Achievers Dream - O & A Level Chemistry Tuition In Singapore
Categories
01 Free Studying Resources 07 Junior College 1 Chemistry

Chapter 03: Chemical Bonding

Chapter 03: Chemical Bonding

VSEPR Theory

The Valence Shell Electron pair Repulsion theory (VSEPR) theory is one of the most important concepts in chem bonding as this theory helps us to determine the shape and bond angles of covalent molecules.

The 3 basic principles that students should take note to remember are,

  • Electron pairs repel each other and arrange themselves as far apart as possible to maximise stability and minimise electrostatic repulsion
  • Orbitals containing lone pairs are larger than those containing bond pairs/odd
    electrons hence they have greater repulsive effect than bond pairs. In other words, lone pair-lone pair repulsion > lone pair-bond pair repulsion > bond pair-bond pair repulsion
  • Repulsion between electron pairs is increased by an increase in the electronegativity of the central atom.

Repulsion between electron pairs is increased by an increase in the electronegativity of the central atom.

Total number of electron pairs Number of bond pairs Number of lone pairs Shape/Bond angle Example
2 2 0 Linear 180° bond-angle N₂O
3 3 0 Trigonal planar 120° bond-angle BF₃
3 2 1 Bent 110° - 120° SO₂
4 4 0 Tetrahedral 109.5° CH₄
4 3 1 Trigonal pyramidal 107° SOCl₂
4 2 2 Bent 105° bond-angle H₂O
5 5 0 Trigonal bipyramidal 120° for atoms lying on the same plane PCl₅
5 4 1 See-saw 120° bond-angle SF₄
5 3 2 T-shaped 90° ClF₃
5 2 3 Linear 180° bond-angle XeF₂
6 6 0 Octahedral 90° bond-angle SF₆
6 5 1 Square pyramidal 90° bond-angle BrF₅
6 4 2 Square planar 90° bond-angle XeF₄

Total number of electron pairs: 2

Number of bond pairs: 2

Number of lone pairs: 0

Shape/Bond angle : Linear 180° bond-angle

Example : N₂O

Total number of electron pairs: 3

Number of bond pairs: 3

Number of lone pairs: 0

Shape/Bond angle : Trigonal planar 120° bond-angle

Example :BF₃

Total number of electron pairs: 3

Number of bond pairs: 2

Number of lone pairs: 1

Shape/Bond angle : Bent 110° - 120° bond-angle

Example : SO₂

Total number of electron pairs: 4

Number of bond pairs: 4

Number of lone pairs: 0

Shape/Bond angle : Tetrahedral 109.5° bond-angle

Example : CH₄

Total number of electron pairs: 4

Number of bond pairs: 3

Number of lone pairs: 1

Shape/Bond angle : Trigonal pyramidal 107°

Example : SOCl₂

Total number of electron pairs: 4

Number of bond pairs: 2

Number of lone pairs: 2

Shape/Bond angle : Bent 105° bond-angle

Example : H₂O

Total number of electron pairs: 5

Number of bond pairs: 5

Number of lone pairs: 0

Shape/Bond angle : Trigonal bipyramidal 120° for atoms lying on the same plane

Example : PCl₅

Total number of electron pairs: 5

Number of bond pairs: 4

Number of lone pairs: 1

Shape/Bond angle : See-saw 120° bond-angle

Example : SF₄

Total number of electron pairs: 5

Number of bond pairs: 3

Number of lone pairs: 2

Shape/Bond angle : T-shaped 90°

Example : ClF₃

Total number of electron pairs: 5

Number of bond pairs: 2

Number of lone pairs: 3

Shape/Bond angle : Linear 180° bond-angle

Example : XeF₂

Total number of electron pairs: 6

Number of bond pairs: 6

Number of lone pairs: 0

Shape/Bond angle : Octahedral 90°

Example : SF₆

Total number of electron pairs: 6

Number of bond pairs: 5

Number of lone pairs: 1

Shape/Bond angle : Square pyramidal 90° bond-angle

Example : BrF₅

Total number of electron pairs: 6

Number of bond pairs: 4

Number of lone pairs: 2

Shape/Bond angle : Square planar 90°

Example : XeF₄

Worked Example 1:
[2020 SAJC H2 Chem Prelim Paper 3 Qn 5(c)(ii)] 

Platinum complexes have been researched extensively due to its anti-cancer properties.
One such platinum complex is cis-platin, which can be synthesised from PtCl₄. The central atom, platinum, in PtCl₄ is surrounded by six pairs of electrons arranged in an octahedral shape.
An octahedral arrangement is shown in Fig. 5.1.

Chemical Bonding

Two different molecular arrangements of PtCl₄ are possible. Draw clear diagrams of these two arrangements. Hence, apply the principles of the VSEPR theory to discuss the relative stabilities of these two possible arrangements.

Chemical Bonding

Solution and Detailed Explanation:

Don’t be intimidated by the application context of the question. As long as you pick out the key information, it is solvable

Part I of the question requires you to draw 2 possible arrangements for the compound PtCl₄. Since there are 4 Cl atoms surrounding the Pt central atom, there should be 4 sigma bond pairs and the remaining 2 electron pairs are lone pairs.
Therefore, the only 2 possible arrangements are:

Part II of the question requires you to discuss the relative stabilities of the above 2 arrangements drawn with the use of VSEPR theory. Recall the 3 Basic principles of VSEPR theory and apply accordingly. Use the diagram drawn in Part I as a visual aid. Do note that the question also requires you to do a comparison, thus you must mention about the stability of BOTH drawn arrangements.

The square planar arrangement/the one on the left is more stable.

Reason: For the one on the left, the two lone pairs of electrons are on opposite sides / 180° away, which minimises lone pair-lone pair repulsion.
For the one on the right, the two lone pairs of electrons are next to each other / 90° away, which experiences

Sigma and Pi bonds:

You may struggle to understand this portion of the chapter due to its abstract nature, as picturing the bonds in your head is not easy. With the help of some pictures we will guide you in understanding sigma and pi bonds better.

Sigma bond (σ) Pi bond (π)
Sigma bond (σ)
  • Atomic orbitals overlap in a head-on manner as shown in the above diagram
  • There can only exist ONE σ bond between two atoms

'side-by-side' (π) overlap

Pi bond (π)
  • Formed when two p orbitals overlap in a side-on manner
  • A π bond can only exist after a σ bond is formed
Sigma bond (σ)
Sigma bond (σ)
  • Atomic orbitals overlap in a head-on manner as shown in the above diagram
  • There can only exist ONE σ bond between two atoms
Pi bond (π)

'side-by-side' (π) overlap'

Pi bond (π)
  • Formed when two p orbitals overlap in a side-on manner
  • A π bond can only exist after a σ bond is formed

Many students tend to think that both the sigma and pi bonds are equal in strength.
But a pi bond is weaker than a sigma bond. This is because of the less effective “side-on” overlap as compared to the “head-on overlap”

Dative bonding:

It is crucial that students remember the conditions for the formation of dative bonds. By using the ammonia-boron trifluoride complex as an example, students can check whether a compound can form a dative bond using the following few steps.

Step 1: The donor atom should have a lone pair of electrons (In this case, the N atom would be the donor atom due to having 2 unused electrons - N has 5 electrons in total in which 3 are used to undergo covalent bonding with H atoms, leaving a lone pair of electron)

Step 2 : The acceptor atom must have an empty orbital in its valence shell (In this case, the B atom has an empty orbital - meaning there is no electrons found in that orbital)
Note: Acceptor atoms are electron deficient species (More information regarding
electron deficient species can be found below)

Step 3 : The donor atom then donates the lone pair of electrons to fill the empty valence orbital of acceptor atom (represented by the arrow between the N and B atom)

Chemical Bonding

Ammonia-borontrifluoride complex

Students are often familiar with the octet rule, where it states that atoms have a tendency to have 8 electrons in their valence shell and carry out bonding in such a way that fulfills the octet rule in order to achieve the electronic configuration of a noble gas.
However, many students tend to forget that there are EXCEPTIONS to this rule, mainly:

1. Odd electron species

Chemical Bonding
  • Such species are known as radicals and are very reactive due to their unpaired electrons.
  • Common examples include NO, NO₂, ClO₂

2. Electron deficient species

  • Such species are molecules with fewer than 8 valence electrons. They hence have the potential to form dative bonds with other molecules
  • Common examples include BF₃, AlCl₃, BeCl₂
Chemical Bonding

3. Species with more than 8 valence electrons

  • Such species have the ability to expand their octet configuration and are elements found in period 3 onward. This is since period 3 elements and onwards have energetically accessible empty d-orbitals to accommodate more than 8 electrons.
  • Common examples are SF₆, PF₅, XeF₄
Chemical Bonding

Intermolecular Forces

The following table is a summary of the intermolecular forces for simple covalent molecules:

Type of Intermolecular Forces

Predominant Type of IMF Present

Instantaneous dipole-induced dipole (id-id) attractions (weakest)

Between all molecules (both non-polar and polar)

Permanent dipole-permanent dipole (pd-pd) attractions

Between polar molecules (Tip: molecules with net dipole moment)

Hydrogen Bonding (strongest)

Between a molecule with H bonded to F, O, or N and a molecule with lone pair of electrons on F, O, or N

Type of Intermolecular Forces

Factors Affecting IMF

Instantaneous dipole-induced dipole (id-id) attractions (weakest)

1. No. of electrons - The more electrons, the larger the atom’s electron cloud and the more easily the electron cloud is polarised,resulting in stronger id-id attractions
(Note: Do not use higher Mr although it implies a larger number of electrons. Use ‘number of electrons’.)

2. Surface area - straight-chain (more open structure) vs branched chain molecules (cluttered) Id-id attractions are more extensive for a straight-chain molecule due to its larger surface area.

Permanent dipole-permanent dipole (pd-pd) attractions

Magnitude of dipole moment.

Hydrogen Bonding (strongest)

1. No. of hydrogen bonds per molecule (which depends on the number of lone pairs of electrons on the F, O or N atom.The more lone pair the electronegative atom has, the stronger the hydrogen bonds.)

2. Electronegativity Difference.

Type of Intermolecular Forces Instantaneous dipole-induced dipole (id-id) attractions (weakest) Permanent dipole-permanent dipole (pd-pd) attractions Hydrogen Bonding (strongest)
Predominant Type of IMF Present Between all molecules (both non-polar and polar) Between polar molecules (Tip: molecules with net dipole moment) Between a molecule with H bonded to F, O, or N and a molecule with lone pair of electrons on F, O, or N
Factors Affecting IMF 1. No. of electrons - The more electrons, the larger the atom’s electron cloud and the more easily the electron cloud is polarised,resulting in stronger id-id attractions
(Note: Do not use higher Mr although it implies a larger number of electrons. Use ‘number of electrons’.)


2. Surface area - straight-chain (more open structure) vs branched chain molecules (cluttered) Id-id attractions are more extensive for a straight-chain molecule due to its larger surface area.
Magnitude of dipole moment. 1. No. of hydrogen bonds per molecule (which depends on the number of lone pairs of electrons on the F, O or N atom.The more lone pair the electronegative atom has, the stronger the hydrogen bonds.)

2. Electronegativity Difference.

Special Things to Note:

  • Do take note that polar molecules (molecules with net dipole moment) consist of both id-id and pd-pd attractions. The boiling point of polar molecules is only higher than non-polar molecules when they have a comparable number of electrons.
  • When intramolecular hydrogen bonds are present, the intermolecular hydrogen bonds in the molecules are less extensive.
Chemical Bonding

Continue Reading on Chapter 4: The Gaseous State