TUTORIAL #7 – Chemical Bonding II (with answers)
1.
Fill in the table below with the correct answers:
(Hint: First, draw the Lewis structures, then; use the structures as a guide to determine the
electron pair and the geometry)
Species
Class
No. of
bonding
pair
electron
No. of lone
pair
electron
Arrangement
of electron
pairs
Molecular
geometry
(shape)
SeF6
IF3
TeF4
XeF5+
ClF2–
XeO3
2.
Iodine forms three compounds with chlorine: ICl, ICl3 and ICl5. Draw the Lewis structures
and determine the molecular geometry (shape) of these molecules.
3.
*Which of these molecules are polar? For each polar molecule, indicate the direction of the
net dipole. (Clue: Draw Lewis structure first, then determine the molecular shape, check the
surrounding atoms, be careful if there are more than two types of surrounding atoms! Check
electronegativity!)
(a) H2O
(d) ClF
(g) CH3Cl
(b) NH3
(e) CCl4
(h) SO3
(c) CO2
(f) HBF2
4.
*State the values of the indicated bond angles below.
(a) O—S—O angle in SO2
(b) F—B—F angle in BF3
(c)
(d)
1
5.
Describe the hybridization around the central atom and the bonding in PF5.
6.
*Acrylonitrile is polymerized to manufacture carpets and wigs. Its Lewis structure is as
follows:
(a) How many sigma (σ) bonds are in the molecule?
(b) How many pi (π) bonds are in the molecule?
(c) What is the hybridization of the carbon atom that is bonded to nitrogen?
7.
*Draw the Lewis structures and designate which are sigma (σ) and pi (π) bonds in each of
these molecules:
(a) OCS
(b) NH2OH
(c) CH2CHCHO
(d) CH3CH(OH)COOH
8.
*Complete the table below:
Compound
Arrangement
of electron
pairs
Molecular
geometry
(shape)
Hybridization
on central
atom
The bond
angle on
the central
atom
Polar or
non-polar?
PF3
XeCl4
NCl3
AsH3
BCl3
2
ANSWERS :
1.
Class
No. of
bonding
pair
electron
No. of lone
pair
electron
Arrangement
of electron
pairs
Molecular
geometry
(shape)
AB6
6
0
Octahedral
Octahedral
IF3
AB3E2
3
2
Trigonal
bipyramidal
T-shaped
TeF4
AB4E
4
1
Trigonal
bypiramidal
XeF5+
AB5E
5
1
Octahedral
ClF2–
AB2E3
2
3
Trigonal
bypiramidal
Linear
XeO3
AB3E
3
1
Tetrahedral
Trigonal
pyramidal
Species
SeF6
2.
Distorted
tetrahedral
(see-saw)
Square
pyramidal
Answer:
Linear
T-shaped
3.
Square pyramidal
Answer:
(a) polar
(d) polar
(g) polar
(b) polar
(e) non-polar
(h) polar
(c) non-polar
(f) polar
4.
(a) 120o (b) 120o (c) ∡1 = 120o , ∡2 = 180o (d) ∡1 = 109.5o , ∡2 = 120o
5.
Answer:
 In PF5, there are 5 electron pairs around phosphorous, all
bonding pairs.
 This requires 5 hybrid orbitals which can be produced by sp3d
hybridization.
 The five P—F sigma bonds are formed by head-to-head
overlap of the sp3d hybrid orbitals on P with 2p orbital on F.
 PF5 has an arrangement of trigonal bipyramidal and the
shape is also trigonal bipyramidal.
3
6.
(a) Six sigma (σ) bonds
(b) Three pi (π) bonds
(c) sp hybridization
7.
8.
(a)
(b)
(c)
(d)
Complete the table below:
Compound
Arrangement
of electron
pairs
PF3
Tetrahedral
XeCl4
Octahedral
NCl3
Tetrahedral
AsH3
Tetrahedral
BCl3
Trigonal
planar
Molecular
geometry
(shape)
Trigonal
pyramidal
Square
planar
Trigonal
pyramidal
Trigonal
pyramidal
Trigonal
planar
The bond
angle on
the central
atom
Polar or
non-polar?
Sp
107.3o
Polar
sp3d2
90o
Non-polar
3
107.3o
Polar
Sp
3
107.3o
Polar
sp2
120o
Non-polar
Hybridization
on central
atom
3
Sp
4