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2010 DHS Year 5 H2 Chemistry
TUTORIAL 8 : CHEMICAL BONDING 1 Describe the formation of each of the following chemical bonds with appropriate reference
to calcium, sodium oxide and hydrogen chloride: (a) ionic bond (b) covalent bond (c) metallic bond Draw dot–and–cross diagrams to illustrate the bonds in (a) and (b), showing only the outer (valence) electrons for each atom.
1(a) Ionic bonds are formed in Na2O. To achieve stable noble gas configuration, atoms of Na lose one valence electron each to form positively
charged Na+ cations
atoms of O gain two donated electrons each to form negatively charged O2– anions
electrostatic forces of attraction exist between the oppositely charged Na+ and O2– ions in a giant ionic lattice structure
+2 Na
–
O
2
(b) Covalent bonds are formed in HCl. To achieve stable noble gas configuration, atoms of H and Cl share one valence electron each to form
HCl molecules
electrostatic forces of attraction exist between the nuclei and the shared pair of electrons between H and Cl atoms in a simple molecular structure
(c) Metallic bonds are formed in Ca when: atoms of Ca lose valence electrons to form positively
charged Ca2+ cations valence electrons are loosely held and delocalised, i.e. free
to move within lattice electrostatic forces of attraction exist between the Ca2+
cations and sea of delocalised electrons in a giant metallic lattice structure
H Cl
3
*2 W, X, Y and Z represent elements of atomic numbers 9, 13, 19 and 34 respectively. State the formula and the type of bonding for the compound which you expect to occur between
(a) W and X (b) W and Y (c) W and Z (d) Y and Z (e) X and Z
9W: Grp VII (non–metal) 13X: Grp III (metal) 19Y: Grp I (metal) 34Z: Grp VI (non–metal) (a) Compound: XW3 (ionic bonds present) (b) Compound: YW (ionic bonds present) (c) Compound: ZW2 (covalent bonds present) (d) Compound: Y2Z (ionic bonds present) (e) Compound: X2Z3 (ionic bonds present)
3 Use the following table format for each of the following cases to draw the dot–and–cross
diagram, describe the shape (by drawing the Lewis structure and stating the shape) and state the polarity of each molecule (where appropriate):
Molecule/ion Dot–and–cross
diagram
Diagram of shape (Lewis
structure)
Shape Polarity (polar/non–
polar)
(a) CO
(a)* CH3
+ (h)* CH3–
(b) SCl2 (i) SbF6–
(c) AsCl3 (j)* N2O4 (d)* ClO3
– (k)* NO2 (e) POCl2
+ (l) * C2H6 (f)* ClF4
– (m)* C2H2
4
SpeciesDot–and–Cross
Diagram Lewis Structure Shape Polar?
(a)* CH3+
Trigonal planar
(b) SCl2
Bent Polar
3
S
ClCl ClCl
S
C
H
H H+
CHH
H +
5
SpeciesDot–and–Cross
Diagram Lewis Structure Shape Polar?
(c) AsCl3
Trigonal Pyramidal
Polar
(d)* ClO3
–
Trigonal Pyramidal
3
As Cl Cl
Cl
ClO O
O
––
Cl O O
O
As
ClCl
Cl
6
SpeciesDot–and–Cross
Diagram Lewis Structure Shape Polar?
(e) POCl2
+
Trigonal planar
(f)* ClF4–
Square planar
3
P Cl O
Cl
+
POCl
Cl
+
Cl F F
F
– FF
ClF
F F
–
7
SpeciesDot–and–Cross
Diagram Lewis Structure Shape Polar?
(h)* CH3–
Trigonal pyramidal
(i) SbF6–
Octahedral
3
C
H
H H–
Sb F
–
F F F
F F
CH
HH
–
–
SbF
F
F
FF
F
8
SpeciesDot–and–Cross
Diagram Lewis Structure Shape Polar?
(j)* N2O4
N NOO
O O
Trigonal planar
about each N
non–polar
(k)* NO2
O
NO
Bent Polar
3
N O
N
O
OO
ONO
9
SpeciesDot–and–Cross
Diagram Lewis Structure Shape Polar?
(l)* C2H6
C C
H
H
HH
H
H
Tetrahedral about each
C
non–polar
(m)* C2H2
CH C H
Linear about each
C
non–polar
3
H
H C
H
H
C
H
H
H C C H
10
4 Draw the dot–and–cross diagram and Lewis structure for each of the following species and indicate their shapes with reasoning.
(a) NO3– and NCl3
NO3–
There are 3 bond pairs and no lone pair around N To minimise repulsion, the 3 electron pairs are directed to
the corners of an equilateral triangle NO3
– is trigonal planar NCl3
There are 3 bond pairs and 1 lone pair around N To minimise repulsion, the 4 electron pairs are directed to
the corners of a tetrahedron bond pair–lone pair repulsion > bond pair–bond pair repulsion NCl3 is trigonal pyramidal
N
O
O O
N Cl
Cl
Cl
NO
OO
– –
ClN
ClCl
11
(b)* POCl3 and IOCl3
POCl3
There are 4 bond pairs and no lone pair around P To minimise repulsion, the 4 electron pairs are directed to
the corners of a tetrahedron POCl3 is tetrahedral
IOCl3
There are 4 bond pairs and 1 lone pair around I To minimise repulsion, the 5 electron pairs are directed to
the corners of a trigonal bipyramid bond pair–lone pair repulsion > bond pair–bond pair repulsion IOCl3 is distorted tetrahedral
P Cl
Cl
O
Cl
I Cl
Cl
O
Cl
P
Cl
ClClO
I
ClCl
ClO
12
5 Draw the dot–and–cross diagram and Lewis structure for each of the following molecules and indicate whether they are polar or not.
(a)* H2S and NCl3
H2S
There are 2 bond pairs and 2 lone pairs around S, so H2S is bent
Dipole moments associated with the polar bonds (S–―H+) do
not exactly cancel out, giving rise to a non–zero overall dipole moment
H2S is polar NCl3
There are 3 bond pairs and 1 lone pair around N, so NCl3 is trigonal pyramidal
Dipole moments associated with the polar bonds (N–―H+) do
not exactly cancel out, giving rise to a non–zero overall dipole moment
NCl3 is polar
H S H
N Cl
Cl
Cl
HH
S
ClN
ClCl
13
(b) CH3Cl and N2H4
CH3Cl
There are 4 bond pairs and no lone pair around C, so CH3Cl
is tetrahedral Dipole moments associated with the polar bonds (Cl–―C+)
do not exactly cancel out, giving rise to a non–zero overall dipole moment
CH3Cl is polar
N2H4
There are 3 bond pairs and 1 lone pair around each N, so
N2H4 is trigonal pyramidal about each N Dipole moments associated with the polar bonds (N–―H+) do
not exactly cancel out, giving rise to a non–zero overall dipole moment
N2H4 is polar
H
H N
H
N HN
H H N
HH
C Cl
H
H
H
14
6 Explain the following observation as fully as you can with the aid of dot–and–cross diagrams and Lewis structures (where appropriate):
(a)* SF4 exist but not OF4
SF4 OF4
For SF4 and OF4 to exist, There would be 4 bond pairs and 1 lone pair around S or O 10 electrons surrounding central atom S or O S, being an element in period 3, has availability of vacant
low–lying d orbitals to accommodate electrons and therefore can expand beyond octet structure
SF4 exists O, being an element in period 2, has absence of vacant low–
lying d orbitals to accommodate electrons and therefore cannot expand beyond octet structure
OF4 does not exist
S F
F
F
F
O F
F
F
F
15
(b)* NO2+ is linear but not NO2
–
NO2+
There are 2 bond pairs and no lone pair around N To minimise repulsion, the 2 electron pairs are directed
opposite to each other NO2
+ is linear NO2
–
There are 2 bond pairs and 1 lone pair around N To minimise repulsion, the 3 electron pairs are directed to
the corners of an equilateral triangle bond pair–lone pair repulsion > bond pair–bond pair repulsion NO2
– is bent
N O O
+
N O O
–
N O O
+
O
ON
–
16
(c) PCl3 has a smaller bond angle than CCl4
PCl3
There are 3 bond pairs and 1 lone pair around P To minimise repulsion, the 4 electron pairs are directed to
the corners of a tetrahedron bond pair–lone pair repulsion > bond pair–bond pair repulsion PCl3 is trigonal pyramidal with bond angle of 107 CCl4
There are 4 bond pairs and 0 lone pair around C To minimise repulsion, the 4 electron pairs are directed to
the corners of a tetrahedron CCl4 is tetrahedral with bond angle of 109.5
P Cl
Cl
Cl
C Cl
Cl
Cl Cl
C
Cl
ClClCl
ClP
ClCl
17
(d) SO2 is polar but not XeF4
SO2
There are 2 bond pairs and 1 lone pair around S, so SO2 is
bent Dipole moments associated with the polar bonds (O–═S+) do
not cancel out, 6to a non–zero overall dipole moment SO2 is polar
XeF4
There are 4 bond pairs and 2 lone pairs around Xe, so XeF4 is square planar
Dipole moments associated with the opposite polar bonds
(F–―Xe+) exactly cancel out, giving rise to a zero overall dipole moment
XeF4 is non–polar
S OO
Xe
F F
F F
OOS
XeF
F F
F
18
*7 Draw the Lewis structure for each of the following ions based on the given shape. By considering the number of bond pairs and lone pairs of electrons around the central atom, deduce the total number of electrons around the central atom, charge of ion and oxidation number of the central atom:
(a) WF2a+ (bent shape) where W is an element in Group V
Electron pairs around W: 2 bond pairs and 1 lone pair
Charge of ion = 5 – 4 = +1
Oxidation number of W = 1 – 2(–1) = +3 (b) XH2
b– (linear shape) where X is an element in Group VII
Electron pairs around X in ion: 2 bond pairs and 3 lone pairs
Charge of ion = 7 – 8 = –1
Oxidation number of X = –1 – 2(1) = –3
WF
F
a+
X H H b–
Why not
W
FF
a+
??
Why not
XH Hb-
??
19
(c) ZCl3d– (T–shape) where Z is an element in Group VI
Electron pairs around Z in ion: 3 bond pairs and 2 lone pairs Charge of ion = 6 – 7 = –1 Oxidation number of Z = –1 – 3(–1) = +2 8 State the conditions necessary for high degree of covalent character in ionic bonds and high degree of
ionic character in covalent bonds.
Using your knowledge of ionic radius and electronegativity, suggest with reasoning, each of the following cases:
(a) *the compounds with the smallest and highest degree of covalent character in their ionic bonds NaCl, Na2S, AlBr3 and Al2O3
High degree of covalent character in an ionic bond arises from a high degree of polarisation when cation has high positive charge and small ionic radius anion has high negative charge and large ionic radius High degree of ionic character in a covalent bond arises from a high degree of electron displacement when covalently bonded atoms have a large electronegativity
difference between them For NaCl, Na2S, AlBr3 and Al2O3 : o charge: Na+ < Al3+ Cl– < S2– Br
– < O2–
Z
Cl
Cl
Cl
d–
20
ionic radius: Na+ > Al3+ Cl– S2– Br– > O2–
compound with lowest degree of covalent character in ionic
bond: NaCl compound with highest degree of covalent character in ionic
bond: AlBr3
Extra Practice: FeBr3, Fe2S3, BaCl2 and BaO
charge: Fe3+ > Ba2+ Br– < S2– Cl– < O2– ionic radius: Fe3+ < Ba2+ Br– > S2– Cl– > O2– compound with lowest degree of covalent character in ionic
bond: BaO compound with highest degree of covalent character in ionic
bond: FeBr3
(b) the compounds with the smallest and highest degree of ionic character in their covalent bonds HCl, Cl2 and ICl
electronegativity: Cl > I > H electronegativity difference: Cl & Cl < I & Cl < H & Cl compound with lowest degree of ionic character in covalent
bond: Cl2 compound with highest degree of ionic character in covalent
bond: HCl
Smallest diff in electronegativity
Largest diff in electronegativity
21
Extra Practice: CO2, CCl4 and CH4
electronegativity: O > Cl > C > H electronegativity difference: C & H < C & Cl < C & O compound with lowest degree of ionic character in covalent
bond: CH4 compound with highest degree of ionic character in covalent
bond: CO2 9 There are three main types of chemical bond (i.e. ionic, covalent and metallic) and six main types
of chemical structure (i.e. giant metallic, giant ionic, giant molecular, giant molecular layered, simple molecular with intermolecular van der Waals’ forces of attraction, simple molecular with intermolecular hydrogen bonds).
Use the following table format to name the type of chemical bond and structure in each of the following substances (do not consider aqueous state) and state the nature of electrostatic forces of attraction (eg. between oppositely charged ions) that must be overcome during melting/boiling.
Substance Chemical bond Chemical structure Electrostatic forces of attraction overcome during melting/boiling
(a) Mg
(a)* Mg (h)* C(graphite) (b) S8 (i)* HCl (c) Si (j) P4 (d)* H2O (k) CH3OH (e)* SiO2 (l) CH3F (f)* SiCl4 (m)* FeSO4 (g)* Ar (n)* (CH3)3N
Subs-
tance Chemical bond
Chemical structure Electrostatic forces of attraction overcome during melting/boiling
(a) * Mg metallic giant metallic between cations and sea of delocalised electrons
(b) S8 covalent simple molecular intermolecular van der Waals’ forces of attraction
(temporary dipole–dipole interactions)
(c) Si covalent giant molecular between nuclei of atoms and shared pair of electrons
(covalent bonds)
22
Subs-tance
Chemical bond
Chemical structure Electrostatic forces of attraction overcome during melting/boiling
(d) * H2O covalent simple molecular intermolecular hydrogen bonds
(e) * SiO2 covalent giant molecular between nuclei of atoms and shared pair of electrons
(covalent bonds)
(f) * SiCl4 covalent simple molecular intermolecular van der Waals’ forces of attraction
(temporary dipole–dipole interactions)
(g) * Ar ― simple molecular van der Waals’ forces of attraction between atoms (temporary dipole–dipole
interactions)
(h) * C graphite
covalent giant molecular layered
between nuclei of atoms and shared pair of electrons
(covalent bonds)
(i) * HCl covalent simple molecular intermolecular van der Waals’ forces of attraction
(permanent dipole–dipole interactions)
(j) P4 covalent simple molecular intermolecular van der Waals’ forces of attraction
(temporary dipole–dipole interactions)
(k) CH3OH covalent simple molecular intermolecular hydrogen bonds
(l) CH3F covalent simple molecular intermolecular van der Waals’ forces of attraction
(permanent dipole–dipole interactions)
(m)* FeSO4 ionic giant ionic between oppositely charged ions
(n) * (CH3)3N covalent simple molecular intermolecular van der Waals’ forces of attraction
(permanent dipole–dipole interactions)
23
*10 Aluminium fluoride (m.p. = 1291oC) and phosphorous trifluoride (m.p. = –151oC) have very different
physical properties.
(a) Draw the dot–and–cross diagrams to illustrate the bonding of these compounds, and hence explain the difference in their melting points in terms of chemical bonding and structure.
Aluminium fluoride
has a giant ionic structure
It has a high melting point as a large amount of energy is needed to overcome the strong electrostatic forces of attraction between oppositely charged ions during melting
Phosphorus trifluoride
has a simple molecular structure
It has a low melting point as a small amount of energy is needed to overcome the weak intermolecular van der Waals’ forces of attraction during melting
3+3 Al
–F
P F F
F
24
10(b) Suggest one further physical property in which they might differ, specifying the difference in the property.
Aluminium fluoride is expected to be a good conductor of electricity in the molten or aqueous state but not phosphorus trifluoride (c) A 0.450 g sample of gaseous aluminium chloride takes up a volume of 51.2 cm3 at a temperature of
100oC and a pressure of 1.02 105 Pa. By applying the ideal gas equation, what do these data suggest about the molecular state of aluminium chloride at this temperature? Draw the suggested structure.
(c) Using pV = M
mRT M = 65 1051.2101.02
3738.310.450
= 267 g mol–1
Aluminium chloride
has an apparent Mr which is twice its expected theoretical Mr of 133.5
exists as dimers through dative bonds under the given
conditions
AlCl
Cl
ClAl
Cl
Cl Cl
25
*11 Nitrogen and boron combine to form boron nitride, with empirical formula BN, which has a graphite–like structure.
(a) Draw the structure of boron nitride in which the boron and nitrogen atoms alternate in a graphite–like structure.
Boron nitride has graphite–like structure with alternate B and N atoms.
(b) By considering the electron distribution of your above structure, suggest with reasoning the likely electrical conductivity of boron nitride.
non–conductor of electricity when perpendicular and parallel to layers: absence of free mobile ions or delocalised electrons to conduct electricity
Note: Lone pair electrons on N are not able to delocalise due to high electronegativity. Hence cannot conduct electricity. Compared to graphite, the non-bonding electrons of C in graphite can delocalise, hence able to conduct electricity parallel to layers. (c) Suggest one probable industrial application of boron nitride.
It could be used as a machinery lubricant.
B N
N B
B N
B N
B
N
B
B
N
N
B
N
N B
B N
N B
N B
N
B
N
N
B
B
N
B
Only need to draw 2 plates per layer (Those i
26
12 Explain each of the following observations as fully as you can in terms of chemical bonding and structure:
(a)* ammonia has a higher boiling point than phosphine (PH3)
Phosphine
has a simple molecular structure It has a low boiling point as a small amount of energy is
needed to overcome the weak intermolecular van der Waals’ forces of attraction during boiling
Ammonia
has a simple molecular structure It has a higher boiling point as a larger amount of energy is
needed to overcome the relatively stronger intermolecular hydrogen bonds during boiling
(b) diamond is a poor conductor of electricity but not graphite
Diamond has a giant molecular structure
non–conductor of electricity: absence of free mobile ions or delocalised electrons to conduct electricity
Graphite has a giant molecular layered structure
non–conductor of electricity when perpendicular to layers: absence of free mobile ions or delocalised electrons to conduct electricity
good conductor of electricity when parallel to layers: the non–bonding valence electrons of the carbon atoms are delocalised over each plane of layers to conduct electricity
(c)* aluminium conducts electricity both in the solid and liquid states
Aluminium has a giant metallic structure
27
good conductor of electricity in both solid and liquid states: presence of delocalised electrons which are free and mobile to conduct electricity
(d) lithium fluoride (LiF) is a non–conducting solid at room temperature which can undergo melting to form
a liquid that conducts electricity
Lithium fluoride has a giant ionic structure
non–conductor of electricity in solid state: ions can only vibrate about fixed positions and are not mobile to conduct electricity
good conductor of electricity in molten state: presence of free mobile ions to conduct electricity
(e)* sodium chloride is brittle but sodium is malleable
Sodium chloride has a giant ionic structure
brittle: a stress applied on an ionic lattice causes sliding of layers of ions; ions of similar charges come together and resultant repulsion shatters the ionic structure
Sodium has a giant metallic structure
malleable (i.e. easily bent): a stress applied on a metallic lattice causes mere sliding of layers of cations in a sea of delocalised electrons without breaking the metallic bonds; the lattice structure is not shattered
28
(f) carbon dioxide is a gas at room temperature but silicon dioxide is a solid with high melting point
Carbon dioxide
has a simple molecular structure has a low boiling point as a small amount of energy is
needed to overcome the weak intermolecular van der Waals’ forces of attraction (temporary dipole-dipole interactions) during boiling, so exists as a gas at room temperature
At room temperature, there is sufficient energy to overcome these intermolecular forces.
Silicon dioxide
has a giant molecular structure has a high melting point: a large amount of energy is needed
to overcome the strong covalent bonds between atoms in a giant extensive network during melting
At room temperature, there is insufficient energy to overcome these bonds.
(g)* ethanol (CH3CH2OH) is miscible in water but bromoethane (CH3CH2Br) is not
Ethanol
has a simple molecular structure soluble in water: favourable hydrogen bonds between
ethanol and water molecules are formed in the solvation process
Bromoethane
has a simple molecular structure with intermolecular van der Waals’ forces of attraction
insoluble in water: no favourable interactions between bromoethane and water molecules can be formed as the weak intermolecular van der Waals’ forces of attraction in
29
bromoethane are not able to displace the strong intermolecular hydrogen bonds in water for solvation/hydration to occur
(h)* butane (CH3CH2CH2CH3) has a higher boiling point than 2–methylpropane (CH(CH3)3)
Both compounds are simple molecular with intermolecular van
der Waals’ forces of attraction Butane is a straight chain alkane while 2–methylpropane has a
branched chain However,
Extent of surface area of contact:: Butane > 2–methylpropane
Strength of intermolecular forces of attraction: Butane > 2–methylpropane
Energy required to overcome these forces: Butane > 2–methylpropane
Hence, boiling point of butane is higher than that of 2–methylpropane
30
(i) butane (CH3CH2CH2CH3) has a lower boiling point than butanol (CH3CH2CH2CH2OH)
Butane
has a simple molecular structure has a low boiling point as a small amount of energy is
needed to overcome the weak intermolecular van der Waals’ forces of attraction (temporary dipole-dipole interactions) during boiling
Butanol
has a simple molecular structure has a higher boiling point as a larger amount of energy is
needed to overcome the relatively stronger intermolecular hydrogen bonds during boiling
*13 Explain each of the following observations on physical property in terms of chemical bonding
and structure:
Substance Solubility in water methanol (CH3OH) soluble
sodium chloride soluble Iodine insoluble
Methanol has a simple molecular structure soluble in water: favourable hydrogen bonds between
methanol and water molecules are formed in the solvation process
Sodium chloride has a giant ionic structure soluble in water: favourable ion–dipole interactions
between ions and water molecules are formed and this process releases energy which facilitates the detachment of ions from the ionic lattice for hydration to occur
(a)
31
Iodine has a simple molecular structure insoluble in water: no favourable interactions between
iodine and water molecules can be formed as the weak intermolecular van der Waals’ forces of attraction in iodine are not able to displace the strong intermolecular hydrogen bonds in water for solvation to occur
Substance Electrical conductivity
copper high magnesium chloride nil in solid state
high in molten state phosphorous nil
silicon carbide (SiC) nil
Copper has a giant metallic structure good conductor of electricity in both solid and liquid states:
presence of delocalised electrons which are free and mobile to conduct electricity
Magnesium chloride has a giant ionic structure non–conductor of electricity in solid state: ions can only
vibrate about fixed positions and are not mobile to conduct electricity
good conductor of electricity in molten state: ions are free and mobile to conduct electricity
Phosphorus has a simple molecular structure non–conductor of electricity: absence of free mobile ions
or delocalised electrons to conduct electricity
Silicon carbide has a giant molecular structure non–conductor of electricity: absence of free mobile ions
or delocalised electrons to conduct electricity
13(b)
32
*14 A student commented that “sodium chloride (NaCl) has a very high melting point while tetrachloromethane (CCl4) boils below 100 C, so ionic bonds are much stronger than covalent bonds.”
Comment on the validity of the above statement using your knowledge of chemical bonding and structure.
Sodium chloride has a giant ionic structure has a high melting point: a large amount of energy is needed
to overcome the strong ionic bonds (electrostatic forces of attraction between oppositely charged ions) during melting
Tetrachloromethane has a simple molecular structure has a low boiling point: a small amount of energy is needed
to overcome the weak intermolecular van der Waals’ forces of attraction during boiling
boiling a simple molecular substance does not involve
supplying energy to overcome the covalent bonds between atoms within the molecules, so the comment that “ionic bonds are much stronger than covalent bonds” is invalid
33
*15 With reference to the type and extent of relevant intermolecular forces, explain as fully as you can the expected order of the following substances in increasing boiling point:
hydrogen (H2), fluorine (F2), hydrogen fluoride (HF), sodium fluoride (NaF) and chlorofluoride (ClF).
Hydrogen, fluorine
have simple molecular structures with weak intermolecular van der Waals’ forces of attraction due to temporary dipole-dipole interactions
However, Relative molecular mass and hence no. of electrons:H2 < F2 Strength of intermolecular forces of attraction: H2 < F2 Energy required to overcome these forces: H2 < F2 chlorofluoride
Has a simple molecular structure with stronger intermolecular van der Waals’ forces of attraction due to permanent dipole-dipole interactions
Large amount of energy is thus needed during boiling Hydrogen fluoride has a simple molecular structure with relatively stronger
intermolecular hydrogen bonds Larger amount of energy is needed during boiling Sodium fluoride has a giant ionic structure
has the highest boiling point: a very large amount of energy is needed to overcome the very strong ionic bonds (electrostatic forces of attraction between oppositely charged ions) during boiling
Expected order in increasing boiling point:
H2 < F2 < ClF < HF < NaF
34
*16 The equation below shows the reaction between boron trifluoride and a fluoride ion.
BF3 + F– BF4–
(a) Draw diagrams to show the shape of the BF3 molecule and the shape of the BF4–ion. In each case,
name the shape. Account for the shape of the BF4– ion and state the bond angle present.
Trigonal planar Tetrahedral
For BF4
-
There are 4 bond pairs and 0 lone pair around B
To minimise repulsion, the 4 electron pairs are directed to the corners of a tetrahedron
BF4- has a bond angle of 109.5
(b) In terms of the electrons involved, explain how the bond between the BF3 molecule and the F– ion is
formed. Name the type of bond formed in this reaction.
B atom in BF3 is electron-deficient (only 6 electrons around it),
hence it accepts one lone pair electrons from F– ion via a dative bond and thus achieve stable octet configuration
B
F
FF
F
–
B FF
F
B
F
F F
F–
35
Multiple–Choice Questions 1.
2.
3.
4.
5.
6.
7.
36
8.
9.
10.
11.
12.
13.
14.
15.
37
16.
17.
18.
19.
20.
21.
22.
Answers : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
D B D A A B D D D D D B C A D D D C C D A C