Uol Ch1200 (General Chemistry) - Topic B Revision

Front 1

Chemical Bond

Back 1

The respective forces that hold groups of atoms together:

• Covalent - the overlap of AO.

• Ionic - electrostatic attraction between oppositely charged ions.

• Metallic - electrostatic attraction between M⁺ˣ and delocalised e⁻.

Front 2

Classification of Bonding

Back 2

Bond triangle:

Based on the electronegativities of the involved atoms.

Front 3

Covalent Bonding

Back 3

Is the sharing of e⁻ between atoms to reach a lower E. state. Resulting in AO overlap of similar E. lvls. with the same symmetry around the bond axis

Front 4

Ionic Bonding

Back 4

Chemical bonding that arises from the electrical attraction between an electro +VE atom (cation) and an electro

-VE atom (anion).

Front 5

Metallic Bonding

Back 5

Metallic bonding results from the attraction between M⁺ˣ and the surrounding delocalised e⁻ (sea of electrons).

Front 6

Degree of Bond Polarity

Back 6

Degree of unequal sharing of e⁻ between respective atoms.

•Covalent: Equal share of electrons (homonuclear)

•Polar covalent: unequal share e⁻. The EN atom has more e⁻ density (heteronuclear)

•Ionic: transfer of electron from a low EN atom to a high EN to form a cation and anion

Front 7

Bonding and states of matter in covalent

Back 7

Can be 3D solids but not required

This bonding is found with small molecules

Front 8

Bonding and states of matter in metallic/ionic

Back 8

3D network of interactions

Solids or liquids but never as gas state

Front 9

Octet rule

Back 9

A general trend that atoms try to satisfy for stability. A filled valance orbital - 8 e⁻.

Front 10

Hypervalency

Back 10

Central atoms which break the Octet rule of having 8 val. e⁻ around an atom, e.g. SF₆.

Period 3 and subsequent elements show deviation from the Octet rule, compared to period 2. As a result of low-lying unfolded d-orbitals which are able to accommodate the extra e⁻.

Front 11

Lewis Acid

Back 11

e⁻ pair acceptor- electron-deficient (electrophile).

E.g. BF₃

Front 12

Lewis Base

Back 12

e⁻ pair donator - electron rich (nucleophile).

E.g. NH₃

Front 13

Dative Covalent Bond

Back 13

An e⁻ lone pair which is provided by 1 respective atom only.

Front 14

Working out Lewis Structure

Back 14

1.Calculate Σval. e⁻

2.Arrange atoms in order of connectivity

3.Distribute in pairs (2 for each bond)

4.Assign double bonds/LP until each atom is satisfied.

Note:

•For cations subtract e⁻.

•For anions add e⁻.

Front 15

Brønsted-Lowry Acid

Back 15

[H⁺] donor

Front 16

Brønsted-Lowry Base

Back 16

Accepts [H⁺]

Front 17

VSEPRT (Valence Shell Electron Pair Repulsion Theory)

Back 17

1. Decide on central atom (most electro +VE).

2. Deduce Σval. e⁻ around the central atom inc. adjacent bonding e⁻.

3. Distribute Σval. e⁻ as BP / LP.

4. Identify geometry.

Note:

•BP and LP of e⁻ around centre atom are positioned as far apart as possible to reduce repulsion.

•Double bonds count similar to single bonds.

Front 18

VSEPR - Multiple Bonds

Back 18

Dealing with molecules with double/triple Bonds (π bonds):

•The π bonds occupy the same space as the σ e⁻. So double bonds count similar to single bonds. I.E. a double bond contains 2 e⁻ to count for.

Front 19

VSEPR Example

POCl₃

Back 19

Central atom: P (grp. 15)

•P: 5 val. e⁻

•Cl (x3): 3 val. e⁻

•O (2x1): 2 val. e⁻ (double bond)

Σval. e⁻ = 10.

5 e⁻ pairs but 2 e⁻ in a π bond so 10 - 2 = 8 e⁻ for σ bonding.

∴ Tetrahedral arrangement

Front 20

VSEPRT - Ions

Back 20

•For anions add relevant e⁻ on the central atom based on the charge.

•For cations subtract relevant e⁻ on the central atom based on the charge.

Front 21

VSEPR Example

SO₄²⁻

Back 21

Central atom: S (grp. 16)

•S: 6 val. e⁻

•O (2x4): 8 val. e⁻

•Add 2 from –VE charge

Σval. e⁻ = 16

but 8 e⁻ in π bonds so 16 - 8 = 8 electrons for σ
bonding.

∴ Tetrahedral arrangement

Front 22

LP Equatorial Vs. LP Axial?

Back 22

The more stable structure will have minimised e⁻ - e⁻ repulsion.

•LP – LP > LP – BP > BP – BP. For most repulsion.

Thus, (A) is preferred

Front 23

VSEPR Example 4

SF₄

Consideration:

Think about the position of the e⁻ pair on the trigonal bipyramidal. Is it equatorial or axial site?

Back 23

Central Atom S ( grp.6)

•S: 6 valence electrons

•F (x4): 4 valence electrons

Σval. e⁻ = 10, i.e. 5 e⁻ pairs

•4 bond pairs

•1 lone pair

∴ Trigonal bipyramid

Front 24

Linear

Back 24

No. of e⁻ pairs: 2

No. of BP pairs: 2

No. of LP pairs: 0

θ = 180°

Front 25

VSEPR Example

BeCl2

Back 25

Central atom Be (group 2)

•Be: 2 val. e⁻

•Cl (x2): 2 val. e⁻

Σval. e⁻ = 4 i.e. 2 pairs

•2 bond pairs only

∴ shape is linear

Front 26

Trigonal planar

Back 26

•No. of e⁻ pairs: 3

•No. of BP pairs: 3

•No. of LP pairs: 0

•θ =120°

Front 27

Tetrahedral

Back 27

•No. of e⁻ pairs: 4

•No. of BP pairs: 4

•No. of LP pairs: 0

•θ =109.5°

Front 28

Trigonal bipryamidal

Back 28

•No. of e⁻ pairs: 4

•No. of BP pairs: 5

•No. of LP pairs: 0

•θ = 90° (axial) and 120° (equatorial)

Front 29

VSEPR Example:

PCl₅

Back 29

Central atom P (grp. 15)

•P - 5 val. e⁻

•Cl (x5) - 5 val. e⁻

Σval. e⁻ = 10, i.e. 5 pairs

•5 BP -

∴ shape is trigonal bipyramidal

Front 30

Octahedral

Back 30

No. of e⁻ pairs: 6

No. of BP pairs: 6

No. of LP pairs: 0

θ = 90°

Front 31

Bent geometry

Back 31

No. of e⁻ pairs: 3

No. of BP pairs: 2

No. of LP pairs: 3

θ = <120°

Front 32

Triggonal Pyramidal

Back 32

No. of e⁻ pairs: 4

No. of BP pairs: 3

No. of LP pairs: 1

θ = <109.5°

Front 33

VSEPR Example:

NH₃

Back 33

Central Atom N (grp. 15)

•N - 5 valence electrons

•3 x H - 3 valence electrons

Σval. e⁻ = 8 i.e. 4pairs

•3 bond pairs

•1 lone pair

∴ arranged as a trigonal pyramidal (distorted tetrahedral).

Front 34

Bent (distorted tetrahedral)

Back 34

No. of e⁻ pairs: 4

No. of BP pairs: 2

No. of LP pairs: 2

θ = <109.5°

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See-Saw

Back 35

No. of e⁻ pairs: 5

No. of BP pairs: 4

No. of LP pairs: 1

θ = <120° (equatorial), 90° (axial).

Front 36

Square Pyramidal

Back 36

No. of e⁻ pairs: 6

No. of BP pairs: 5

No. of LP pairs: 1

θ = 90°

Front 37

T-Shaped

Back 37

No. of e⁻ pairs: 5

No. of BP pairs: 3

No. of LP pairs: 2

θ = 120° (equatorial), 90° (axial)

Front 38

Square Planar

Back 38

No. of e⁻ pairs: 6

No. of BP pairs: 4

No. of LP pairs: 2

θ = 90°

Front 39

Confronting Molecular Orbitals

Back 39

Front 40

MO theory

Back 40

MO theory allocates e to MO. Formed by AO with comparable E. and same symmetry with respect to the bond axis.

It can be summed up in 4 main points

Front 41

MO Principle 1:

Σ No. MO = Σ No. AO

contributed by the atoms that have combined.

Back 41

When atomic orbital “A” and “B” both bond together ---> x2 MO. E.G. take H₂, both AO is a 1s orbital. If 2 were to join there would be 2 MO (bonding and anti-bonding).

Front 42

MO Principle 2:

In terms of E.

bonding orbitals < parent orbitals.

Whilst,

anti-bonding orbital > than the parent orbital.

Back 42

Bonding MO are constructive (in-phase) interactions between AO in a molecule. e in bonding orbitals help stabilize the system

Antibonding electrons formed by deconstructive (out-of-phase) combination of AO. These destabilized system.

Front 43

Principle 3: Electrons are assigned to orbitals of successively higher E. (Aufbau principle)

Back 43

Electrons fill bonding orbitals first since lower E than the anti bonding (Aufbau principle).

No more than 2 electrons can go into each orbitals and are singularly filled before pairing ( Hund's rule).

Front 44

Principle 4

AO ---> MO most effectively when the AO is of similar E.

Back 44

MO likely to form when AO is of similar E.

E.G. 1s AO combinining with another 1s AO. Forming a stable MO.

Front 45

Calculating BO

Back 45

Stability of a molecule can be determined via calculating BO.

I.E:

0 - no bond

0.5 - half-bond

1 - 1 bond

2 - 2 bonds

3 - 3 bonds