Section 4b Bonding And Structure Ii (covalent Bonding)

Modern College AL Chemistry Notes (2009 – 10)

Section 4B

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Modern College AL Chemistry Notes (2009 – 10)

Prepared by Mr. Chau Chi Keung, Richard

Section 4B

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Modern College AL Chemistry Notes (2009 – 10)

Section 4B

4.5 Covalent Bonding Revisited 4.5.1. Formation of covalent bond – Overlap of atomic orbitals 

In simple molecules, electrons might be shared in pairs so that each atom could attain the stable noble gas electronic configuration.

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Strong, directional electrostatic attraction between the nuclei of the bonding atoms and the shared pair of electrons (bond pair).

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There exists an overlap between the atomic orbitals of the bonding atoms. Take hydrogen molecule (H2) as an example:

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In a hydrogen molecule, each 1s orbital containing an unpaired electron can overlap an orbital of another atom containing an unpaired electron to form a covalent bond.

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The overlapping of atomic orbitals results in overlapping of electron clouds which permits the exchange of electrons between the atoms.

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The overlapping of electron clouds leads to a greater electron density between the two nuclei, indicating that the electrons are localized there with greater probability (Figure (a)).

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The electrostatic attraction arises from the electrostatic attractive forces between bonding electrons and nuclei which outweighs the electrostatic repulsive forces between two nuclei and two electrons. The overall potential energy of the system thus decreases (Figure (b)).

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The energy drop is called the bond dissociation enthalpy ( 鍵 離 解 焓 ). It is the energy Prepared by Mr. Chau Chi Keung, Richard

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Modern College AL Chemistry Notes (2009 – 10)

Section 4B

required to break the bond (+436 kJmol–1 for H2 molecule). 

Besides, the bond length is the equilibrium distance between two nuclei of the two bonding atoms. Its value also depends on the balance of the repulsion and the attraction (0.104 nm for H2 molecule.)

(a)  

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(b)

Helium atom has the electronic configuration 1s2. The two valence electrons of helium atom fill its valence orbital (1s orbital). The completely filled orbital of a helium atom cannot be occupied by the valence electrons of the other helium atom because each orbital can accommodate a maximum of two electrons only; otherwise, it will violate Pauli’s Exclusion Principle. Therefore, helium forms no covalent bonds and is a monoatomic molecule.

Example 1 Write the structural formulae of the following compounds, showing the lone pair electrons of the central atom (if any). Hydrogen fluoride (HF) Water (H2O) Ammonia (NH3)

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Modern College AL Chemistry Notes (2009 – 10) Boron trifluoride (BF3)

Phosphorus pentachloride (PCl5)

Section 4B Sulphur hexafluoride (SF6)

4.5.2. Octet rule and its limitations 

 

Octet Rule (八隅體規則): In forming chemical bonds, atoms tend to achieve the stable noble gas electronic configuration with 8 electrons in the valence shell (價電子層)(except helium which only has 2 electrons in the valence shell) by gaining, losing or sharing of electrons. However, the phosphorus atom in PCl5 is actually surrounded by 10 electrons and the sulphur atom in SF6 is surrounded by 12 electrons. P and S can “violate the octet rule” because their valence shell is the electron shell of n = 3, which can contain 18 electrons at most. P and S have low-lying vacant d-orbitals so that further bond formation can be  allowed (i.e. Their valence shell can contain more than 8 electrons) (P 和 S 具有 低 能 階和沒有填滿的 d 軌態，所以在形成化學鍵時，它們的價電子層可以容納超過 8 粒電子。) Take P as an example: Before 3s

3p

3d

After

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 



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3s 3p 3d Phosphorus can expand its octet by promoting its 3s electrons to the low-lying vacant d-orbitals in order to form covalent bonds with 5 chlorine atoms. Hence, phosphorus can form two chlorides – PCl3 and PCl5. In contrast, since nitrogen does not have those low-lying vacant d-orbitals, it only has a maximum of three half-filled p orbitals to form three bonds. Hence, nitrogen is only able to form one chloride – NCl3.

There also exist species which violate the octet rule due to an “electron deficiency” Example: Boron trifluoride (BF3)  The central boron atom only has 6 electrons (In this case, that B atom is said to be  Prepared by Mr. Chau Chi Keung, Richard

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Modern College AL Chemistry Notes (2009 – 10)

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Section 4B

“electron deficient”). Since the atom size of boron is small, the I.E. required for B to form a cation (B3+) will be very high. Therefore BF3 is a covalent compound rather than being ionic.

4.6 Dative Covalent Bonding (配位共價鍵)  

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In some cases, one atom will donate both the electrons involved in the covalent bond formation between two atoms. This type of covalent bond is called dative covalent bond. A dative covalent bond is formed by the overlapping of an empty orbital of an atom with an orbital occupied by a lone pair of electrons of another atom (配 位 共 價 鍵 是 由 一 個 原子的空軌態，與另一個原子中被一對孤偶電子佔據的軌態重疊而形成的。). The atom that supplies the shared pair of electrons is known as the donor (給予體) while the other atom involved in the dative covalent bond is known as the acceptor (受體). Surely, the donor must have at least one lone pair of electrons for bond formation while the acceptor must have an empty orbital to take up that pair of electrons. Dative covalent bond and covalent bond are identical once they have been formed. There is no real distinction between them except that in dative covalent bond only one of the bonded atoms provides the two electrons which are shared. Dative covalent bond is sometimes designated by drawing an arrow from the donor to the acceptor atom. Example 1: NH3BX3 molecule (X = F or Cl) 

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Example 2: Ammonium ion (NH4+)

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Modern College AL Chemistry Notes (2009 – 10)

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Section 4B

Example 3: Hydroxonium ion (H3O+)

Aluminium chloride (AlCl3) and nitric acid are another two noteworthy examples of dative covalent bond. Similar to its Group III counterpart boron, the atomic size of aluminium is relatively  small. The I.E. required for Al to form a cation of +3 charge will be very high. Therefore  AlCl3 is a covalent compound rather than being ionic.

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Being in Period 3, Al also has low-lying vacant d-orbitals to accept the lone pair electrons from Cl. Therefore, dative covalent bonds can be formed and this explains why AlCl3 can form Al2Cl6 – a dimer (二聚物), particularly in liquid and gas phases.

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Modern College AL Chemistry Notes (2009 – 10)

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Section 4B

For nitric acid, the bonding will be like this:

Based on this structure, it is predictable that dative covalent bond also exists in nitrate ions (the detailed structure of nitrate ions will be discussed later in this part).

4.7 Energetics of Covalent Bonding 4.7.1. Bond enthalpy (鍵焓 ) 

Bond enthalpy is the energy associated with a chemical bond. When a chemical bond is broken or formed, a certain amount of energy is absorbed from or released to the surroundings.

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Take the complete combustion of methane as an example:

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In particular, the bond dissociation enthalpy (鍵 離 解 焓 ) is the energy required to break one mole of a particular covalent bond in molecules under standard conditions (∴It must have a positive value.) 

Take hydrogen molecule as an example: H2(g) → 2H(g) ♦

♦

ΔHθ = + 436 kJmol–1

It means that the bond dissociation enthalpy of the H–H bond in H2 molecule is 436 kJmol–1. Do not mix this up with the enthalpy change of atomization, which refers to the

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Modern College AL Chemistry Notes (2009 – 10)

Section 4B

energy required to produce one mole of gaseous atom. For a diatomic molecule, it is half of the value of the bond dissociation enthalpy. 

For hydrogen chloride, we have, H – Cl(g) → H(g) + Cl(g) ♦

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ΔHθ = + 431 kJmol–1

In this case, the bond dissociation enthalpy of the H–Cl bond in HCl molecule is 431 kJmol–1.

A methane molecule has 4 C–H bonds. However, the values of bond dissociation enthalpy of all these 4 C–H bonds are not identical with one another.

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CH4(g) → CH3(g) + H(g)

∆ Hθ = +422 kJ mol–1

CH3(g) → CH2(g) + H(g)

∆ Hθ = +480 kJ mol–1

CH2(g) → CH(g) + H(g)

∆ Hθ = +425 kJ mol–1

CH(g) → C(g) + H(g)

∆ Hθ = +335 kJ mol–1

For the complete dissociation (in other words, atomization of methane), we have, CH4(g) → C(g) + 4H(g)

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∆ Hθ = +1662 kJ mol–1

Divide this value by 4 (∵4 C–H bonds are broken), we have 1648 = +415 .5 kJmol–1 4

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The value obtained is called the average bond enthalpy (平均鍵焓) of the C–H bond (denoted as E(C–H))

In general, the average bond enthalpy is defined as the average of the bond dissociation enthalpies required to break a particular chemical bond (denoted as E(X–Y)). 

 

For example, E(C–H) is the average value of bond dissociation energies of C–H bond in all kinds of hydrocarbons. E(X–Y) can be deduced from the energetic data of many compounds. Do not mix up with the bond dissociation enthalpy, which refers specifically to a particular bond of a particular compound.

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By convention, “bond enthalpy” refers to “average bond enthalpy” if not specified.

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Some bond enthalpy values are shown in the table below: Bond C–C C=C H–H O–O O=O N–N N=N N≡N F–F Cl–Cl

Bond Enthalpy / kJmol–1 348 612 436 146 496 163 409 944 158 242

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Modern College AL Chemistry Notes (2009 – 10) Br–Br I–I C–H N–H O–H H–F H–Cl H–Br H–I

Section 4B

193 151 416 388 463 565 431 364 297

4.7.2. Estimation of average bond enthalpies 

Bond enthalpies can be determined by thermochemical methods. Thermochemical methods involve the calorimetric determination of the enthalpy change of formation, atomization and other processes.

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The thermochemical data obtained can then be used to calculate bond enthalpies by applying the Hess’s law.

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Reminder: Both the reactants and products involved in the bond enthalpy calculation must be in the gaseous state.

Example 2 Given that: Standard enthalpy change of formation of methane = –75 kJ mol–1 Standard enthalpy change of atomization of graphite = +718 kJ mol–1 Standard enthalpy change of atomization of hydrogen = +218 kJ mol–1 Construct an enthalpy cycle and determine the average bond enthalpy of the C–H bond. 4E (C–H) C(g) + 4H(g)

∆ Hθf [CH4(g)]

∆ Hθatom [C(graphite)] + 4∆ Hθatom [H2(g)]

C(graphite) + 2H2(g) 4E (C–H) = {∆ Hθatom [C(graphite)] + 4∆ Hθatom [H2(g)]} – ∆ Hθf [CH4(g)] ∴ E (C − H ) =

718 + 4( 218 ) − ( −75 ) = +416 kJmol 4

−1

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Modern College AL Chemistry Notes (2009 – 10) ♦

Section 4B

No. of H atom = No. of C–H bonds within the molecule. ♦

♦

The enthalpy change involved in the breakdown of methane molecule into its constituent atoms (i.e. 4E (C–H)) is also called the standard enthalpy change of atomization of methane (∆ Hθatom [CH4(g)]). More specifically, atomization of a compound means the breaking down of one mole of the gaseous compound into its constituent atoms in the gaseous state. For methane, we have: CH4(g) → C(g) + 4H(g)

∆ Hθatom [CH4(g)] = + 1665 kJmol–1

Example 3 Construct an enthalpy cycle and determine the bond enthalpy of the C–C bond. Given that: Standard enthalpy change of formation of ethane = – 85 kJ mol–1 Standard enthalpy change of atomization of graphite = + 718 kJ mol–1 Standard enthalpy change of atomization of hydrogen = + 218 kJ mol–1 Average bond enthalpy of C–H bond = + 416 kJ mol–1

(Answer: + 333 kJmol–1)

Example 4 Given that: Standard enthalpy change of formation of carbon dioxide = – 393 kJ mol–1 Standard enthalpy change of atomization of graphite = +718 kJ mol–1 Standard enthalpy change of atomization of oxygen = +249 kJ mol–1 Construct an enthalpy cycle and determine the bond enthalpy of the C=O bond.

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Modern College AL Chemistry Notes (2009 – 10)

Section 4B

(Answer: +805 kJmol–1)

Example 5 Calculate the bond enthalpy of the H–Br bond from the following data: ΔH / kJmol–1 1 218 2 H2(g) → H(g) Br2(l) → Br2(g) 30.6 1 112 2 Br2(g) → Br(g) 1 1 –36.3 2 H2(g) + 2 Br2(g) → HBr(g)

(Answer: +381.6 kJmol–1)

Example 6 Given that the standard enthalpy change of atomization of butane and pentane are +5165 kJmol–1 and +6340 kJmol–1 respectively. Calculate E (C–H) and E (C–C).

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Modern College AL Chemistry Notes (2009 – 10)

Section 4B

(Answer: E (C–C) = +347.5 kJmol–1, E (C–H) = +412.5 kJmol–1)

4.7.3. Use of average bond enthalpies to estimate enthalpy changes of reactions 

General rules of thumb:  Bond breaking: Endothermic  Bond forming: Exothermic Enthalpy change of reaction ≈ Energy for bond breaking – Energy for bond forming 

  

We can use the bond enthalpies to estimate the enthalpy change of certain chemical reaction. Small discrepancies will obviously arise since average bond enthalpies are used. Take the reaction between hydrogen and ethene as an example:

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Section 4B

Example 7 Given that at 298 K, the bond enthalpies of the Cl–Cl, H–H and H–Cl bonds are 242, 436 and 431 kJ mol–1 respectively. Calculate the enthalpy change of formation of hydrogen chloride gas.

(Answer: –92 kJmol–1)

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Modern College AL Chemistry Notes (2009 – 10)

Section 4B

Given the following bond enthalpies: Bond E (X–Y) / kJmol–1 C–C 348 C–H 412 C=O 743 O=O 496 O–H 463 Calculate the enthalpy change of combustion of propane.

(Answer: –1690 kJmol–1)

Example 9 Given the following bond enthalpies: Bond E (X–Y) / kJmol–1 C–C 348 C–H 412 C–Br 284 Calculate the enthalpy change of atomization of 1-bromobutane.

(Answer: +5036 kJmol–1)

Example 10 Given the following thermochemical data Enthalpy Change of vaporization of cyclohexane (ΔHvap [C6H12(l)]) Enthalpy Change of atomization of carbon Enthalpy Change of dissociation of H2 molecules Enthalpy Change of combustion of carbon Prepared by Mr. Chau Chi Keung, Richard

ΔH / kJmol–1 +33 +718 +436 –393 Page 15

Modern College AL Chemistry Notes (2009 – 10)

Section 4B

Enthalpy Change of combustion of hydrogen Average bond enthalpy of C–C bond Average bond enthalpy of C–H bond Calculate the enthalpy change of combustion of cyclohexane, C6H12(l).

–286 +348 +416

(Answer: –3885 kJmol–1)

4.7.4. Bond enthalpy, bond strength and bond length 

The bond enthalpy is a measure of bond strength. A larger value implies a stronger bond.

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However, it is not proportional to the bond order (鍵級) 

Generally, Bond order = No. of covalent bonds formed =

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No. of bonding electrons 2

Bond length is one of the major factors affecting the strength of covalent bond.  

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Bond length refers to the distance between the two bonded nuclei The larger the atoms joined by a particular bond, the longer the bond length. Larger atoms tend to have more electrons than the smaller ones. This results in an increase in both screening effect and electron cloud repulsion. Both of these effects contribute to a weakening of the bond, as seen for the hydrogen halides and halogens in the following two tables: Hydrogen halide H–F H–Cl H–Br H–I

Bond length / nm 0.109 0.135 0.151 0.171

Bond enthalpy / kJmol–1 565 431 364 297

Halogen F–F Cl–Cl

Bond length / nm 0.142 0.198

Bond enthalpy / kJmol–1 158 242

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Modern College AL Chemistry Notes (2009 – 10) Br–Br I–I  

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0.228 0.266

193 151

In general, bond length is inversely related to bond strength. From the bond length values given, the expected value for the F – F bond enthalpy in would be about 300 kJ mol–1. In fact the measured value of E (F–F) is surprisingly low. This can be attributed to the high degree of repulsion between lone pair electrons in such a short bond. However, if F is bonded to an atom with no lone pair, due to the small size of F, the bond would be very strong. This explains why F has a larger variety of compounds that the other elements because the compound formed are energetically more stable than the others (High bond enthalpy ⇒ Large amount of energy released during bond formation).

Bond strength is also affected by the number of bonding electrons / bond order. 

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Section 4B

A double bond consists of two pairs of shared electrons, and therefore there is a greater electron density between the nuclei than the corresponding single bond. It would result in greater forces of attraction between the nuclei. This can be reflected by a shorter bond length and a greater bond enthalpy.

It can be generalized that the more electrons that constitute a bond, the greater the strength of the bond. Triple bonds would be expected to be stronger than double bonds which, in turn, should be even stronger than single bonds. Bond C–C C=C C≡C C–O C=O C≡O N–N N=N N≡N

Bond length / nm 0.154 0.134 0.121 0.143 0.122 0.113 0.146 0.120 0.110

Bond enthalpy / kJmol–1 348 612 837 358 745 1070 163 409 944

4.7.5. Bond length and covalent radii  

Covalent radius is defined as half of the internuclear distance between two atoms of the same kind held together by covalent bond. In other words, it is half of the bond length of homoatomic covalent molecules (where identical atoms are bonded together, like H2, O2, Cl2, etc.)

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Modern College AL Chemistry Notes (2009 – 10)

Section 4B

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Covalent radii (in nm) of some elements are shown below:

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The variation of covalent radius shows the following general trends:  Increases down a group as the number of completed electron shells in the atoms increases. Decreases across a period as the increase in effective nuclear charge causes a reduction  of the atomic size. Covalent radii are roughly additive, that is, Bond length of A – B ≈ Covalent radius of element A + Covalent radius of element B

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or A − B ≈ 

A −A + B −B 2

We may make use of this relationship to predict bond length of various molecules. Some examples are shown in the table below: Bond

Calculated bond length / nm

C–C C–H C–O C–F C–Br C–Cl H–Cl

0.077 + 0.077 = 0.154 0.077 + 0.037 = 0.114 0.077 + 0.073 = 0.150 0.077 + 0.072 = 0.149 0.077 + 0.114 = 0.191 0.077 + 0.099 = 0.176 0.037 + 0.099 = 0.136

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Experimentally determined bond length / nm 0.154 0.109 0.143 0.138 0.193 0.177 0.128 Page 18

Modern College AL Chemistry Notes (2009 – 10) N–Cl Br–Cl I–Cl 

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0.074 + 0.099 = 0.173 0.114 + 0.099 = 0.213 0.133 + 0.099 = 0.232

Section 4B 0.174 0.214 0.233

Such predictions are particularly accurate for the bonding atoms have no (or small) difference in electronegativity. (e.g. for N–Cl bond, Difference in electronegativity = 3.0 – 3.0 = 0). Larger discrepancies arise for the cases which atoms with a great electronegativity difference are joined by a covalent bond (e.g. for C–O bond, Difference in electronegativity = 3.5 – 2.5 = 1.0). This is because the bonding electrons are drawn closer towards the more electronegative atom, causing a shortening of internuclear distance. In fact, the bond between atoms of different electronegativities is not purely covalent but carries some ionic character (will be discussed in the later part of Section 4).

4.8 Shapes of Covalent Molecules and Polyatomic Ions – The VSEPR Theory (價層電子對相斥學說) 4.8.1. Prologue: Introduction and general considerations    

VSEPR Theory: Valence Shell Electron Pair Repulsion Theory The shapes of molecules or polyatomic ions refer to the geometric arrangement of atoms within the molecules or ions. The shape of a molecule was related to the number of electrons in the outermost electron shell (i.e. valence shell) of the central atom. The electron arrangement around the central atom will determine the overall shape of the molecule. Some general rules must be considered when explaining the molecular geometry: Electron pairs in the valence shell of the central atom of a molecule will stay as far  apart as possible to minimize the electrostatic repulsion between electron pairs in the valence shell. The electron pairs are arranged in space with the maximum separation in space so as  to minimize the electrostatic repulsion of electron clouds. (由於 電子 對之 間 存 在

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斥力，因此各最外層的電子對在空間上的排佈，必需使它們之間的斥力減 至最低，故各電子對會盡量遠離對方。) The repulsion between two lone pairs must be greater than that between a lone pair and bond pair which is, greater than that between two bond pairs. Lone pair–lone pair > Lone pair–bond pair > Bond pair–bond pair (孤偶電子對─孤偶電子對 > 孤偶電子對─鍵合電子 > 鍵合電子─鍵合電子)

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Modern College AL Chemistry Notes (2009 – 10)

Section 4B

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An “electron pair” is regarded as electrons occupying localized orbitals. It can be:  A single bond  A double bond  A triple bond  A lone pair, or  An unpaired electron

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Starting from this part, when drawing the structure of molecules, you are expected to know how to express the 3-D structure of a molecule. Here are some conventional 3-D expressions:

(or bonds lying on the same plane)

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As mentioned previously, when considering the shape of a molecule or polyatomic ion, both the number of bond pair and lone pair should be considered. In general, we may use the formula shown below:

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The sum of m + n equal the total number of electron pairs around the central atom and this determines the shape of molecule or ion. The number of lone pair in a molecule (n) can be predicted by the following formula:

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n=

No. valance e − in A − (Sum of e − required to fill the valance shell of each X) 2

4.8.2. Species with 2 electron pairs 

Common form: AX2 Shape: Linear (直線形)  Bond angle: 180°  Examples: Beryllium chloride (BeCl2), carbon dioxide (CO2) 

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Modern College AL Chemistry Notes (2009 – 10) Beryllium chloride (BeCl2)

Section 4B

Carbon dioxide (CO2)

4.8.3. Species with 3 electron pairs 

Case 1: AX3 Shape: Trigonal planar (平面三角形) – Basic shape  Bond angle: 120°  Examples: Boron trifluoride (BF3), sulphur trioxide (SO3)  Boron trifluoride (BF3) Sulphur trioxide (SO3)

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Case 2: AX2E Shape: Angular / bent / V-shaped (V 形)  Bond angle: ≈120°  Examples: Sulphur dioxide (SO2)  Sulphur dioxide (SO2)

4.8.4. Species with 4 electron pairs 

Case 1: AX4 Shape: Tetrahedral (正四面體形) – Basic shape  Bond angle: 109.5°  Prepared by Mr. Chau Chi Keung, Richard

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Modern College AL Chemistry Notes (2009 – 10) 

Section 4B

Examples: Methane (CH4), ammonium ion (NH4+) Methane (CH4) Ammonium ion (NH4+)

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Case 2: AX3E Shape: Trigonal pyramidal (三角錐體形)  Bond angle: ≈107°  Examples: Ammonia (NH3), phosphorus trichloride (PCl3), hydroxonium ion (H3O+)  Due to the presence of a lone pair in the central N or P atom, there will be an increased  electron pair repulsion, resulting in an compression (壓縮) of bond angle. (Remember: Lone pair – bond pair > Bond pair – bond pair) Ammonia (NH3) Phosphorus trichloride Hydroxonium ion (H3O+) (PCl3)

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Case 3: AX2E2 Shape: Angular / bent / V-shaped  Bond angle: ≈ 104.5°  Examples: Water (H2O), hydrogen sulphide (H2S), amide ion (NH2–)  The electron pair repulsion will be even greater since there are 2 lone pairs in the  central atoms. Thus the compression of bond angle will be more prominent. (Remember: Lone pair – lone pair > Lone pair – bond pair > Bond pair – bond pair) Water (H2O) Hydrogen sulphide (H2S) Amide ion (NH2–)

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Modern College AL Chemistry Notes (2009 – 10)

Section 4B

4.8.5. Species with 5 electron pairs 

Case 1: AX5 Shape: Trigonal bipyramidal (三角雙錐體形) – Basic shape  Bond angle: 120°(equatorial / horizontal bonds) (平鍵), 90°(axial bonds) (軸鍵/直鍵)  Examples: Phosphorus pentachloride (PCl5)  Phosphorus pentachloride (PCl5)

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Case 2: AX4E Shape: Seesaw shape / irregular tetrahedral (蹺蹺板形/變形四面體形)  Bond angle: ≈120°(equatorial bonds), <90°(axial bonds)  Examples: Sulphur tetrafluoride (SF4)  Sulphur tetrafluoride (SF4)

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Case 3: AX3E2 Shape: T-shaped (T 形)  Bond angle: 90°(axial bond)  Examples: Chlorine trifluoride (ClF3)  Chlorine trifluoride (ClF3)

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Case 4: AX2E3 Shape: Linear  Prepared by Mr. Chau Chi Keung, Richard

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Modern College AL Chemistry Notes (2009 – 10)  

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Section 4B

Bond angle: 180° Examples: Triiodide ion (I3–) Triiodide ion (I3–)

Hint: Based on the basic shape, the other variation in shape for 5 electron pairs can be drawn by deleting the bonds on the equatorial plane one by one (i.e. The lone pairs must lie on the equatorial plane).

4.8.6. Species with 6 electron pairs 

Case 1: AX6 Shape: Octahedral (正八面體形) – Basic shape  Bond angle: 90° for both equatorial bonds and axial bonds  Examples: Sulphur hexafluoride (SF6)  Sulphur hexafluoride (SF6)

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Case 2: AX5E Shape: Square pyramidal (正方錐體形)  Bond angle: 90° for both equatorial bonds and axial bonds  Examples: Iodine trifluoride (IF5)  Iodine trifluoride (IF5)

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Modern College AL Chemistry Notes (2009 – 10)

Section 4B



Case 3: AX4E2 Shape: Square planar (正方平面形)  Bond angle: 90°  Examples: Xenon tetrafluoride (XeF4) (四氟化氙)  Xenon tetrafluoride (XeF4)



Hint: Unlike the case for 5 electron pairs, the other variations in shape for 6 electron pairs should be drawn by deleting the axial bonds one by one.

4.8.7. Species with 7 electron pairs (Extra) 

Case: AX7 Shape: Pentagonal bipyramidal (五角雙錐體)  Bond angle: 72°(equatorial bonds), 90°(axial bonds)  Examples: Iodine heptafluoride (IF7)  Iodine heptafluoride (IF7)

Example 11 For each of the following species, draw a three-dimensional structure showing the bond pairs and lone pairs (if any) on the central atom. State also the molecular geometry in each case.

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(a) ICl2

(b) XeOF4

(c) SiCl4

(d) OF2

(e) SCl2

(f) BrO3–

(g) ICl4–

(h) BrF5

(i) XeO4

Example 12 Explain why in phosphine (PH3), the bond angle between the two P–H bonds is about 94° whereas the bond angle between the two N–H bonds in ammonia (NH3) is about 107° although both nitrogen and phosphorus are in Group V of the Periodic Table. This is because the lone pair on the phosphorus is much more diffuse, owing to the larger size of P atom when compared with N atom. Therefore the two P–H bond pairs in PH3 are repelled more by that lone pair than in the case of the two N–H bond pairs in NH3. As a result, the two P–H bond pairs are pushed closer together, causing a reduction of the H–P–H bond angle. On the other hand, since N is more electronegative than P, the bonding electrons in NH3 are drawn closer to the central atom. Because of this, the bonding electrons in NH3 experience greater repulsion than those in PH3. Hence H–N–H bond angle is greater.

4.8.8. Shapes of molecules and polyatomic ions – An Epilogue 

Why the shapes of molecules are so important? Prepared by Mr. Chau Chi Keung, Richard

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Modern College AL Chemistry Notes (2009 – 10)





Chemical basis of odor and taste



Chemical basis of vision: 11-cis-retinal → 11-trans-retinal



Double-helical structure of DNA



Explain solubility of substances



Explain polarity (極性) of molecules



Explain melting and boiling points



Responsible for intermolecular interactions

Section 4B

The key concepts of VSEPR Theory and molecular geometry are summarized as follows. 



Electron pairs in the valence shell of the central atom of a molecule will stay as far apart as possible to minimize the repulsion between them In terms of electrostatic repulsion: Lone pair – lone pair > Lone pair – bond pair > Bond pair – bond pair



Hint: When determining the molecular shape: 

First, think about the Lewis structure, which shows the bonding and lone pairs of electrons in a molecule. This can help you to count the number of electron pairs around the central atom.



Start from the basic shape



If there is lone pair, derive the variation using the basic shape Total no. of electron pairs 2 3 4 5

6

No. of bond pairs

No. of lone pairs

Shape

2 3 2 4 3 2 5 4 3 2 6 5 4

0 0 1 0 1 2 0 1 2 3 0 1 2

Linear *Trigonal planar V-shape *Tetrahedral Trigonal pyramidal V-shape *Trigonal bipyramidal Seesaw shape T-shape Linear *Octahedral Square pyramidal Square planar

*Basic shape

4.9 Theory of Orbital Hybridization (雜化軌態理論) 4.9.1. Introduction 

Hybridization means mixing. It refers to the process of combining the orbitals.



In 1931, Linus Pauling first proposed that the wave functions for the s and p atomic orbitals can be mathematically combined to form a new set of equivalent wave functions called Prepared by Mr. Chau Chi Keung, Richard

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Section 4B

hybrid orbitals (雜化軌態 / 混成軌態) . 

Later, it was found that such theory can be used to explain phenomena which could not be explained by traditional theory of chemical bonding. For example: 





The strength of C = C bond (612 kJmol–1) is not twice of the strength of C – C bond (348 kJmol–1) bond. (i.e. the two bonds in C = C double bond are not equivalent.) The s and p orbitals in carbon can give rise to molecules with different shapes and bond angles in three-dimensional space. A carbon atom in ground state (1s22s22p2) does not have enough unpaired electrons for the formation of 4 single bonds like it does in methane.



The four single bonds in CH4 are equivalent and the shape is tetrahedral.



Violation against the octet rule (e.g. PCl5 and SF6).



Besides, it was also found that hybridization theory was also able to explain the molecular geometry. For example, the s and p orbitals in carbon can give rise to molecules with different shapes and bond angles in three-dimensional space. (to be discussed later).



During the hybridization processes, electron is promoted to the higher orbital first. (e.g. from 2s to 2p) (電子會先由原本的軌態躍遷至較高能階的軌態). This requires energy.



Since bond formed will be stronger with hybridization, this will be paid off by the energy released during bond formation.



Bear in mind that no matter what type of hybridization is concerned, only the orbitals with similar energy can be hybridized (e.g. 2s → 2p, 3p → 3d) (涉 及 的 軌 態 的 能 階 不 可 以 相差太遠). For example, a 2s orbital cannot hybridize with a 3s orbital.



The whole hybridization process can be summarized as follows:

Some examples of hybridization will be elaborated below: 4.9.2. sp3 hybridization (sp3 雜化 ) – Methane , ammonia and water as examples 

sp3 hybridization often occur in compounds of period 2 elements.



One of the two 2s electrons is first uncoupled from pairing and then promoted/excited to a Prepared by Mr. Chau Chi Keung, Richard

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vacant 2p orbital (一個電子會由 2s 軌態躍升至空置的 2p 軌態). 

This results in mixing of orbitals, which then leads to the formation of four new equivalent orbitals called sp3 hybrid orbitals (4 unpaired electrons) (形成四個不成對的電子).



In addition, these four sp3 hybrid orbitals are in a tetrahedral (正 四 面 體 ) arrangement in three-dimensional space.

Example 1: Methane (CH4) 

Carbon can give four sp3 hybrid orbitals to form four covalent bonds. In fact, the carbon atom prefers this hybridized tetravalent state (i.e. being able to form 4 bonds) to the unhybridized divalent state.



This is because the energy required for electron promotion (excitation) can be outweighed by the energy released by forming two extra covalent bonds.



The tetrahedral set of four sp3 hybrid orbitals of C atom are used to share electron with four 1s orbitals of the H atoms to form four C–H bonds (∴ Methane is tetrahedral in shape).



It should be noted that the four hybrid orbitals have the same shape and same energy level. (四個 sp3 雜化軌態的能量及形狀均相同。)



Thus the four C–H bonds are more or less the same in terms of length and strength.

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Example 2: Ammonia (NH3) 

In ammonia molecule, one of the four sp3 hybridized orbitals is occupied by an electron pair from nitrogen. This is the lone pair of ammonia.

Example 3: Water 

In water molecule, two of the four sp3 hybridized orbitals is occupied by an electron pair from oxygen. These are the two lone pairs of water.

4.9.3. sp2 hybridization (sp2 雜 化 ) – Ethene and boron trifluoride as examples 

Similar to sp3 hybridization, one of the two 2s electrons is first uncoupled from pairing and then promoted to a vacant 2p orbital.



However, only 3 hybrid orbitals (sp2 orbitals) will be formed this time. One 2p electron will remain unpaired (In other words, one 2p orbital remain unused / unhybridized). (2s 軌態與兩 個 2p 軌態雜化，形成三個 sp2 雜化軌態。剩下一個 2p 軌態未有參與雜化，也就是說還 有一粒未成對的電子。) Prepared by Mr. Chau Chi Keung, Richard

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Section 4B



The three sp2 orbitals will lie on the same plane at angles of 120 °, called a trigonal planar geometry (平面三角).



The unhybridized 2p orbital is at a right angle to the plane of the three sp2 hybrid orbitals

Example 1: Ethene (C2H4)  The unused 2p orbital on C atom is perpendicular to the plane of the three sp2 orbitals.



In forming the ethene molecule, one of the three sp2 orbitals on C atom will overlap with a sp2 orbital from another C atom, while the other two will overlap with two 1s orbitals of the H atoms. The covalent bonds formed in these two processes are called σ bonds. (兩個碳原子 的一個 sp2 雜化軌態重疊，形成 C–C 鍵。而另外兩個 sp2 雜化軌態則各與一個氫原子的 1s 軌態重疊，形成 C–H 鍵。) 

The overlapping involved here is called “head-on overlap”( 迎 頭 重 疊 ) or “linear overlap”

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Section 4B

Actually the four C–H bonds in methane are formed by the same way



On the other hand, the extra 2p orbital from each C atom which is at right angle to the plane of C and H atoms will overlap laterally above and below the plane, resulting in another type of covalent bond called π bond.



The overlapping involved here is called “sideway overlap” ( 側 面 重 疊 ) or “lateral overlap”.



Hence, the C=C double bond in ethene is a combination of one σ bond and one π bond.



It should be noted that the lateral overlap between the two unused 2p orbitals is less effective than the linear overlap which results in formation of σ bonds. (軌態以側面重疊的成鍵方式， 作用遠比軌態以迎頭重疊的成鍵方式遜色。) 

Therefore, π bond is weaker than σ bonds. (σ 鍵遠比 π 鍵為強)



In other words, the bond energy of π bond is weaker than σ bonds.



Hence, it can be deduced that, E (C=C) ≠ 2 × E (C–C) (In fact, E (C=C) < 2 × E (C–C))





This explains why the two bonds in C=C double bond are not equivalent.

In addition, since the sp2 orbitals in both C atoms adopt a planar geometry and the π bond is not free to rotate, ethene has a planar structure.

Example 2: Boron trifluoride (BF3)



The three sp2 hybridized orbitals in B atom overlap with 2p orbitals of the three F atoms to form 3 σ bonds.



The sp2 hybridized orbitals consists of three lobes with trigonal planar geometry.



It should be noted that there is still an empty 2p orbital in B atom and so it does not fulfill Prepared by Mr. Chau Chi Keung, Richard

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Section 4B

octet. Thus, as mentioned earlier, BF3 is an electron deficiency species and ready to accept electron from other molecule.

4.9.4. sp hybridization (sp 雜化 ) – Ethyne (乙炔 ) as an example 

At first, one 2s electron is promoted to 2p orbital.



Then, one 2s orbital and one 2p orbital undergo hybridization to form two sp hybrid orbitals. In this case, there will be two 2p orbitals remain unhybridized. (一個 s 軌態與一個 p 軌態混 合，形成兩個 sp 雜化軌態，而剩餘的兩個 2p 軌態則沒有雜化。)



The two sp hybrid orbitals are two lobes with linear geometry. (兩個 sp 雜化軌態呈直線排 列，軌態間的角度為 180°。)



One sp hybrid orbital overlaps with the 1s orbital of H to form a C–H σ bond, while another one overlaps with the sp hybrid orbital of another C atom to form a C–C σ bond. (兩個碳原 子的一個 sp 雜化軌態迎頭重疊，形成 C–C σ 鍵。而剩餘的 sp 雜化軌態各與一個氫原子 的 1s 軌態迎頭重疊，形成 C–H σ 鍵。)



The two unhybridized p orbitals on one C atom overlap with the two unhybridized p orbitals of another C atom laterally to form 2 C–C π bonds. (各個碳原子上兩個未經雜化的 2p 軌 態則會側面重疊，形成兩個 π 鍵。)



Hence, the C≡C triple bond in ethyne is a combination of one σ bond and two π bonds. Prepared by Mr. Chau Chi Keung, Richard

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Section 4B

Based on the fact that π bond is weaker than σ bond, it can be deduced that, E (C≡C) ≠ 3 × E (C–C) (In fact, E (C≡C) < 3 × E (C–C))

4.9.5. Expansion of octet – sp3d and sp3d2 hybridization 

As mentioned previously, P and S can form compounds which violate octet rule. This is called the expansion of octet and is actually done hybridization.



Both P and S have low-lying vacant d orbitals. The hybridization process involves the promotion (excitation) of electron from 3s orbital to 3d orbital.



As a result, P and S can fully utilize their valence electrons to form a wide variety of compounds or polyatomic ions. 

P can form 5 covalent bonds at most. Examples include PO43–, P2O5 and PCl5.



S can form 6 covalent bonds at most. Examples include SO3, SO42– and SF6.





It should be noted that such hybridization process is impossible for elements in period 2, as there is no low-lying vacant d orbital available. 



This also explain why P and S can have a wide range of possible oxidation numbers (e.g. For S: –2 to +6)

The energy required to promote 2s or 2p electrons to any vacant orbitals in the electron shell n = 3 will be so large that it cannot be compensated by the subsequent bond formation process.

Another interesting example is the compound of xenon (氙) ([Kr] 4d105s25p6), a noble gas in period 5 with atomic number 54. 



Not all noble gases can form compound. As the principle quantum number increases, the energy gap between the orbitals gets smaller. For heavy noble gas atom like xenon, electrons from lower energy level may be promoted to higher energy level to yield unpaired electrons for bonding formation. Besides, since the size of Xe atom is large, the attraction between the nucleus and the outermost electrons in a Xe atom will be so weak that they can be easily attracted by

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highly electronegative atoms like F and O. 

As a result, xenon can form compounds with highly electronegative elements like oxygen and fluorine. (e.g. XeF2, XeF4, XeF6, XeO3, XeO4 and XeOF4 etc.)

4.9.6. Epilogue: Importance of the hybridization theory 

Hybridization theory can help us to address some problems which cannot be solved by the traditional theory on covalent bonding and VSEPR theory. For example: 

Unequal strength of the 2 bonds in a C=C double bond



Number of unpaired electrons available for bond formation (violation of the octet rule)







There are not enough unpaired electrons in a C atom in ground state (1s22s22p2) for the formation of 4 single bonds, but we all know that carbon can bond with four H atoms to form methane. The four single bonds in CH4 are equivalent and the shape is tetrahedral.

In fact, when applying the VSEPR theory, the hybridization state of the central atom is also a crucial factor when determining the shape of molecules or polyatomic ions. 



Hybridization state: Determines the number of electron pairs electron pairs around the central atom. VSEPR Theory: Determines the spatial orientation (立體結構) of hybrid orbitals. Hybridization state of the central atom sp sp2

Orientation of the hybrid orbitals Linear Trigonal planar

sp3 sp3d sp3d2

Tetrahedral Trigonal bipyramidal Octahedral

Prepared by Mr. Chau Chi Keung, Richard

Examples H–C ≡ C–H H2C = CH2 BX3 (X = F or Cl) CH4, CX4 (X = F, Cl or Br) PX5 (X = F, Cl or Br) SF6

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4.10 Resonance Structure (共振結構) 

As mentioned in Section 4.7, there exist a simple additive relationship (加和性) of covalent radii, that is, Bond length of A–B

≈

or A − B ≈

Covalent radius of element A + Covalent radius of element B A −A + B −B 2



However, such additivity was found not applicable for certain species such as benzene molecule, nitrate ion and carbonate ion.



They were expected to have more than one type of covalent bond – single and double bond within the same molecule or ion. In other words, they should have more than one value of bond length.



However, experimental data show that all bond lengths are equivalent in each of these substances (see the table below). Species Benzene (C6H6) Nitrate ion (NO3–)



Bond length (nm) Calculated value Experimental value 0.154 (C–C) 0.139 0.134 (C=C) 0.136 (N–O) 0.121 0.115 (N=O)

This can be explained by a phenomenon called “delocalization of electrons” or “resonance of electrons” (or resonance effect).

4.10.1. Benzene ring 

The search for the actual structure of benzene was once a complex and controversial issue in modern Chemistry.



In 1865, German chemist August Kekulé suggested the following structure, which was a ring structure having alternating single and double bonds.

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Modern College AL Chemistry Notes (2009 – 10) 

 

Electron and X-ray diffraction studies show that the benzene molecule is planar and that all the carbon-carbon bond lengths are the same (0.139 nm). This indicates that all the bonds are similar and have intermediate character between carbon-carbon double and single bonds. It can be called “partial double bonds” (∵ In terms of bond strength, they are stronger than C–C bonds, but weaker than C=C bond). In fact, benzene may be regarded as being a resonance hybrid of two structures:





Section 4B

Although it does not completely match with the reality, you may imagine that the single and double bonds are shifting back and forth.

The real structure of benzene will be like this:



 







In benzene molecule, all six carbon atoms are sp2 hybridized. They are joined together in a ring by overlap of carbon sp2 hybrid orbitals, forming 6 C–C σ bonds. 6 C–H σ bonds are formed by further overlap with hydrogen 1s orbitals. The unhybridized 2p orbital of each carbon atom will overlap laterally with each other to form π bonds. This results in an extensive π electron cloud (also called a “π system”) lying above and below the plane of C and H atoms.

In this “π system”, the π electrons are not localized at fixed positions. Instead, they will move freely around the whole ring (苯的 π 電子離域至整個分子). Such phenomenon is called “delocalization of electrons” or “resonance of electrons” making all the 6 C–C bonds equivalent.

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Modern College AL Chemistry Notes (2009 – 10)  



Section 4B

The delocalization of π electrons gives extra stability to benzene. Beside, based on the resonance model, the C–C bonds will be equivalent instead of an arrangement of alternating single and double bonds.

The real structure of benzene can be further verified by energetic studies: 



The standard enthalpy of formation of benzene from its constituent atoms, assuming it to be a cyclic triene (the Kekulé structure), can be calculated from the relevant bond enthalpies. The calculated value is +221 kJmol–1. However, the actual measured value is only +52 kJmol–1, which suggests that an extra 169 kJ of energy is released when 1 mole of benzene molecule is formed.

Enthalpy / kJ mol–1 +221 kJmol–1

–169 kJmol–1

+52 kJmol–1 6C(graphite) + 3H2(g) 





In other words, the actual delocalized structure is more stable by 169 kJmol–1 when compared with the hypothetical cyclic triene structure. This energy is known as delocalization energy, and it is the amount by which the delocalized structure is energetically more stable than a cyclic triene.

Another possible energetic approach is the comparison between the enthalpy change of hydrogenation of benzene and that of the hypothetical cyclic triene structure: 





Enthalpy change of hydrogenation (加氫焓變): The enthalpy change when one mole an unsaturated compound reacts with an excess of hydrogen to become fully saturated. Consider the hydrogenation of benzene

The theoretical value of enthalpy change of hydrogenation can be calculated by using bond enthalpy terms. Bond

E (X–Y) / kJmol–1

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C–C 348 C=C 612 C–H 413 H–H 436 ΔH = Σ(Energy absorbed for bond breaking) – Σ(Energy released for bond forming) = [3 × E (C=C) + 3 × E (H–H)] – [3 × E (C–C) + 6 × E (C–H)] = [3(612) + 3(436)] – [3(348) + 6(413)] = – 375 kJmol–1 

However, the experimental value was found to be less negative (only – 209 kJmol–1).

= –166

Delocalization energy kJmol–1

–1

Enthalpy / kJ mol –375 kJmol–1 –209 kJmol–1







It was found that less heat is released than the calculated value. The difference between these values indicates that the actual structure of benzene is more stable than Kekulé structure by (–209) – (–375) = 166 kJmol–1. This extra stability can be attributed to the delocalization of the bonding electrons over all 6 carbon atoms. As the π electrons delocalize around the ring, the 6 C–C bonds become equivalent. Such distribution minimizes the repulsion among the electrons and gives a more stable structure.

4.10.2. Nitrate ion 

The structure of nitrate ion can be regarded as a resonance hybrid of the following three structures:

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Modern College AL Chemistry Notes (2009 – 10)

 



The arrows show the movement of electron pairs. It should be noted that the real structure of the nitrate cannot be represented by any individual resonance structure. Furthermore, the electrons are actually not shifting back and forth in the structure. Instead, they are evenly distributed in the structure.

The actual structure will be like this:







Section 4B

Since one π bond is shared among three N–O bonds, the bond order of the N–O bond in nitrate ion is 1⅓. Orbital representation would be a better way to represent the actual structure.

Points to note: 

Resonance structures exist only on paper.



In writing resonance structures, only electrons are allowed to move.





The actual molecule or ion will be better represented by a hybrid / combination of all the resonance structures involved. The energy of the actual molecule is lower than the energy that might be estimated for any contributing structure.

4.10.3. Carbonate ion 

The structure of carbonate ion can be regarded as a resonance hybrid of the following three structures: Prepared by Mr. Chau Chi Keung, Richard

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Modern College AL Chemistry Notes (2009 – 10)



Section 4B

Bond order = 1⅓ (Intermediate between C–C and C=C bond).

4.10.4. Conjugated system (共軛體系 ) in some dienes 

A conjugated system occurs in an organic compound which has alternating single and double bonds (單鍵雙鍵交替排列).



Example: Buta-1,3-diene(丁-1,3-二烯)





Molecular formula: H2C=CH–CH=CH2



Delocalization of π electrons often occurs conjugate system

Consider the hydrogenation (加氫反應) of buta-1,3-diene: H2C=CH–CH=CH2(g) + 2H2(g) → CH3CH2CH2CH3(g)



Since resonance occurs in buta-1, 3-diene, the experimental value of the enthalpy change of hydrogenation is less negative (–239 kJmol–1) than the calculated value based on bond enthalpy terms (–250 kJmol–1).



Reference: HKALE 1996 Paper II Q.1

4.11 Multiple Bonds   

Single bond: A covalent bond with two shared electrons Multiple bond: More than two shared electrons in a bond (e.g. a double bond involves the sharing of 4 electrons between two bonding atoms). For bonds between the same pair of elements, multiple bonds are always stronger and shorter than single bonds (but not proportional ⇒ E (C≡C) ≠ E (C–C) × 3). Bond C–C C=C C≡C

Bond Order 1 2 3

Bond length / nm 0.154 0.134 0.121

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Bond enthalpy / kJmol–1 348 612 837 Page 41

Modern College AL Chemistry Notes (2009 – 10) C–O C=O C≡O N–N N=N N≡N 

1 2 3 1 2 3

Section 4B

0.143 0.122 0.113 0.146 0.120 0.110

358 745 1070 163 409 944

When using the VSEPR Theory, you should always bear in mind that both double bond and triple bond are regarded as one electron pair (∴CO2 is linear in shape, not tetrahedral).

4.12 Covalent Crystals – Giant Covalent Structures 4.12.1. Diamond 

A crystal of diamond contains millions of carbon atoms. Each carbon atom is sp3 hybridized and bonded tetrahedrally to four neighboring carbon atoms by identical and directional strong C–C covalent bonds (all are σ bonds).



This gives a three-dimensional covalent network structure. Such C–C bonding pattern explains the stability, extreme hardness and very high melting point of diamond (In fact, diamond is the hardest natural substance in the world).



For a diamond to melt or undergo a chemical change, many strong C–C bonds within the crystalline structure must be broken.



Because of the incredible hardness and high melting point (3550°C), applications of diamond are endless: scratch proof cookware, watch crystal, lifetime drill bits and razor blade.



Furthermore, the localization of the electrons within C–C bonds prevents them from moving freely in an applied electric field and thus diamond does not conduct electricity.

4.12.2. Graphite 

Graphite is made up of hexagonal arrays of carbon atoms arranged in layer.



Within each layer, each carbon atom in graphite is sp2 hybridized and forms bonds with three other carbon atoms (C–C σ bonds).



As a result, network of coplanar hexagons is formed, with a C–C bond length of 0.142 nm.



Weak van der Waals’ forces hold the layers together and distance between adjacent layers is Prepared by Mr. Chau Chi Keung, Richard

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0.335 nm. 

The weak van der Waals’ forces of attraction between the layers allow one layer to slide easily over another layer. This explains why graphite is soft and slippery and can be used as a lubricant (潤滑劑), especially at high temperatures.



The unhybridized p orbital of each carbon atom will overlap laterally with each other. Therefore, the non-bonding electrons of the carbon atoms can be delocalized within the layers. Hence, graphite is a conductor of electricity.



It should be noted that the C–C bond in graphite has the partial double bond character due to the delocalization of electrons. Thus the C–C bond length in graphite (0.142 nm) is smaller than that in diamond (0.154 nm).



For this reason, the C–C bonds in graphite are stronger than that in diamond.

4.12.3. Quartz  



Quartz is a three-dimensional lattice of silicon(IV) oxide, SiO2. In the quartz lattice, each silicon atom is bonded tetrahedrally to four neighboring oxygen atoms by identical and directional strong Si–O covalent bonds. Thus the coordination number of Si is 4. Each oxygen atom is bonded to 2 silicon atoms and so the coordination number of O is 2.

Prepared by Mr. Chau Chi Keung, Richard

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Modern College AL Chemistry Notes (2009 – 10)

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Section 4B

It is very difficult to break a covalent crystal of quartz, because this would involve breaking many Si – O covalent bonds. As a result, quartz is hard with high melting points (1700℃). Furthermore, the localization of the electrons within Si–O bonds prevents their moving freely in an applied electric field and hence quartz does not conduct electricity. Quartz is one of the major components of the glass and ceramics.

Prepared by Mr. Chau Chi Keung, Richard

Page 44

Modern College AL Chemistry Notes (2009 – 10)

Prepared by Mr. Chau Chi Keung, Richard

Section 4B

Page 45