A.S. 91390
Chemistry 3.4
5 external credits

• Electrons are found in shells or energy levels around the nucleus.
• Energy levels are divided into subshells (s,p,d,f)
• Each subshell can have a number of s,p,d and f orbitals (place where e- is most likely to be found)
• E.g. 2 nd energy level has an
S and a P subshell, there are
3 P orbitals in the P subshell
• Note: f orbitals are not in the exam

• Electrons in the same subshell have similar energies
• Each orbital can hold 2 electrons
• S,p,d and f orbitals have characteristic shapes
• Electrons in each orbital have characteristic energies.

• 2,8,18,32
• Electrons fill orbitals with the lowest energy
• This is usually the one nearest to the nucleus
BUT…
• 4s sublevel is further from the nucleus but has a lower energy level than 3d, therefore it is filled first
• The electron configuration is written grouping orbitals in the same energy level together
• Manganese 2,8,13,2 is written
• 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 2
Or
[Ar]3d 5 4s 2
Exceptions
• Copper [Ar]3d 10 4s 1
• Chromium[Ar]3d 5 4s 1
They become stable by promoting an electron from the 4s subshell

• Electrons are represented by or
• Put 1 electron into each orbital before spin pairing them
• Draw the diagram to represent Boron,
Nitrogen, Neon and
Scandium

• Valence electrons of the transition metals are found in 4s and 3d
• The 4s electrons are further from the nucleus so they are lost first
• E.g. Mn [Ar]3d 5 4s 2 becomes Mn 2+ [Ar]3d 5
• Transition metals form coloured compounds because of the incomplete d sublevel

• Atomic radius decreases across the period because the valence electrons are pulled closer to the nucleus
• Positive charge of the nucleus increases
• Electrons are added to the same energy level
• Electrostatic attraction between valence electrons and nucleus increases
• Atomic radius increases down the group because of the additional electron shells and the increased shielding by inner electrons, even though there are more protons in the nucleus the effective electrostatic attraction decreases .

• Positive ions are formed when atoms lose valence electrons
• Fewer electrons means less repulsion between them so the ionic radius is smaller than the atomic radius
• Loss of all the valence electrons results in the removal of an energy level

• Negative ions are formed when atoms gain valence electrons
• More electrons means more repulsion between them so the ionic radius is bigger than the atomic radius

• IE is the minimum energy, measured in kJmol -1 , needed to remove an electron from each atom in a mole of gaseous atoms.
• M
(g)
M +
(g)
+ e -
• It will be endothermic as energy is taken in to break the electrostatic attraction of electron and nucleus
• Down a group IE decreases
• It is easier to remove the electron because it is further from the nucleus and the inner shells shield it so the electrostatic attraction is less.
• Across a period IE increases
• It is harder to remove electrons because the positive nuclear charge increases but the electrons are added to the same energy level, there is no extra shielding so the electrostatic attraction increases

• The small decrease in IE represents a subgroup change (s p)
• The electron is further from the nucleus so less energy is needed to remove it
• 3 unpaired p subshell electrons are more stable as there are fewer repulsions occurring
• Group 1 metals have the lowest IEs because they react to lose electrons
• Non-metals have high IEs
• Removing subsequent electrons from an atom will require successively more energy because the attraction exerted by the nucleus is effectively stronger
ESA pg 70 Q2,3

• EN is the attraction of the nucleus to a bonded pair of electrons and depends upon:
– Positive nuclear charge
– Atomic radius
– Shielding
• EN decreases down a group , despite the increase in protons, because of increased distance from the nucleus and the additional shielding effect of inner shells
• EN increases across a row because electrons are in the same energy level the atomic radius is the same and there is no additional shielding, but there are more protons so the electrostatic attraction increases

Molecular formula
Atom placement
Sum of valence e -
Remaining valence e -
Lewis structure
For NF
3
Place atom with most incomplete shell in the centre
: F :
N
: :
F :
N 5e -
F 7e X 3 = 21e -
Total 26e -
Draw single bonds.
Subtract 2e for each bond.
Give each atom 8e -
(2e for H)

• Follow the same process remembering to add or subtract the additional electrons
E.g. H
3
O +
• Enclose the diagram in square brackets and show the charge
O = 6e -
H 1e X 3 = 3e -
1 + ion so - e -
Total 8e -
+
H H
O
H

• Atoms with atomic number >14 can use vacant d orbitals to have
5 or 6 pairs of electrons around the central atom
(P and S are commonly used in exams)
• Draw the Lewis diagram for PCl
6
, XeF
4 and POCl
(P is the central atom)
3

• Base shape is dependent on the areas of electron density around the central atom (maximum repulsion)
• Actual shape depends on the number of atoms joined to the central atom.
• What shape would the molecule be if there were 3 areas of electron density around the central atom, but 2 were bonded and 1 was non-bonded?
• V- shape e.g. SO
2
,NO
2
linear trigonal planar
Examples:
Cl
2
, CO
2
, HCN
Examples:
SO
3
, BF
3
, CO
3
2-

Molecular Shapes
If there are 4 areas of electron density around the central atom, all will repel maximally to give a tetrahedral base shape with a bond angle of 109.5
• If all 4 regions are bonded the shape is tetrahedral
• If 3 regions are bonded and 1 is non-bonded the shape is trigonal pyramid e.g. NH
3
• If 2 regions are bonded and 2 are non-bonded the molecule is V- shaped e.g. H
2
O
• In each case the bond angle remains 109.5
tetrahedral

Molecular Shapes
If there are 5 areas of electron density around the central atom, all will repel maximally to give a trigonal bipyramid base shape with bond angles of 120 and 90
• If all 5 regions are bonded the shape is trigonal bipyramid e.g. PCl
5
• If 4 regions are bonded and 1 is non-bonded the shape is see-saw e.g. SF
4
• If 3 regions are bonded and 2 are non-bonded the molecule is T- shaped e.g.
ICl
3
• Add non-bonding electrons to the equatorial plane first

Molecular Shapes
If there are 6 areas of electron density around the central atom, all will repel maximally to give an octahedral base shape with bond angles of 90
• If all 6 regions are bonded the shape is octahedral e.g. PCl
6
• If 5 regions are bonded and 1 is non-bonded the shape is square based pyramid e.g. SF
5
• If 4 regions are bonded and 2 are non-bonded the molecule is square planar e.g. ICl
4

Non-Polar covalent
• Bonding electrons are shared equally
• No difference in EN
• Usually seen when identical atoms bond
• E.g. H
2
Polar covalent
• Bonding electrons are shared but not equally
• Small difference in
EN (0.5-2)
• Usually seen when 2 different non-metals bond
• E.g. HCl
Increasing E.N. difference between atoms
Ionic
• Electrons are transferred
• big difference in
EN (>2)
• Seen when a metal and a nonmetal bond
• E.g. NaCl

•
2
2
•

•
2
•
• This is known as a bond dipole .
• The green atom is slightly negative and the blue one is slightly positive

• Molecular polarity occurs if the individual bond dipoles do not cancel and so give the whole molecule a slight charge.
• Bond dipoles cancel if they are arranged symmetrically .
-water is a polar molecule because… oxygen is more electronegative than hydrogen, and therefore bond electrons are pulled closer to oxygen. The shape of the molecule means the bond polarity is not symmetrical , so dipoles will not cancel .

Draw the shape, find the EN values, determine the polarity of the bonds and whether the molecule is polar or not.
(a) Ammonia, NH
3
The shape is trigonal pyramidal.
molecule is asymmetrical so bond dipoles don’t cancel and molecule is polar
H
N
H
H
EN
N
= 3.0
EN
H
= 2.1
H
N
H
H bond polarity
H
N
H
H molecular polarity

(b) Boron trifluoride, BF
3 has 24 valence e and all electrons around the B will be involved in bonds.
F
The shape is trigonal planar.
Bond angle?
Bond polarity?
EN
F
EN
B
= 4.0
= 2.0
F
B
120 0
F
B and F have different EN, hence the bonds are polar. The shape is symmetrical so the dipoles cancel and the molecule is non polar

• 3D lattice of metallic atoms surrounded by a delocalised sea of mobile valence electrons
• Held together by the electrostatic attraction of the metal cations for the valence electrons
( metallic bonding )
• Going down a group decreases the attraction between nucleus and valence electrons so the strength of the metallic bond is weaker.
• M.P. Li > Na > K > Rb
• The more valence electrons present in the ”sea”, the stronger the metallic bond
• M.P. Na < Mg < Al

• Melting Point
– Varies, but often high
(lots of energy needed to break metallic bonds)
• Solubility
– Not soluble. Strong bonds are not broken by the attraction of solvent molecules
• Conductivity
– Very good (due to the free moving electrons)
• Hardness
– Can be deformed without breaking (due to the nondirectional bonds)
– Metals can be hammered into shape (malleable) and drawn into wires (ductile)

• 3D Lattice of positive (cations) and negative (anions)
• Held together by strong electrostatic attraction between the ions called ionic bonding
• E.g. MgF
2
F is a lattice of Mg in a 1:2 ratio
2+ and
• Strength of the bond depends on ionic charge and radius so CaO
(Ca 2+ and O 2) has a higher M.P. than NaCl (Na + and Cl )
• LiF has a higher M.P. than LiCl
(ionic radii F = 0.133nm, Cl =
0.181nm)
N.B. Name the ions in exams

• Melting Point
– High (lots of energy needed to break strong bonds)
• Solubility
– Most dissolve in water
(strong attraction of polar water molecules overcomes the ionic bond)
• Conductivity
– Do not conduct electricity
(ions are not free to move)
– If dissolved or molten they will conduct
• Hardness
– Hard but brittle (a force can cause layers of ions to shift so they repel each other)

• As atomic size increases the strength of the covalent bond decreases. This is due to lower electrostatic attraction between nucleus and bonding electrons
• Bond dissociation enthalpy is the energy required to break a mole of bonds in the gaseous state.
• Single bonds are longer and weaker than double and triple bonds
• Intramolecular bonds are made and broken when chemicals react
• Intermolecular bonds determine the physical properties (MP, BP)
• Intramolecular forces are always stronger than intermolecular ones.

(van der Waals)
• Caused by molecular dipoles which can be temporary or permanent.
• Temporary dipole-dipole attractions occur because at any moment in time electrons may gather on one part of the molecule making it slightly negative.
• The temporary dipole in one molecule can induce a dipole in an adjacent molecule which results in them being attracted to each other
• Greater molecular mass means more electrons therefore greater chance for temporary dipoles and higher
MP/BP

(van der Waals)
• Permanent dipole- dipole attractions occur in polar molecules
• Temporary dipole-dipole attractions will also still occur so more energy will be required to break the intermolecular forces and MP/BP will rise
• Hydrogen bonding occurs in molecules which contain H-F, H-O or
H-N.
• The big EN difference allows a strong attraction between the positive H and non-bonding electrons from an adjacent molecule
• The small size of H allows it to get close to adjacent molecules

Explain the trend in the boiling points of the hydrogen halides .
• HF has hydrogen bonding aswell as temporary and permanent dipoles so needs a lot of energy to break the intermolecular forces which explains the high BP. HCl, HBr and HI are all polar and HCl will have the greatest EN difference so will have the strongest permanent dipoles, however it has the lowest molecular mass so it has fewer temporary dipoles than HBr and HI. The increase in BP for HCl<HBr<HI is due to greater molecular mass and therefore temporary dipoles rather than permanent.

• Ice has lots of hydrogen bonds forming a lattice network
• Adding energy makes the molecule vibrate until enough is present to break some intermolecular forces and disrupt the lattice.
• The energy added goes into breaking bonds so the average temperature remains constant as the state changes (melting/ fusing)
• Latent heat of fusion is the energy needed to convert 1 mole of solid into 1 mole of liquid Δ fus
H o positive (bond breaking is is always endothermic) as energy is absorbed
H
2
O
(s)
H
2
O
(l)
H
2
O
(l)
Δ fus
H o = +6 kJmol -1
Write the equation for the freezing of water
H
2
O
(s)
ΔH o = -6 kJmol -1

• Heating causes the liquid particles to move faster (E k increases)
• Adding energy makes the molecules vibrate until enough is present to break more intermolecular forces and form a gas.
• The energy added goes into breaking bonds so the average temperature remains constant as the state changes (vaporising)
• Latent heat of vaporisation is the energy needed to convert 1 mole of liquid into 1 mole of gas Δ vap
H o is always positive (bond breaking is endothermic) as energy is absorbed
H
2
O
(l)
H
2
O
(g)
H
2
O
(g)
Δ vap
H o = +41 kJmol -1
Write the equation for the condensing of water
H
2
O
(l)
ΔH o = -41 kJmol -1

• Strong intermolecular forces mean:
• High MP/BP
• High latent heat of fusion and vaporisation
• Volatile liquids have low
Δ vap
H o so the easily become gaseous
• Some substances (I
2
, CO
2 have a lower BP than MP therefore thy have no liquid state (at 1 Atm), they sublime Δ sub
H o
)

• Enthalpy is the energy in a substance
Enthalpy of products and reactants cannot be measured, but… enthalpy change during a reaction can be measured as energy is released or absorbed from the surroundings in standard conditions
• ∆H is the difference between the energy needed to break the bonds in the reactants, and the energy given out when new bonds are made in the products. It is measured in Kj/mol.
∆H reaction
= ∑E bonds broken
- ∑E bonds made

• Energy is given out
(exits) from the chemical bonds
• The products have less energy than the reactants
• Surroundings gain heat energy
• Enthalpy Change is negative ( ∆H )
Combustion and neutralisation reactions are exothermic

• Energy is taken in
(enters) the chemical bonds
• The products have more energy than the reactants
• Surroundings lose heat energy
• Enthalpy Change is positive ( ∆H + )
• Photosynthesis or gas liquid solid

Surroundings
Bonds made
Energy
Exothermic reaction ∆H -
Energy
Bonds made
Solid
Bonds broken
Liquid
Bonds broken
Gas
Energy
Endothermic reaction ∆H +
Energy

• Allows a reaction to follow a different path which requires less activation energy . This means a greater proportion of particles have sufficient energy to overcome the activation barrier and therefore there is a higher frequency of effective collisions per unit time .

• 0.01g Mg
(s) reacts with HCl and the temperature rises producing 100.8J. Calculate the enthalpy of reaction in kJmol -1 Moles of Mg n=m/M n= 0.01/24 n= 4.167 x 10 -4 moles
• ∆H = ∆E /n
= 100.8/4.167 x 10 -4
= -242kJmol -1

• Δ f
H o is the energy released or absorbed when 1 mole of product is formed under standard conditions
• Na
(S)
+ 0.5Cl
2(g)
NaCl
(S)
Δ f
H o
• Write the equation for the formation of H
-411kJmol -1
2
SO
4(l)
• H
2(g)
+ S
(s)
+ 2O
2(g)
H
2
SO
4(l)
• In aqueous systems the energy change of an exothermic or endothermic reaction is transferred into the water .
• This heat energy transferred in the reaction (q) can be calculated in a calorimeter using: q= m x c x ΔT
• C = specific heat capacity of water = 4.18Jg
-1
• m = mass in g
• T = temperature in o C
Don’t multiply out the 0.5 as this would produce 2 mol of product

• 0.4g of hexane was burned to heat 150ml of water. The temperature changes from 22 o C to 47 o
C. Calculate the value of ∆H.
• M
C6H14
= 86.2gmol
-1 water is 4.18Jg
-1o C -1 and specific heat capacity of
• q= m x c x ΔT
• q = 150 x 4.18 x 25 = 15675J
• m/M = n 0.4/86.2 = 4.64 x10 -3
Mass of the water 1ml =1g
• ∆H = 15.675/4.64 x10 -3 = 3380kJmol -1
Don’t forget to add the sign, is it exothermic or endothermic?

• The enthalpy change due to a chemical reaction is independent of the pathway taken.
• N
2
(g) + 3H
2
(g) → 2NH
3
(g) ΔH f
= - 92.4 kJ mol -1 reversing the reaction changes the sign of the enthalpy change
• 2NH
3
(g) → N
2
(g) + 3H
2
(g) ΔH r
= + 92.4 kJ mol -1

• Calculate Δ
-394kJmol f
-1
H o C
2
H
5
OH
(l) given Δ and Δ f
H o H
2
O
(l) c
H o C
2
H
=-286kJmol
5
-1
OH=-1367kJmol -1 , Δ f
H o CO
2(g)
=
• Write the equation for the formation of ethanol
• 2C
(s)
+ 3H
2(g)
+ 0.5O
2(g)
C
2
H
5
OH
(l)
• Write the equation for the combustion of ethanol
• C
2
H
5
OH
(l)
+ 3O
2(g)
3H
2
O
(l)
+ 2CO
2(g)
Δ c
H o =-1367kJmol -1
• 3H
2
O
(l)
+ 2CO
2(g)
C
2
H
5
OH
(l)
+ 3O
2(g)
Δ c
H o =+1367kJmol -1
• Write the equation for the formation of carbon dioxide
• C
2(g) 2(g)
CO
2(g) f
H o =-394kJmol -1 -1 x2
• Write the equation for the formation of water
• H H
2 (l) 2
Δ O f
H o Δ H o =-858kJmol -1
• Add together equations and enthalpies
• 3H
2
O
(l)
3O
2(g)
+ 2CO
+ 2CO
2(g)
2(g)
+ 2C
(s)
+ 3H
2
O
+ 2O
2(g)
+ 3H
2(g)
(l)
+ 1.5O
2(g) x3
C
2
H
5
OH
(l)
+ reverse
Δ f
H o C
2
H
5
OH
(l)
= -788kJmol -1 + -858kJmol -1 +1367kJmol -1 = -279kJmol -1

•
• The enthalpy change for a reaction can be calculated if you know the Δ f
H o of the reactants and the products
Δ f
H o for elements is always 0
• Given
∆ r
H o = ∑Δ f
H o
(products)
- ∑Δ f
H o
(reactants)
•
•
•
• Calculate Δ r
• Δ
Δ
Δ
Δ
– Δ f
H o NH
3(g)
– Δ f
H o NO
(g)
= -46kJmol -1
= +90kJmol -1
– Δ f
H o H
2
• 4NH
3(g)
O
(g)
= -286kJmol -1
H o for this reaction:
+ 5O
2(g) r
H o = 360 + (-1716) - (-184) r
H o = 544 + (-1716) r
H o = -1172 kJmol -1
4NO
(g) r
H o = 4(90) + 6(-286) - 4(-46) + 0
+ 6H
2
O
(g)
Don’t forget this value is per mole

• Entropy is a measure of disorder in a system
• The universe tends towards disorder.
• Entropy increases with increasing temperature
• Increasing the particles in a system also increases the entropy because there are more ways in which the particles can be arranged
• ∆S positive or negative?
• I
2(s)
• 2NO
2(g)
I
2(g)
N
2
O
4(g)
Positive, gases are more disordered negative, fewer moles of gas therefore more ordered

• ∆S o = ∑S o
(products)
- ∑S o
(reactants)
• Spontaneous reactions proceed towards disorder so ∆ r
• ΔG is the Gibbs free energy which depends on:
S o >0
– enthalpy (ΔH)
– entropy (ΔS)
– absolute temperature (T, in kelvin).
Calculations based on
Gibbs free energy are not in the exam
∆G = ∆H - T∆S
• When ΔG is negative, a process or chemical reaction proceeds spontaneously in the forward direction.
• When ΔG is positive, the process proceeds spontaneously in reverse.
• When ΔG is zero, the process is already in equilibrium, with no net change taking place over time.
• Is it spontaneous?
• C
2
H
5
OH
(l)
+ 3O
2(g)
2CO
2(g)
+ 3H
2
O
(l)
∆ c
H o = -1369kJmol -1
• It is spontaneous because entropy increases (+∆S) and it is exothermic so enthalpy is negative (-∆H) so ∆G must be negative

• ∆H 0 r
• ∆H 0 fus
= enthalpy change for any reaction
= enthalpy change for solid liquid
• E.g. H
2
O(s) H
2
O(l)
• ∆H 0 vap
= enthalpy change for liquid gas
• E.g. H
2
O(l) H
2
O(g)
• ∆H 0 c
= enthalpy change for combustion reaction
• E.g. CH
4
+ 2O
2
CO
2
+ 2H
2
O
• ∆H 0 f
= enthalpy change for formation of a product from its atoms
• E.g. C + 2H
2
CH
4