Lecture 4 - Periodic trends in bonds and bonding Flashcards
(21 cards)
Learning objectives
4.1 Explain the relative strengths of single vs multiple bonds
4.2 Exemplify importance of multiple bonding in first row
4.3 Discuss how homonuclear bond strengths vary through periodic table
4.4 Discuss how heteronuclear bond strengths vary due to size and electronegativity differences
4.5 Discuss patterns in oxidation states of elements
4.6 Describe the continuum from ionic to covalent to metallic bonding as electronegativity of elements changes
4.7 Discuss bonding type in terms of concepts of Van Arkel-Ketelaar triangle
Single vs multiple bonds
Energy KJ/mol for:
O-O: 142
O=O: 498
S-S: 263
S=S: 431
Use these numbers to explain why O2 is more favourable and S8 is more favourable
So for X2 (ie X=X)
The amount of energy released per atom:
O: 498/2 = 249
S: 431/2 = 215.5
for X8 (where all are single bonds and attached to 2 X’s)
O: ( 8 × 142)/8 = 142
S: (8 × 263)/8 = 263
( you multiply by 8 as there are 8 atoms but divide by 8 as its per atom)
As you can see:
For O: E(O=O) > 2 × E(O-O) so double bonds are favoured
For S: O: E(S=S) > 2 × E(S-S) so single bonds are favoured
The reason for this is that the O=O bond is shorter than the S=S so there is better pi overlap in O=O
What is the general rule for whether single or multiple bonds are favoured?
If double bond bond energy > 2 x single bond bond energy, you should expect multiple bonding
Bond enthalpy / homonuclear bond strength trend summary
Down the group bonds get weaker
Across a row bonds get stronger
N-N, O-O and F-F are all weaker than expected
Note H2 is strongest
Bond enthalpy / homonuclear bond strength down the group
Down the group bonds get weaker
- atoms get bigger
- orbitals are more diffuse (bigger + “floppy) basically means weaker bonding
- less overlap
Bond enthalpy / homonuclear bond strength explaining N-N, O-O and F-F
The lower values are due to the antibonding effect
pi* more antibonding then pi and bonding
Most marked for 2p elements where overlap is strongest
In other words antibonding orbitals are more antibonding than bonding orbitals are bonding - check slide 15 for diagram
Bond enthalpy / homonuclear bond strength across the row
Bond strength increases
- Zeff increases
- atoms get smaller
- orbitals get less diffuse (more contracted + closer to nucleus)
Decreases down the table ( ie smaller gradient across period 2 to compare to period 4) so effect more marked for 2p than 4p series
Look at F2 MO diagram on slide 16
Explain why F-F is weak
|E1| < |E2| as in energy of |pi|>|pi|
pi population therefore makes bond partially weak
Effect is more pronounced for F than Cl
so F-F is weaker than expected
Heteronuclear (A-B) bond strength considers 4 key parts
What are they?
Also memorise eq???
- Size of element
- Lone pairs
Both effect E(A-A) x E(B-B) - EN difference
- “Extra pi bonding”
Both effect (EN(A) - EN(B))^2
Heteronuclear equation
√( E(A-A) x E(B-B) ) + 96.5( (EN(A) - EN(B) )^2
x = average E
- = difference in EN
Compare C-F and C-I to work out which has a stronger bond
(EN can also mean X when given values in exams)
F: small, with high EN
I: big with low EN
As size increases, bond gets weaker
As EN difference gets greater, bond gets stronger
Therefore C-F must be the stronger bond
Compare C-X to Si-X
C: small with high EN
Si: bigger
C has a higher EN than Si and as EN difference increases, bonds get stronger
Second effect “wins” in this case but can be reverted
What is “extra pi-bonding”
Look at slide 18 for a diagram
For instance
∩ ∩ (supposed to imitate p orbitals pretend they
X — X are the same sizes) overlap between filled and
U…..Ü empty p-orbital isn’t great
\ // ∩ (empty Dxz orbital on bottom right of S
S — F filled px orbital
// \ Ü leads to partial pi bond formation due to proximity
Oxidation state vs charge
Oxidation states are listed like +2,+3 etc
Charges are listed like 2+,3+ etc
Oxidation states are all to do with EN, they are a formalism
What are common oxidation states for groups 1,2,13,14,15,16,17 and 18
1: always +1
2: always +2
13: +1 or +3
14: +2 or +4
15: -3, +3 or +5
16: -2, +2, +4, +6
17: -1, +1, +3, +5, +7 (chlorate for eg is +7)
18: 0, +2, +4, +6
Common oxidation states differ by 2
There are many exceptions but this is a general idea
Group 13-15 often show oxidation states 2 lower than group oxidation state - “inert pair effect”
Transition metal oxidation states differ by 1 unit
Inert pair effect examples - (it is to do with the 6s2 e-s)
+3 BCl3 ✓ :BCl ✗
| AlCl3 ✓ :AlCl ✗
| GaCl3 ✓ :GACl ✓
| InCl3 ✓ :InCl ✓
+1 TlCl3 ✗ :TlCl ✓
As you can see for group 13 it favours +3 and as you go down it begins to favour +1
✓ for both is known as reflex oxidation states
Chlorine oxidation states for compounds containing Cl and O
Cl- (-1), ClO-(+1), ClO₂- (+3), ClO₃- (+5), ClO₄-(+7)
Main group oxidation states usually differ by 2 units in compounds
Cl (O), Cl=O (+2) and O=Cl=O (+4) are super unstable (basically don’t exist) as they are radicals so don’t get these oxidation states
Predicting bonding type in compounds eg compound AB using their EN values
1)A low B high (eg CsF) is ionic | top of triangle
2)A low B low (eg CsSn) is metallic | bottom left corner
3)A high B high (eg ClF) is covelant | bottom right corner
Covelant bonding
High EN and more electrons favour localised covelant bonding through more compact orbitals
What is the Van Arkel-ketelaar triangle
Plots the average EN: (EN(A) + EN(B))/2 |x axis
Against difference in EN: (EN(A) - EN(B)) |y axis
Metal —> non-metal along bottom
Metal —> ionic along left hand edge
Ionic —> covalent down right hand edge
elements are Aₙ with zero EN difference ie only made of themselves eg H2, Be etc. pure elements along the bottom
100% metallic = Cs, 100% ionic = CsF, 100% covenant = F2
BeBr2 is almost completely in the center of the triangle - good to just have a rough idea of entire triangle
