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HYBRIDIZATION REVIEW WORKSHEET
1. What is the number of sigma (s) and pi (p) bonds and the hybridization of the carbon atom in
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Sigma Pi Hybridization A. 4 1 sp2 B. 4 1 sp3 C. 3 2 sp3 D. 3 1 sp2 2. What is the best description of the carbon-oxygen bond lengths in ?
A. One short and two long bonds
B. One long and two short bonds
C. Three bonds of the same length
D. Three bonds of different lengths
3. Which allotropes contain carbon atoms with sp2 hybridization?
I. Diamond
II. Graphite
III. C60 fullerene
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
4. Which of the following species is (are) planar (has (have) all the atoms in one plane)?
I.
II.
III.
A. I only C. I and II only
B. II only D. II and III only
5. What is the molecular shape and the hybridization of the nitrogen atom in NH3?
Molecular shape Hybridization A. tetrahedral sp3 B. trigonal planar sp2 C. trigonal pyramidal sp2 D. trigonal pyramidal sp3
6. Which statement about sigma and pi bonds is correct?
A. Sigma bonds are formed only by s orbitals and pi bonds are formed only by p orbitals.
B. Sigma bonds are formed only by p orbitals and pi bonds are formed only by s orbitals.
C. Sigma bonds are formed by either s or p orbitals, pi bonds are formed only by p orbitals.
D. Sigma and pi bonds are formed by either s or p orbitals.
7. NO is trigonal planar and NH3 is trigonal pyramidal. What is the type of hybridization of N in each of these species?
N in NO N in NH3 A. sp2 sp3 B. sp2 sp2 C. Sp3 sp2 D. sp3 sp3 8. In ethanol, C2H5OH (l), there are covalent bonds, hydrogen bonds and van der Waals?? forces. Which bonds or forces are broken when ethanol is vaporized?
A. only hydrogen bonds
B. covalent bonds and hydrogen bonds
C. covalent bonds and van der Waals?? forces
D. hydrogen bonds and van der Waals?? forces
9. Which statement best describes the attraction present in metallic bonding?
A. the attraction between nuclei and electrons
B. the attraction between positive ions and electrons
C. the attraction between positive ions and negative ions
D. the attraction between protons and electrons
10. Which statement is correct about multiple bonding between carbon atoms?
A. Double bonds are formed by two ?? bonds.
B. Double bonds are weaker than single bonds.
C. ?? bonds are formed by overlap between s orbitals.
D. ?? bonds are weaker than sigma bonds.
Short Answers:
1. (i) List the following substances in order of increasing boiling point (lowest first).
CH3CHO C2H6 CH3COOH C2H5OH
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(2)
(ii) State whether each compound is polar or non-polar, and explain the order of boiling points in (c)(i).
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(8)
2. Atomic orbitals can mix by hybridization to form new orbitals for bonding.
Identify the type of hybridization present in each of the three following molecules.
Deduce and explain their shapes.
(i) OF2
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(3)
(ii) H2CO
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(3)
(iii) C2H2
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(3)
3. (i) Use the VSEPR theory to predict and explain the shape and the bond angle of each of the molecules SCl2 and C2Cl2
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(ii) Deduce whether or not each molecule is polar, giving a reason for your answer.
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(3)
4. (i) Explain the meaning of the term hybridization.
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(ii) Discuss the bonding in the molecule CH3CHCH2 with reference to
· the formation of ?? and ?? bonds
· the length and strength of the carbon-carbon bonds
· the types of hybridization shown by the carbon atoms
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(6)
5. The boiling points of the hydrides of group 6 elements increase in the order
H2S < H2Se < H2Te < H2O.
Explain the trend in the boiling points in terms of bonding.
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................................................................................................................................................ (6)
Answers:
1) A 2) C 3) C 4) C 5) D 6) C 7) A 8) D
9) B 10) D
Short Answers:
1) Lowest Highest
Ethane Ethanone Ethanol Ethanoic acid
Van der Waals dipole-dipole H-bonding H-bonding
with larger mass - therefore, greater Van der Waal??s
2) i) bent ii) trigonal planar iii) planar
- SCl2; bent (1070) - polar
C2Cl2: planar (1800) - non-polar
5) As size increases down Family 16, the electronegativity decreases. As a result, the H-bonding progressively weakens, making for easier breaking of the intermolecular forces.
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