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MOLECULAR SHAPE AND POLARITY

Subtopic 4.2

1

LEARNING OUTCOMES (covalent bonding)

1. Draw the Lewis structure of covalent molecules (octet rule such as

N𝐻3, 𝐶𝐶𝑙4, 𝐻2O, 𝐶𝑂2, 𝑁2𝑂4, and exception to the octet rule such

as 𝐵𝐹3, NO, 𝑁𝑂2, 𝑃𝐶𝑙5, 𝑆𝐹6)

2. Explain the concept of overlapping and hybridisation of the s and

p orbitals such as 𝐵𝑒𝐶𝑙2, 𝐵𝐹3, 𝐶𝐻4, 𝑁2, HCN, 𝑁𝐻3, 𝐻2O

molecules

3. Predict and explain the shapes of and bond angles in molecules

and ions using the principle of valence valence shell electron pair

repulsion, e.g. linear, trigonal planar, tetrahedral, trigonal

bipyramid, octahedral, v-shaped, seesaw and pyramidal

4. Explain the existence of polar and non-polar bonds (including C-

Cl, C-N, C-O, C-Mg) resulting in polar or/and non-polar molecules

LEARNING OUTCOMES (covalent bonding)

5. Relate bond lengths and bond strengths with respect to single,

double and triple bonds

6. Explain the inertness of nitrogen molecule in terms of its strong triple

bond and nonpolarity

7. Describe typical properties associated with ionic and covalent

bonding in terms of bond strength, melting point and electrical

conductivity

8. Explain the existence of covalent character in ionic compounds such

as 𝐴𝑙2𝑂3, 𝐴𝑙𝑙3, and Lil

9. Explain the existence of coordinate (dative covalent) bonding such as

𝐻3𝑂+, N𝐻4+

, 𝐴𝑙2𝐶𝑙6 , and [Fe (𝐶𝑁)6]³ˉ

MOLECULAR GEOMETRY

Molecular

Geometry

The 3D

arrangement of

atoms in a

molecule Affects

physical and

chemical

properties

Is predicted by

using Valence

Shell Electron

Pair Repulsion

(VSEPR) model

VSEPR THEORY

1. Valence Shell Electron Pair Repulsion theory

2. It states that each group of valence electrons

around a central atom is located as far away

as possible from the others in order to

minimize repulsion

3. To predict the molecular shape from the Lewis

structure * Kalau dekat, akan repulse each other

VSEPR

THEORY

Two type of electron pair around

the central atom:

bonding pairs and lone pairs

Electron pairs around the central

atom will repel one another and

arrange themselves as far apart

as possible from each other.

VSEPR THEORY

Why? To minimise the electron

pair-electron pair repulsion

around the central atom

ELECTRON GROUP

1. Electrons pair of lone electron that occupy localized

region around the central atom

2. The repulsion may occur either between :

Bonding pair (single bond or double bond or

triple) with bonding pair

Lone pair (odd electron) with bonding pair

Lone pair with lone pair

Bonding pair electron

Lone pair

electron

Note:

The electron pairs repulsion will determine the orientation of atoms in space

ELECTRON GROUP

Bonding pair-bonding pair repulsion

Lone pair-lone pair repulsion

Lone pair-bonding

pair repulsion

> >

Decrease of the repulsion force

FIVE BASIC MOLECULAR SHAPES

1. Linear (180°)

2. Trigonal planar (120°)

3. Tetrahedral (109.5°)

4. Trigonal bipyramid (90°, 120°)

5. Octahedral (90°)

ELECTRON GROUP ARRANGEMENT

1. Determined by number of electron groups around the

central atom

2. Can be one of the 5 basic shapes :

2 electron group/bonding pair : linear

3 electron group/bonding pair : trigonal planar

4 electron group/bonding pair : tetrahedral

5 electron group/bonding pair : trigonal bipyramidal

6 electron group/bonding pair : octahedral

1. Double bond and triple bonds can be treated

like single bond (approximation)

One electron group

2. Order of electron group repulsion

Lone pair – lone pair > lone pair – bonding pair

> bonding pair – bonding pair

SOME RULES WHEN USING VSEPR THEORY

SHAPE OF MOLECULE

1. Basic shapes are based on the repulsion between the

bonding pairs.

2. Tips to determine the molecular shape :

Step 1 : Calculate total number of valence electron

Step 2 : Draw Lewis structure of the molecule

Step 3 : Calculate the number of bonding pairs or

electron groups (Place bonding pairs as far as

possible to minimize repulsion)

Step 4 : Predict the shape using VSEPR theory

A. MOLECULAR SHAPE WITH 2 BONDING PAIR

Example: BeCl2

Lewis structure

shape

Linear

180° Be : 2e

2Cl : 2 X 7e = 14e

Total : 16 e

Cl .. : .. Cl Be

.. : ..

Electron group arrangement : linear

Molecular shape : linear (bond angle : 180°)

A𝑋2

B. MOLECULAR SHAPE WITH 3 BONDING PAIR

Example: BCl3 Lewis structure

B: 3e

3Cl : 21e

Total: 24e

B .. : ..

Cl

Cl

Cl .. :

..

..

: ..

Repulsive forces

between pairs are the same

120°

Trigonal planar

A𝑋3

Electron group arrangement : trigonal planar

Molecular shape : trigonal planar (bond angle : 120°)

C. MOLECULAR SHAPE WITH 4 BONDING PAIR

Example: CH4

Lewis structure

C

H

H H

H

Equal repulsion between bonding pairs – equal angle

109.5°

Tetrahedral

C: 4e

H : 4e

Total: 8e

Electron group arrangement : tetrahedral

Molecular shape : tetrahedral (bond angle :109.5°)

Pyramid is not tetrahedral

AX4

D. Molecules with 5 bonding pairs

Example: PCl5 Lewis structure

P

Cl

Cl

Cl

Cl

Cl

..

: ..

..

: ..

..

: ..

..

: ..

Shape:

120°

Trigonal bipyramidal

90°

Electron group arrangement : trigonal bipyramidal

Molecular shape : trigonal bipyramidal

bond angle :120°, 90°

E. Molecules with 6 bonding pairs Example: SF6

Lewis structure

S : 6e 6F : 42e Total : 48e

Octahedral

S

F

F

FF

F

F

90o

90o

Electron group arrangement : octahedral

Molecular shape : octahedral

bond angle : 90°

19

Class of molecules

Number of bonding

pairs

Number of lone pairs

Shape

AX2E

e.g. SnCl2

Sn: 4e

2Cl : 14e

Total : 18e

2 1

Bent / V-shaped

Bond angle : < 120o

Shape of molecules which the central atom has one or more

lone pairs (3 ELECTRON PAIRS)

Example: SnCl2 Lewis structure

Sn: 4e

2Cl : 14e

Total: 18e

AX2E


Molecular shape : bent or v-shaped (bond angle < 120°)

Sn

.. : ..

Cl Cl .. :

:

:

1. Calculate total number of valence

electron

2. Draw the Lewis structure

3. Calculate the number of bonding

pairs or electron groups (Place

bonding pairs as far as possible to

minimize repulsion)

4. Predict the shape using VSEPR theory

21

Class of molecules

Number of bonding

pairs


Shape

AX3E

e.g. N𝐻3

3 1

Trigonal pyramidal Bond angle : < 109.5o

4 electron pairs in the valence shell of central

atom:

N : 5e

H : 3e

Total : 8e

22

Class of molecules

Number of bonding

pairs


Shape

AX2E2

e.g. 𝐻2O

2 2

Bent / V-shaped Bond angle : < 109.5o

4 electron pairs in the valence shell of central atom:

O : 6e

H : 2e

Total : 8e

23

Class of molecules

Number of bonding

pairs


Shape

AX4E

e.g. S𝐹4

4 1

Distorted tetrahedral (see-saw) Bond angle : < 90o, < 120o


S : 6e

F : 28e

Total : 34e

24

Class of molecules

Number of bonding

pairs


Shape

AX3E2

𝑒𝑔. 𝐵𝑟𝐹3

3 2

T-shaped Bond angle : < 90o


atom:

Br : 7e

F : 21e

Total : 28e

25

Class of molecules

Number of bonding

pairs


Shape

AX2E3

eg. 𝐼3−

2 3

Linear Bond angle : 180o


atom:

I : 7e X 3

Total : 21e

26

Class of molecules

Number of bonding

pairs


Shape

AX5E

e.g. I𝐹5

5 1

Square pyramidal Bond angle :90o and

180o


atom:

I : 7e

F : 35e

Total : 42e

27

Class of molecules

Number of bonding

pairs


Shape

AX4E2

e.g. Xe𝐹4

4 2

Square planar Bond angle : 90o


Xe : 8e

F : 28e

Total : 36e

KEEP IN MIND !

Elements period 3 or higher can form compounds that have both

octet and expanded octet valence shell

Example :

𝑃𝐶𝑙3 + 𝐶𝑙2 → 𝑃𝐶𝑙5

octet expanded octet

*Period 3 : Na, Mg, Al, Si, P, S, Cl, Ar

Never “expand” atoms of period 1 and 2 such as C and O.

They must be octet!

********************************************************

KEEP IN MIND !

Elements period 3 or higher can form compounds that have both

octet and expanded octet valence shell

Example :

𝑃𝐶𝑙3 + 𝐶𝑙2 → 𝑃𝐶𝑙5

octet expanded octet

*Period 3 : Na, Mg, Al, Si, P, S, Cl, Ar

Never “expand” atoms of period 1 and 2 such as C and O.

They must be octet!

Example :

********************************************************

Predict the geometry molecular shape for the

following molecules and ions by using VSEPR theory:

(a) PCl5 (b) IF

4+

(c) ClF3 (d) I

3-

(e) IOF5 (f) BrF

5

(g) XeF4 (h) O

3

EXERCISE 1

ClF3

I3

-

1. Valence electron

2. Lewis structure

3. Lone and bond pair

4. Predict the shape

ANSWER

32

Step 1 : Write the Lewis structure to see the number of electron groups (Bonding and lone pairs at the central atom)

Step 2 : Determine the electron group arrangement

Step 3 : Predict the ideal bond angle and any deviation caused by lone pairs

Step 4 : Draw and name the molecular shape by counting the bonding groups

GUIDELINES FOR APPLYING VSEPR THEORY

POLAR AND NON POLAR BONDS

1. Atoms with different electronegative from polar bonds (difference in EN)

2. Depicted as polar arrow :

3. Example :

C - Cl

BOND POLARITY

𝛿 − 𝛿 + C - Cl

# Polar bond

POLAR BOND

H F

electron rich

region electron poor

region

Polar bond

36

NON – POLAR BOND

F F

Non – polar bond

37

Example :

C – F

C – C (no difference of electronegativity)

BOND POLARITY

𝛿 + 𝛿 − C - F

# Polar bond

# Non polar bond

ELECTRONEGATIVITY

Electronegativity of an atom is the ability of an atom that is

covalently bonded to another atom to attract electrons to itself

An atom with a high electronegativity will attract electrons

towards itself and away from the atom with a lower

electronegativity

Electronegativity of an atom is inversely proportional to its size.

The smaller the atomic size, the stronger the attraction for the

bonding electrons and the higher electronegativity.

1. For covalent bond that consists of two identical atoms, the bonding electrons are shared equally between the two atoms and are attracted equally to both the atoms. This type of bond is called non-polar bond.

2. Diatomic molecules such as 𝐻2, 𝑂2, 𝐶𝑙2, 𝑁2, have non polar bonds because the atoms in the molecules have same electronegativity. These molecules are called non polar molecules.

NON POLAR BONDS

1. In covalent bond that contains two atoms that are not identical. The bonding electrons will be attracted more strongly by the more electronegative element and this result in unsymmetrical distribution of electrons.

2. Ex : 𝐻 − 𝐶𝑙 . The more electronegative chlorine atom attracts the bonding electron pair more strongly than hydrogen does.

3. HCl is a polar molecule and it contains polar bond.

4. The separation of charge in polar covalent bond like H-Cl is called polarisation.

5. When two electrical charge of opposite signs are separated by small distance, a dipole is established. Thus, HCl has a dipole because the H-Cl is a polar.

POLAR BOND

DIPOLE MOMENT (𝜇)

A quantitative measure of the polarity of a bond is its dipole moment ( µ ).

µ = Q r

Where : µ = dipole moment

Q = the product of the charge from

electronegativity

r = distance between the charges

Dipole moments are usually expressed in debye units(D)

1D (Debye) = 3.36 x 𝟏𝟎−𝟑𝟎Cm

RESULTANT (NET) DIPOLE MOMENT

1. Determined by molecular shape and bond polarity

Resultant dipole moment > 0 (polar molecule)

Resultant dipole moment = 0 (non-polar molecule)

2. Example :

∆𝐸𝑁 = 0 (charge in electronegativity)

𝜇 = 0

𝐼2 is a non polar molecule

I - I


3. Example :

∆𝐸𝑁 ≠ 0 (charge in electronegativity)

𝜇 ≠ 0

𝐼2 is a polar molecule

CI - I


4. Example :

𝐶𝑂2 shape =linear

The two bond dipoles cancel each other (molecule is symmetrical)

Resultant dipole moment, 𝜇 = 0

𝐶𝑂2 is a non polar molecule

𝐶𝑂2 𝑂 = 𝐶 = 𝑂


5. Example :

𝑂𝐶𝑆 shape =linear

The two bond dipoles do not cancel each other

Resultant dipole moment, 𝜇 ≠ 0

𝑂𝐶𝑆 is a polar molecule

OCS 𝑂 = 𝐶 = 𝑆


6. Example :

𝐵𝐹3 shape =trigonal planar

The two bond dipoles cancel each other

Resultant dipole moment, 𝜇 = 0

𝐵𝐹3 is a NON polar molecule

𝐵𝐹3

B

F

F

F


7. Example :

𝐵𝐹3𝐵𝑟 shape =trigonal planar

The three bond dipoles do not cancel each other

Resultant dipole moment, 𝜇 ≠ 0

𝐵𝐹3𝐵𝑟 is a polar molecule

𝐵𝐹2Br

B

Br

F

F

Carbon tetrachloride, CCl4

- molecular geometry : tetrahedral

- Chlorine is more electronegative than carbon,

- Dipole moment can cancel each other

- has no net dipole moment (µ = 0)

- therefore CCl4 is a nonpolar molecule.


- Cl is more electronegative than C, C is more

electronegative than H

- Dipole moment cannot cancel each other

- has a net dipole moment (µ ≠ 0)

- therefore CH3Cl is a polar molecule.

Chloromethane, CH3Cl

Ammonia, NH3


- N is more electronegative than H,

- Dipole moment cannot cancel each other

- has a net dipole moment (µ ≠ 0)

- therefore NH3 is a polar molecule.

BOND MOMENTS AND RESULTANT DIPOLE MOMENTS

H

1. Polar molecules possess polar bond.

2. A bond is polar when the two atoms that are participating

in the bond formation have different electronegativities. In

polar molecule, all the bonds collectively should produce a

polarity.

3. Though a molecule has polar bonds, it does not make the

molecule polar.

4. If the molecule is symmetric and all the bonds are similar,

then the molecule may become non polar.

5. Therefore, not all the molecules with polar bonds are polar.

DIFFERENCE BETWEEN POLAR BOND AND POLAR MOLECULES

The presence of polar bonds does not always lead to a

polar molecule

C−O is a polar bond

But, 𝐶𝑂2 is a non polar molecule

We have to CONSIDER BOTH (bond polarity and

molecular shape)

KEEP IN MIND!

𝑂 = 𝐶 = 𝑂

1. A molecule will be nonpolar if :

a) The bonds are non-polar

b) No lone pair in the central atom and all the surrounding atoms are the same

A SUMMARY ON HOW TO DETERMINE MOLECULAR POLARITY

CI – CI (non polar)

B

Br

F

F

B

F

F

F

POLAR NON POLAR


c) A molecule in which the central atom has lone pair electron will usually be polar with few exceptions


POLAR

O

H H

N

H H H POLAR


c) A molecule in which the central atom has lone pair electron will usually be polar with few exceptions


POLAR POLAR

Br

F

F

F

S

F

F

F

F

2. Exceptions : (NON POLAR MOLECULES)


LINEAR

A

X

X

A

X X

X X

Xe

F F

F F

SQUARE PLANAR

60

SO2 ; HBr ; SO3 ; CH2Cl2 ; ClF3 ; CF4 ; H2O ; XeF4 ; NF3

Exercises :

Predict the polarity of the following molecules:

1. VSEPR theory : predict molecular shapes by assuming that electron groups tend to minimize their repulsions

2. But, it does not tell how those shapes (which is observed experimentally), can be explained from the interactions of atomic orbitals.

HYBRID ORBITAL OVERLAP AND HYBRIDIZATION

1. Covalent bonds are formed by sharing electrons from

overlapping atomic orbitals

2. Two types of bonds : σ bond and π bond

3. Example :

VALENCE BOND (VB) THEORY

1. Atoms in simple molecules or ions such as 𝐻2, HF, 𝑁2, etc. use pure s and/or p orbitals in forming covalent bonds.

2. Example : 𝐻2 (hydrogen molecules)

DIRECT OVERLAP OF s AND p ORBITAL

Example : HF (Hydrogen Flouride)

H = 1𝑠1

F = 1𝑠2 2𝑠2 2𝑝5

Example : 𝐹2 (Flourine molecules)

F = 1𝑠2 2𝑠2 2𝑝5

DIRECT OVERLAP OF s AND p ORBITAL

1. Mixing of two or more atomic orbitals to form a new

set of equivalent hybrid orbitals in the same energy

level

2. The spatial orientation of the new orbitals is cause

more stable bonds and are consistent with the

observed molecular shape.

3. 3 types of hybridization : sp, s𝑝2, s𝑝3 hybridization

HYBRIDIZATION

4.3.2 Formation Hybrid orbitals

• Overlapping of hybrid orbitals and the pure orbitals occur when different type of atoms are involved in the bonding.

• Hybridization of orbitals:

mixing of two or more atomic orbitals to form a new set of hybrid orbitals

• The purpose of hybridisation is to produce new orbitals which have equivalent energy

• Number of hybrid orbitals is equal to number of pure atomic orbitals used in the hybridization process.

Hybridization

• Hybrid orbitals have different shapes from

original atomic orbitals

• Types of hybridisation reflects the

shape/geometry of a molecule

• Only the central atoms will be involved in

hybridisation

HYBRIDIZATION

s orbital p orbital

sp orbital

TYPES OF HYBRID ORBITALS

Type Examples Electron group Electron group

arrangement

sp Be𝐶𝑙2 2 Linear

s𝑝2 B𝐹3 3 Trigonal planar

s𝑝3 C𝐻4 4 Tetrahedral

1. Draw Lewis structure

2. Predict electron group arrangement using VSEPR model

3. Deduce the hybridization of the central atom by matching

the arrangement of the electron groups with the hybrid

orbitals

4. Use partial the orbital diagram to explain the mixing of

atomic orbitals

DETERMINING HYBRID ORBITALS

Molecular formula

Lewis Structure

Molecular shape and electron group arrangement

Hybrid orbitals

1. Resulting from end to end overlap

2. Has highest electron density along the bond axis

3. Allow free rotation

4. All single bonds are 𝜎 bond

SIGMA (𝜎) BOND

73

+

bond

It formed when orbitals overlap from end to end

Example:

i. overlapping s orbitals

H H H H

bond

74

ii. Overlapping of s and p orbitals

H +

Px orbital

x x H

bond

75

iii. Overlapping of p orbitals

x x + x

bond

1. Resulting from side to side overlap

2. Has two regions of electron density

One above and one below the 𝜎 bond axis

3. One 𝜋 bond hold two electrons that move through both regions of the bond

4. 𝜋 bond restricts rotation

Pi (𝜋) Bond

bond It formed when two p-orbitals of the same orientation overlap

sideways

Double bond consists of one 𝜎 bond and one 𝜋 bond

Example :

y y

+

y y

bond

bond

Example :

𝐶𝑂2 has two 𝜋 bond and one 𝜎 bond

Triple bond always consists of one 𝝈 bond and two 𝝅 bond

𝑂 = 𝐶 = 𝑂 𝜋 𝜋

𝜎 𝜎

𝑁 ≡ 𝑁 𝜋

𝜎

𝜋

No of Lone Pairs +

No of Bonded Atoms Hybridization Examples

2

3

4

sp

sp2

sp3

BeCl2

BF3

CH4, NH3, H2O

How do I predict the hybridization of the central atom?

Count the number of lone pairs AND the number

of atoms bonded to the central atom

Two equivalents sp hybrid orbitals that lie 180° apart

2 electron groups (from VSEPR theory)

Electron group arrangement : linear

Molecular shape : linear

FORMATION OF sp HYBRIDIZATION

sp sp

180°

Lewis structure

Valence electron configuration

Hybridisation

𝑩𝒆𝑪𝒍𝟐

FORMATION OF sp HYBRIDIZATION

2s 2p

O1 :

O2 :

2s 2p

Cl−𝐵𝑒 − 𝐶𝑙 𝜎 𝜎

Be ↑↓

↑ ↑

↑ ↑

sp orbitals

sp sp 2p 2p

Empty 2p orbitals

One s orbital + one p orbital → two equivalent sp orbitals

Three equivalents 𝑠𝑝2 hybrid orbitals that lie 120° apart



Molecular shape : trigonal planar

FORMATION OF 𝑠𝑝2 HYBRIDIZATION

sp2

sp2

sp2

120°

Lewis structure


Hybridisation

𝑩𝑭𝟑


2s 2p

O1 :

O2 :

2s 2p

𝜎

𝜎

B ↑↓

↑ ↑

↑ ↑

𝑠𝑝2 orbitals

𝑠𝑝2 𝑠𝑝2 2p

Empty 2p orbitals

One s orbital + two p orbital → three equivalent 𝑠𝑝2 orbitals

B

Br

F

F

𝜎 ↑

↑

↑

𝑠𝑝2

TRY DRAW THEM 𝑠𝑝2 hybrid!!

Bond angle : 109.5°


Electron group arrangement : tetrahedral

Molecular shape : tetrahedral


sp3

sp3

sp3

sp3

109.5°

Lewis structure


Hybridisation

𝑪𝑯𝟒


2s 2p

O1 :

O2 :

2s 2p

𝜎

𝜎

B ↑↓

↑ ↑

↑ ↑

𝑠𝑝3 orbitals

𝑠𝑝3 𝑠𝑝3 𝑠𝑝3

Empty 2p orbitals

One s orbital + three p orbital → four equivalent 𝑠𝑝3 orbitals

𝜎

↑

↑

↑

𝑠𝑝3

↑

↑

↑

C H

H

H

H

𝜎

TRY DRAW THEM 𝑠𝑝3!!

Example : Methane, CH4

Ground state : C : 1s2 2s2 2p2

C H H H

H Lewis Structure

1s 2s 2p

Excitation: to have 4

unpaired electrons

Excited state : 1s 2s 2p

sp3 hybrid

shape: tetrahedral

sp3

sp3

sp3

sp3

H H

H

H

C

91

sp3-Hybridized C atom in CH4

sp3

sp3

sp3

sp3 1s

1s

1s

sp3 hybrid

• Mixing of s and three p orbitals

sp3

sp3

Other Example:

1. BF3

• Lewis structure :

• Valence Electron configuration :

F :

B ground state :

B hybrid :

• Molecular geometry

• Orbital overlap/Hybridisation:

Example: BF3

sp2

sp2 sp2

F : 1s22s22p5

Shape : trigonal planar

Pure p orbital

Other Example :

2. NH3

Lewis structure :

Valence orbital diagram ;

H :

N ground state :

N hybrid :

Orbital Overlap :

Molecular Geometry :

96

sp3

sp3

sp3

sp3

1s

1s

1s

97

Other Example:

3. H2O

• Lewis structure :

• Valence orbital diagram;

O ground state :

O hybrid :

• Molecular geometry

• Orbitals overlap:

HYBRIDISATION IN MOLECULES

CONTAINING DOUBLE AND

TRIPLE BONDS

Lewis structure

(ETHANE)


Hybridisation

𝑪𝟐𝑯𝟒

FORMATION OF HYBRIDIZATION

2s 2p

O1 :

O2 :

2s 2p

C ↑↓

↑ ↑

↑ ↑

𝑠𝑝2 orbitals

𝑠𝑝2 𝑠𝑝2 2𝑝

UNHYBRIDIZED 2p orbitals


↑

↑

↑

𝑠𝑝2

↑

↑

↑

𝜎

𝜋

𝐶 = 𝐶

H H

H H

HOW TO DRAW ?

101

bond

bonds

Lewis structure

(ACETYLENE)


Hybridisation

𝑪𝟐𝑯𝟐


2s 2p

O1 :

O2 :

2s 2p

C ↑↓

↑ ↑

↑ ↑

𝑠𝑝 orbitals

𝑠𝑝 2𝑝 2𝑝


One s orbital + one p orbital → two equivalent 𝑠𝑝 orbitals

↑

↑

↑

𝑠𝑝

↑

↑

↑

𝜎

𝜋

𝐶 ≡ 𝐶 H H

𝜋

HOW TO DRAW ?

Lewis structure

(BENZENE)


Hybridisation

𝑪𝟔𝑯𝟔


2s 2p

O1 :

O2 :

2s 2p

C ↑↓

↑ ↑

↑ ↑

𝑠𝑝2 orbitals

𝑠𝑝2 𝑠𝑝2 2𝑝



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𝑠𝑝2

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BENZENE ???? (Look at the notes!)

108

ANSWER:

For each of the following, draw the orbital

overlap to show the formation of covalent bond

a) 𝐻2𝑂

b) N2

c) 𝐶𝐻3 Cl d) Al𝐶𝑙3

INERTNESS OF NITROGEN

MOLECULE 1. Nitrogen is a very electronegative element.

2. It is an inert (unreactive) element.

3. Inertness due to 2 factors :

a) Strong triple bond

b) Non polarity of 𝑁2

4. Bond energy 𝑁 ≡ 𝑁 is very high due to triple bond.

5. This strong bond must be broken first, then it can form with other

compounds.

6. A lot of energy needed to break

7. Only at high temperature, nitrogen can react with other elements

to form compounds.

8. Nitrogen molecules is non-polar. The absence of polarity on the

molecule explains why nitrogen is unreactive.

COVALENT CHARACTER IN IONIC

COMPOUNDS

1. Not all compounds are ionic, and not all compounds are covalent.

2. Polarisation of chemical bonds occur not only in covalent

compound but also in ionic compound.

3. Most ionic compounds have covalent character due to incomplete

transfer of electrons.

4. If a small cation with high electric charge approaches a large

anion, the cation will attract the electron cloud from the negative

ion.

5. Polarisation : The distortion of electron cloud of the anion by a

neighbouring cation.

6. Polarising power : The extent to which a cation (positive ion) can

polarise an anion (negative ion)

COVALENT CHARACTER IN IONIC

COMPOUNDS

7. Polarisation (distortion of electron cloud) of the anion produces

covalent character in the ionic bond because the valence

electron is partially shared between the cation and anion.

8. The greater the degree of polarisation of the anion, the greater

the amount of covalent character in the ionic bond.

9. The polarising power of a cation towards an anion is

proportional to the charge density.

𝑐ℎ𝑎𝑟𝑔𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 𝑐ℎ𝑎𝑟𝑔𝑒

𝑖𝑜𝑛𝑖𝑐 𝑟𝑎𝑑𝑖𝑢𝑠 (𝑠𝑖𝑧𝑒)

10. Small and highly charged cations such as 𝐿𝑖+ and 𝐴𝑙3+ have

high charge density, so, high polarising power.

Question 1

Arrange the following chlorides in order of increasing covalent

character. Explain your answer.

NaCl, 𝑀𝑔𝐶𝑙2 , 𝐴𝑙𝐶𝑙3

Answer 1

NaCl < 𝑀𝑔𝐶𝑙2 < 𝐴𝑙𝐶𝑙3

The charge density of the cations increases in the order :

𝑁𝑎+< 𝑀𝑔2+ < 𝐴𝑙3+

Hence, polarising power of the cations towards the 𝐶𝑙_ ion increases

in the same order.

Polarising power of 𝐴𝑙3+ is high so, it is a covalent compound.

Question 2

Arrange two compounds of beryllium order of increasing covalent

character.

𝐵𝑒𝐹2 , 𝐵𝑒𝐼2

ANSWER 2

𝐵𝑒𝐼2 is expected to have a high degree of covalent character.

The size of 𝐼− ion is larger than 𝐹− ion, hence it is easier to be

polarised by the 𝐵𝑒2+ ion

molecular shape and polarity - wordpress.com...2016/07/03 · 5. relate bond lengths and bond...

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