Download CHE 0315 SEM 3, 2013/14 TOPIC 5: CHEMICAL BONDING 1. State

CHE 0315
SEM 3, 2013/14
TOPIC 5: CHEMICAL BONDING
1.
State the concept of bonding.
[1]/5a
Only valence electron is involved in the formation of chemical bonds and the
purpose
of
sharing
or
transfer
of
electrons
is
because
to
achieve
octet
configuration.
2.
(a)
By using the aid of appropriate model, describe the formation of metallic bond.[3]/5c
The metal atoms contribute their valence electrons to form a sea of
delocalized electron and the positive metal ions in an orderly array. A
metallic bond is formed by the electrostatic attraction between the
positively
charged metal and the sea of delocalized electron.
Delocalized electrons
Metal cations
(b)
State why all metals are good conductor.
[1]/5f
The delocalized electrons in metals carry the electrical charge.
(c)
The melting point of aluminum is higher than sodium. Explain.
[2]/5e
Both aluminium and sodium possess metallic bonding. Since aluminium has
three valence electrons and sodium has one valence electron, the metallic
bonds in aluminium are stronger than sodium. Therefore, aluminium has
higher melting point than sodium.
CHE 0315
SEM 3, 2013/14
3.
[5]/5c
Draw the Lewis structures of the following compounds.
(a)
CH2Br2
H
: :
:
: Br
C
Br :
:
H
(b)
NCl3
..
N
..
: Cl
..
..
..
Cl :
:Cl :
..
(c)
H2SO4
(d)
CN-
:C
(e)
N:
-
CH4O (All unshared electrons are on the O)
H
H
C
H
H
..
O
..
H
H
C
C
H
H
..
C
H
H
H
OR
O
..
H
CHE 0315
4.
SEM 3, 2013/14
By using the lewis structures, show the formation of ionic bonding of the following
compounds:
(a)
[8]/5d
MgO
.
: O. : . Mg .
(b)
Na2O
.
.
:
+ O : + Na
.
.
Na
(c)
.
Li
+
..
.O .
..
2
Li
.. 2:O :
..
+
..
.O .
..
..
.O .
..
..
.O .
..
2 Al
3+ .. 23 :O :
..
Al2O3
.
Al .
.
(a)
Na
+
.
Al .
.
5.
.. 2+
:O :
Na
..
+
Li2O
.
Li
(d)
.. 2Mg2+ [ :O:]
..
State two requirements for the formation of a coordinate covalent bond.
i.
The donor atom must have at least a lone pair of electrons
ii.
The acceptor atom must have an empty orbital to accommodate
[2]/5g
the pair of electrons.
(b)
Draw Lewis structures to illustrate the formation of NH4+ from H+ and ammonia, NH3
and show which of the bond the coordinate covalent bond is.
H+
:N
H
H
+
H
H
H
N
[2]/5g
H
H
coordinate covalent bond
CHE 0315
(c)
SEM 3, 2013/14
Draw Lewis structures to illustrate the formation of BF3NH3 from ammonia, NH3
and boron trifluoride, BF3 and show which of the bond the coordinate covalent bond is.
[2]/6g
(a)
H
F
F
B
B
N
F coordinateHcovalent bond
Draw the Lewis structures of the following compounds.
(i)
ICl2-
..
: Cl
..
(ii)
2-
..
..
..
Cl :
I
..
NO2
..
..O
(iii)
..
..
O.
N
..
BBr3
..
: Br :
..
: Br
..
(iv)
..
Br :
..
B
PO43-
..
: O:
..
:O
..
P
3..
..O:
:O :
(v)
BeCl2
..
: Cl
..
(vi)
H
F
..
6.
N
..
H
..
+
H
F
F
H
BCl3
Be
..
..
Cl :
[6]/5c
CHE 0315
SEM 3, 2013/14
..
: Cl :
..
: Cl
..
(b)
7.
(a)
..
..
Cl :
B
State the type of octet rule exception for the following species.
(i)
ICl2-
-Expanded octet
(ii)
NO2
-Odd number of electrons
(iii)
BBr3
-Incomplete octet
(iv)
PO4
-Expanded octet
(v)
BeCl2 -Incomplete octet
(vi)
BCl3
3-
[6]/5l
-Incomplete octet
Rank the bonds in each set in order of increasing bond length and bond energy.
[6]/5h
Compounds
Bond length
Bond energy
C-I, C-F, C-Br, C-Cl
C–F < C–Cl < C–Br < C-I [1]
C-I < C-Br < C-Cl < C-F [1]
C-S, C=O, C-O
C=O < C–O < C–S [1]
C–S < C–O < C=O [1]
N-H, N-S, N-O
N–H < N–O < N–S
N–S < N–O < N–H [1]
(b)
[1]
HI decomposes easily when heated to red heat but HCl does not. State the reason.
[1]/5h
This is due to weaker H-I bond. HCl has stronger bond because HCl bond is
shorter than HI bond.
Draw the resonance structures of:
[2]/5n
O
..
(i)
[3]/5n
N
..
N:
..
-
..
N
..
N
..
N
..
Draw the resonance structure of CO2.
-
..
:
..
(a)
O
N3-
:N
9.
O
N
N
:
(b)
O
..
..
O
:
..
..
..
:O
:
O3
..
..
(a)
:
8.
N
[3]/5n
SEM 3, 2013/14
(ii)
C
..
O
..
..
..
O:
..
:
..
:O
..
..
C
O
O
C
O
Choose the best resonance structures. Explain.
..
O
..
C
:
CHE 0315
[2]/5k
..
.. is the best resonance structure because every atom’s
O
formal charge is zero.
10.
CH4, NH3 and H2O have three electron domains respectively; theoretically all three of
the three molecules will have the same bond angle. But in reality, bond angles of
CH4, NH3 and H2O are 120°, 107.3° and 104.5° respectively. Explain the
observation.
[3]/5o
These three molecules posses different angles because of the different
electron-pair repulsion.[1] CH4 has the weakest electron pair repulsion i.e.
bonding pair repulsion[1] followed by NH3 and the strongest repulsion in
H2O i.e. lone pair repulsion.[1]
11.
State the shape and bond angle of the following molecules and ions by using VSEPR
theory.
(a)
AsH3
[6]/5p
Trigonal pyramidal, 109.5°
H
As
..
H
(b)
I3-
Linear, 180°
.. ..
I
..
..
:I
..
(c)
C2H4
H
..
..I :
Trigonal planar, 120°
H
H
H
C
C
H
CHE 0315
SEM 3, 2013/14
(d)
OF2
Bent, 109.5°
..
:F
..
(e)
SbF5
..
..
O
..
F
.. :
Trigonal bipyramidal, 90° & 120°
..
:F:
..
: ..F
:F :
..
(f)
XeO2F2
..
:
..F
Sb
: F:
..
Seesaw, 90° & 120°
..
:F:
..
O
..
.Xe
.
..
..O :
..
: :
F
12.
(a)
For each of the bonds below:
C—Cl
(i)
N—O
H—O
Use delta notation (δ+ and δ–) to indicate which atom is more electronegative.
[3]/5j
(ii)
C—Cl
N—O
H—O
δ+— δ–
δ+ — δ–
δ+ — δ–
Use the polar arrow to show the polarity of the bond.
C—Cl
(c)
N—O
[3]/5j
H—O
Identify the type of bond described for each of the following as polar covalent,
non-polar covalent, or metallic.
[4]/5j
(i)
The C–O bonds in CO2
Polar covalent
(ii)
The bonds in F2
Non-polar covalent
(iii)
The bonds in Ba
Metallic bond
CHE 0315
SEM 3, 2013/14
(iv)
13.
The bonds in H2O
Polar covalent
A molecule of formula AY3 is found experimentally to be polar. State the possible shapes for
AY3.
[3]/5j
The molecular shapes that have a central atom bonded to three other atoms
are
trigonal
planar,
trigonal
pyramid
and
T-shaped.
The
trigonal
molecules are nonpolar, so it cannot be AY3. Trigonal pyramidal molecules
planar
and T-
shaped molecules are polar, so either could represent AY3.
12.
Identify the statements as TRUE or FALSE. Correct any those are false.
(a)
Two  bonds comprise a double bond.
[1]/5u
False, a double bond is one sigma (σ) and one pi (π) bond.
(b)
Bonds formed from atomic s orbitals are always  bonds.
[1]/5u
True
(c)
A π bond consists of two pairs of electrons.
[1]/5u
False, a π bond consists of one pair of electrons; it occurs after a σ
bond has been previously formed.
13.
(a)
Describe the hybridization process of the central atom in AlI3.
[2]/5s
[1]
Al
Al
[1]
Al
(b)
Draw the orbital overlapping diagram of AlI3.
σ
σ
σ
[3]/5s
CHE 0315
SEM 3, 2013/14
[1] correct shape of sp2
[1]sp2 labelling
[1] σ labelling
14.
In each of the following equations, determine the type of hybridization of the underlined
atom as a reactant and as a product.
(a)
BF3 + NaF
Na+ + BF4-
B changes from sp
2
[2]/5t
to sp . Boron in BF3 has three electron groups with
3
sp2 hybridization.[1] In BF4–, 4 electron groups surround B with sp3
hybridization.[1]
(b)
SO2 + ½ O2
SO3
[2]/5t
No change, S in SO2 is surrounded by three electron groups [1](one single
bond, one double bond and one lone pair) and S in SO3 is also surrounded by
3 electron groups [1](two single bonds and one double bond); both have
sp2 hybridization.