The Chalcogens Flashcards
What is the general formula of an oxyacid
OₘE(OH)ₙ
How can you work out the pKa of an oxyacid?
pKa ~ 8-5ₘ
(ₘ being the number of E=O units)
If you use 8-5ₘ to work out the pKa of an oxyacid, what does this tell us?
1) The strength of an acid is independent of the number of OH groups, n (the OH groups do not allow delocalisation of the charge on the anion)
2) The strength of the acid increases by 10⁵ for each E=O moiety (pKa decreases by 5)
A strong acid has a….
- Large Ka
- Small pKa
Using Pauling’s rules, what is the pKa for the following acid?
ₘ = 1x N=O unit
pKa = 8-5ₘ
8-5(1) = 3
pKa = 3
Using Pauling’s rules, what is the pKa for the following acid
ₘ = 2x N=O units
pKa = 8-5ₘ
8-5(2) = -2
pKa = -2
The following molecule below is an example of a phosphorus oxyacid
Which part of the molecule is ionisable
- The P-OH are ionisable
- The P-H are not ionisable
Phosphorus oxyacids can have more than one ionisable P-OH
But…
Each sucessive ionisation becomes less favourable as the charge is less stabilised (buy the E=O)
e.g. as more protons are lost from H₃PO₄, the pKa increases (i.e. it becomes less acidic)
How would you form sulfurous acid (H₂SO₃)
By adding sulfur dioxide to water
What two isomers of sulfurous acid exist? (H₂SO₃)
Either with 1x S=O or 2x S=O
How can you from Sulfuric acid (H₂SO₄)?
By reacting sulfur trioxide with water
Balance the following equation
Group 17 elements have a wide range of common oxidation state for oxides and halides
Apart from…
Fluorides
Due to being highly electronegative
What are the names of the following oxyacids of group 17 halogens?
HOXO
HOXO₂
HOXO₃
- HOXO - halous; chlorites, bromites
- HOXO₂ - halic; chlorates, bromates, iodates
- HOXO₃ - perhalic; perchlorates, perbromates, periodates
The higher the oxidation states for oxyacids of group 17 halogens, it is less favourable for the halogen due to how electronegative they are
So they are
Good oxidising agents
(can easily be reduced again)
Explain the trend for electron affinity of the group 16 elements
- The electron affinity for O is lower than S because of its very small size, resulting in increased electrostatic repulsion when an electron is added
- S and Se have similar values (despite Se outshell being further away) due to the poor shielding of the d-block (d-block contraction)
- Te and Po have similar values (despute Po outshell being further away) due to poor shielding of the f-block (f-block contraction)
Second period elements tend do have fewer oxidation states due to not having a d-orbital
Do oxygen follow this trend?
No
Oxygen has a range of oxidation states
Higher oxidation states of +6 and +4 are available for S, Se and Te
But only +2 and +4 is available for Po
The inert pair effect
What are the two main allotropes for oxygen?
O₂ and O₃
Considering the electron configuration of oxygen is: 1s² 2s² 2p⁴
Fill in the following molecular orbital diagram for dioxygen
Considering the electron configuration of oxygen is: 1s² 2s² 2p⁴
Fill in the following molecular orbital diagram for dioxygen in the excited state
(Hint: there are two possibilties)
How is O₃ formed?
How does this reaction not use up all of the O₂ in the atmosphere?
O₃ is formed at high electrical potential, at a surface in an ozonizer, and low concentrations under UV irradiation
The back reaction (O₃ + O → 2O₂) is highly exothermic
Other Chalcogens (like oxygen) are pretty reactive
What are the main two ways in which chalcogens react?
- Dissolve in oxidising agents (e.g. sulfur dissolves in nitric acid and forms oxyacids)
- Attacked by halogens (e.g. S₈ + 24F₂ → 8SF₆)
What compounds of sulfur exist and why?
- Sulfur has many allotropes
- Due to S-S single bond being favoured due to the poor pi-orbital overlap
- A range of allotropes exist as the S-S-S bonds distances and angles can vary greatly and cycles are thermodynamically favoured (no “dangling” bonds i.e. an unsatisifed valence on an immobilised atom)
Group 17 elements can form two bonds with hydrogen (i.e. EH₂)
Explain the difference in bond angles shown below
- The Te-H bonds involve p-character from the central ‘E’ atom
- i.e. very little contribution from the valence s orbital
- There are 2 LP on Te in py and s atomic orbitals
Why is H₂S more volatile than H₂O
No Hydrogen bonding in H₂S
How can we form group 17 hydrides?
EH₂ are formed by reacting a chalcogenide with an acid
(the other hydrides are highly toxic and very pungent)
Why are the following values so different for water compared to other group 17 hydrides?
Hydrogen bonding
For the other group 17 hydrides the strongest interaction is dipole-dipole
What happens to the strength of the group 17 E-H bond, going down the group?
- Bond gets weaker down the group (hence bond length also gets longer)
- Due to larger orbitals and hence poor overlap
What happens to the bond angle of the group 17 hydrides, going down the group
- Goes from v-shaped 104°C for water to 90° for S, Se and Te
- Due to little hybrisation S to Te (p-orbitals used in E-H bonds)
What happens to acidity going down group 17 hydrides?
- Acidity increases down the group
- due to weakening of E-H bond
Sulphur can form SF₄ (iv) and SF₆ (vi) fluorides
Why can’t oxygen?
What are the properties of SF₄ and SF₆?
- Because it cannot expand its ocetet
(no d-orbitals) - SF₆ is chemically very inert in contrast to SF₄ - due to SF₄ having a LP
SeF₆ is much more reactive than SF₆
Why?
- Both S and Se have similar electronegativities and hence similar bond strength (electrostatic interactions)
- SF₆ is more sterically hindered due to being smaller and hence less susceptible to attack compared to SeF₆
- SeF₆ has d-orbtials which are available for bonding
- Sulfur being in period 2 does not
- Electrons in the s-orbital penetrate most closely to the nucleus
- Thus, you would expect s-orbitals to always be lower in energy than p, d, or f or hybrids
- So why are sp³ hybrid orbitals used in NH₃ if they are higher in energy than s-orbitals?
For smaller atoms, with small valence orbitals, electrostatic repulsions are more significant than for larger atoms in period 3 and below
Drago’s rules: if the following are true, then there is no need to consider hybridisation for the central atom
1) the central atom has at least one lone pair of electrons
2) the central atom belongs to group 13, 14, 15, 16 and the 3rd to 7th period
3) if the electronegativity of the surrounding element is 2.5 or less
4) Sum of sigma bonds + lone pairs = 4
Which elements does this apply for?
Looking at the table below, work out if the following compounds are hybridised or not
