dispersion force, which leads to high boiling point.
11. Which of the following has other intermolecular forces in addition to London dispersion force?
(a) C6H6
(b) CS2
(c) C2H4
(d) Cl2
(e) HF
Hint: p.p. 453-458. For non-polar molecules, the only intermolecular force existed is London dispersion
force. There are more than one type of intermolecular forces for other molecules.
Types of Intermolecular Forces (IMF)
• Ion - ion forces; Ion-dipole forces; Dipole-dipole forces; Dispersion Forces; Hydrogen Bonds
12. Which of the following is capable of hydrogen-bonding among itself?
(a) HI
(b) KF
(c) B2H6
(d) C2H6
(e) CH3COOH
Hint: p.p.457-458. The hydrogen bond is a special dipole-dipole interaction between the hydrogen atom in
a polar N-H, O-H, or F-H bond and an electronegative O, N, or F atom. .
13. Which of the following is a correct order of boiling points?
(a) CO2 < RbF < CH3OH < CH3Br
(b) CH3Br < CO2 < CH3OH < RbF
(c) CO2 < CH3Br < CH3OH < RbF
(d) RbF < CH3OH < CO2 < CH3Br
(e) CH3OH < RbF < CH3Br < CO2
Hint: p.p. 453-458. Boiling points reflect the strength of intermolecular force. The ion-ion force is much
stronger than other intermolecular forces. Next is the hydrogen bond. Among all the intermolecular forces,
dispersion force is the weakest. Dispersion forces usually increase with molar mass (more electrons), or
size of the atom.
14. Which of the following is responsible for the phenomenon that insect does not drown in the water?
(a) viscosity
(b) unit cell
(c) surface tension
(d) hydrogen bonding
(e) capillary effect
Hint: p. 459.
15. Pt, (195.08 amu), a face-centered cubic unit cell. What is the net number of metal atoms inside the unit
cell? What is the density (g/ cm3) of the metal if the edge of the unit cell is 3.924x10 -10 m.
(a) 1, 4.8
(b) 2, 4.8
(c) 4, 4.8
(d) 2, 21.45
(e) 4, 21.45
Hint:(a) Number of atoms in the unit cell = 8x(1/8) + 6x(1/2) = 4
(b) The density of Pt metal:
First, convert molar mass to mass per atom: 195.08/6.02x10 23) = 3.239x10-22 g Pt/1 Pt atom.
Since there are 4 atoms in a unit cell, the mass per unit cell is
4x(3.239x10-22 g Pt/1 Pt atom) = 1.296x10-21g.
The volume of unit cell is (3.924x10-10 m x 1x102 cm/m)3 = 6.042x10-23 cm3
Density = mass / volume = {1.296x10-21g/(6.042x10-23 cm3 = 21.45 g/ cm3
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