2010 Chem Bond Tut Ans All

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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



(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



(e) POCl2

+

Trigonal planar

(f)* ClF4–

Square planar

3

P Cl O

Cl

+

POCl

Cl

+

Cl F F

F

– FF

ClF

F F

–

7



(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



(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



(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

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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

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(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

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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


NCl3 is polar

H S H

N Cl

Cl

Cl

HH

S

ClN

ClCl

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(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


N2H4 is polar

H

H N

H

N HN

H H N

HH

C Cl

H

H

H

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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

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(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

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*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-

??

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(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

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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)

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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


(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


(k) CH3OH covalent simple molecular intermolecular hydrogen bonds

(l) CH3F covalent simple molecular intermolecular van der Waals’ forces of attraction


(m)* FeSO4 ionic giant ionic between oppositely charged ions

(n) * (CH3)3N covalent simple molecular intermolecular van der Waals’ forces of attraction


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*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

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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

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*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

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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

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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

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(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

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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

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*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

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2010 Chem Bond Tut Ans All