Download Covalent Bonding 5 Practice Problems

Chemistry 11 (HL)
Unit 4 / IB Topic 14.2
Covalent Bonding 5 Practice Problems
Hybridisation and Sigma & Pi Bonds
ANSWERS
1.
hybrid orbitals form when atoms form bonds
a pure s orbital “blends” with 1, 2 or 3 pure p orbitals to make new orbitals with shapes and
energy values that are a mix of the pure orbitals
2.
hybrid
orbital
shape
around
atoms with
these
hybrid
orbitals
description of how it forms
angle
between
hybrid
orbitals
1 pure s orbital mixes with 3 pure p orbitals to make 4
3
identical “sp ” orbitals
sp
3
pure atomic orbitals
p
p
p
tetrahedral
hybrid orbitals
109.5º
(4 NCCs)
blending
sp3 sp3 sp3 sp3
s
1 pure s orbital mixes with 2 (of the three) pure p
2
orbitals to make 3 identical “sp ” orbitals (and leaving
one pure p orbital)
sp
2
pure atomic orbitals
p
p
p
trigonal
planar
120º
hybrid orbitals
blending
p
sp2 sp2 sp2
s
(3 NCCs)*
1 pure s orbital mixes with 1 (of the three) pure p
2
orbitals to make 2 identical “sp ” orbitals (and leaving
two pure p orbitals)
linear
sp
pure atomic orbitals
p
s
p
p
hybrid orbitals
blending
sp sp
p
180º
(2 NCCs)*
p
* the pure p orbitals are
used to make double/triple bonds
3.
What type of hybrid orbitals does the central atom have in these molecules/ions?
a) CF4
C = 4 NCCs = sp
b)
3
BH3
B = 3 NCCs = sp
2
H
B
H
H
p. 1
Chemistry 11 (HL)
Unit 4 / IB Topic 14.2
c) CH2O
d) NH3
C = 3 NCCs = sp
2
N = 4 NCCs = sp
e) CO2
f)
C = 2 NCCs = sp
g) C2H2
BH4
3
H
1-
B = 4 NCCs = sp
3
H
B
H
1-
H
H C C H
C = 2 NCCs = sp
4.
Compare the formation of sigma bonds and pi bonds.
• similarities:
involve sharing of an electron pair
involve the overlap of orbitals from two atoms
• differences:
sigma bond is end on end (or head to head) orbital overlap; pi bond is the
side by side overlap of orbitals
sigma bond electrons lie on the axis between the 2 nuclei; pi bond electrons
are in a plane above and below the nuclei
sigma bonds are stronger than pi bonds
5.
How many sigma and pi bonds are in these molecules?
a) CO2
b) NH3
sigma = 2, pi = 2
d)
C 2H 4
sigma = 3, pi = 0
e) C2H2
sigma = 5, pi = 1
6.
sigma = 3, pi = 2
c) H2CO
sigma = 3, pi = 1
f) O3
sigma = 2, pi = 1
Draw a Lewis structure for ethanoic acid, CH3COOH.
sp
sp2
3
sp2
sp3
sigma = 7, pi = 1
p. 2
Chemistry 11 (HL)
7.
Unit 4 / IB Topic 14.2
Urea, H2NCONH2, is present in animal urine.
O
H
N
C
H
C atom = sp
N
H
H
2
N atoms = sp
3
shape around C atom = trigonal planar
shape around N atom = tetrahedral
bond angle around C atom = 109.5º
bond angle around N atom = 107º
sigma bonds = 7; pi bonds = 1
8.
The structure of ascorbic acid (vitamin C) is shown below.
a)
sigma bonds = 20; pi bonds = 2
(there is a sigma bond between the
O and H atoms in the OH groups on
the molecule)
b)
12 (each O atom has 2 lone electron
pairs)
c)
C atoms with only single bonds = sp
C atoms with double bond = sp
3
2
O atoms with only single bonds = sp
O atoms with double bond = sp
d)
2
Circle the carbon atoms having a
trigonal planar shape.
p. 3
3
Chemistry 11 (HL)
9.
Unit 4 / IB Topic 14.2
Caffeine has the formula C8H10N4O2. A model of caffeine is shown below. Atoms without
symbols are hydrogen atoms.
a)
All N atoms have 1 lone electron pair.
All O atoms have 2 lone electron pairs.
b)
N*
sp
3
107º
tetrahedral
C*
sp
2
120º
trigonal planar
O
sp
2
(omit angles and shape around this atom)
p. 4