Periodicity and Bonding Flashcards
Generally what happens to bonds as we go down a group?
- Generally, the bonds become weaker and longer down a group
- This is due to less of the valence orbital volume overlaps and thus sharing of e- density is less effective
What happens to orbital charge density going down a group?
Larger valence orbital volume but only ever 2e- per orbital, so more diffuse orbitals with lower charge density going down a group
Why are multiple bonds formed predominately for first period elements
e.g. C=C or C=N
- 2p-π orbitals (used in multiple bonding) where there is good overlap - due to small covalent radius of E
- 3p-π orbitals used with larger covalent radius of E - poor p-π overlap and single bonds favoured - too big essentially to form multiple bonds
Why is the C-O bond not as strong as the Si-O bond
- Si is a larger atom, hence has larger valance orbitals BUT there is also a bigger difference in electronegativity
- Silicon is not very electronegative compared to carbon
- So there is a stronger electrostatic interaction between silicon and oxygen
- This means that σ-bonds are favoured for silicon compounds over π
What is the name of the interaction which stabilises multiple bonds?
pπ-pπ overlap
Explain the bonding interactions for a P=O double bond?
- Electron donation from a filled p-orbtial on the oxygen, to an vacant d-orbtial on the phosphrous
- The phosphorus already has 4 bonds but can expand it octet using the d-orbitals
- Resulting in pπ-dπ bonding
When comparing NMe₃ to N(SiH₃)₃
VSEPR theory predict pyramidal geometry (C₃v)
However, the Si-containing molecule is planar (D₃h)
Why?
- Generally believed that planarity of N(SiH₃)₃ comes from a pπ-dπ interaction
- Due to overlap of N p-orbtials with d/σ’ orbitals on Si + enhanced steric factors = trigonal planar
- (NMe₃ cannot do the same due to no d-orbitals)
Define valance
This is the number of valence electrons used by an atom to form bonds to other things
Group 17 halogens form how many bonds to give a full valance ocetet
Halogens have 7 valence e-
Hence will form 1 bond to give a full valance ocetet
Heavier elements can expand their ocetet however
Group 16 will form how many bonds to give a full valance ocetet
Will form two bonds to give a full valence octet
Heavier elements can expand their ocetet
Group 15 will form how many bonds to give a full valance octet
3 bond gives the valance octet (3 bond pairs and 1 lone pair - lewis bases)
Heaier elements again can expand their octet
Group 14 will form how many bonds to give a full valance octet
4 bonds gives full valance octet
Multiple bonding common for C (can be π or σ, so many geometries possible)
Pi-bonding less common down the group
Group 13 will form how many bonds to give a full valance octet
- only 3 valance electrons, so can only have a valance of 3 bonded pairs giving only 6 electrons
- So the triel compounds are electron deficient (Lewis acids)
Generally, bonds get weaker down a group due to larger valance orbitals, less effective overlap/sharing of e- density
Why is period 2 an exception?
The second period is an exception due to greater electrostatic repulsion
Which is pushing the atoms apart, hence making the end weaker
For smaller atoms, N to F, with more charge-dense orbitals
Why is the H-H bond so strong?
H-H is an exception despiting having high electronegativity
Smaller orbitals, good overlap, and effective sharing of electron density
There is no electrostatic repulsion (only 2e- present) = very strong bond
Why does bond strength increase from left to right going across a period
Going from left to right, the atom gets smaller and more electronegative (from 1-14)
Leading to more effective orbital overlap and sharing of electron density, hence stronger bonds
Why is the B-F bond so strong?
- Small orbtials and effective overlap meaning good sharing of electron density
- The bond is also polar covalent so there is additional electrostatic attraction
- There is also pi, donation from lone pairs of F to empty p-orbitals on B is possible
Why is the Al-F bond weaker than the B-F bond?
- Al is a larger atom, hence the valence orbitals will be more diffuse
- Therefore will have poorer sharing of electron density
The heteronuclear bond energies for In-F and Tl-F are relatively the same
However, the main difference is that the oxidation state of the metals is different, why?
- In-F has an O.S. of (iii)
- While Tl-F has an O.S of (i)
- Due to the inert pair effect
- BUT due to the big difference in electronegativity between Tl and F, the bonding will be more ionic which weakens the strength of the bond (hence not really a fair comparision)
The heteronuclear bond energies for B-F is stronger than the C-F bond (485)
Why?
- C-F bond is weaker than the B-F bond
- There is a smaller difference in electronegativity going left to right (lower ΔΧ)
- Hence reduced electrostatic interaction
- (no pi-donation can occur for these period 2 elements as all C orbitals are filled)
The Heteronuclear bond energies for C-F is weaker than for Si-F
Why?
- Si-F stronger than C-F
- Larger difference in electronegativity for Si-F (higher ΔΧ) meaning increased electrostatic interaction
- Silicon has the ability of expanding it’s ocetet due to having d-orbitals, so can undertake pπ-dπ bonding (from F LPs)
The Heteronuclear bond energies for N-F is weaker than for P-F
Why?
- P-F stronger than N-F
- N is highly electronegative, meaning the difference in electronegativity is lower (smaller ΔΧ) and decreased electrostatic interaction
- N and F have LPs and are smaller so there will also be increased electrostatic repulsion
- AND phosphorus has d-orbitals available for pπ-dπ bonding
The Heteronuclear bond energies for P-F is stronger than for S-F
Why?
- P-F is stronger than S-F
- There is a lower difference is electronegativity for S-F (lower ΔΧ), meaning decreased electrostatic interactions
- S and F both have multiple LPs and are small = increased electrostatic repulsion greater than for P
The following graph shows group 14 chlorides
The top row is the ethanlpy of formation
The Bottom row is M-Cl bond strength
Explain the trends
- Enthalpy of formation is negative because energy is released through forming bonds
- Generally there is a decreases as we go down the group = less exothermic as the bonds get weaker
- This is due to larger valance orbitals and hence weaker bonds
- Apart from C-Cl as carbon is weakest due to smaller difference in electronegativity - less electrostatic interactions
The following list shows the preferred oxidation state for group 14 chlorides
Expain why?
- Both carbon and silicon are tetrachlorides due to having only 2e- in the p orbitals (ocetet rule)
- Both Ge and Sn can be tetrachlorides or dichlorides due having d-obitals enabling pπ-dπ interactions
- Pb can also be tetrachlorides or dichlorides but dichlorides are favoured due to inert pair effect and relativistic effects
