P-Block Flashcards
Define ionisation energy and explain the trend in ionisation energy across the p-block.
Ionisation energy is the the energy required to remove completely an electronfrom the gaseous atom or molecule in its ‘ground state’. Across the p-block there is generally an increase from left to right and a decrease as the group is descended. A sharp drop occurs when a new row is started.
What is the definition of Electron affinity (EA)? What is the trend in EA across the p-block?
Electron affinity is the energy released when a gaseous atom, molecule or ion in its ‘ground state’ gains an electron. Across the p-block there is a general increase in electron affinity across a period, with a slight dip for group 15 (most notable for smaller elements)
What is the definition of electronegativity? What is the trend in electronegativity across the p-block?
Electronegativity is the ability of an atom to attract electron density towards itself in a molecule. Across the p-block electronegativity increases left to right and decreases top to bottom.
What are the groupings of principle quantum numbers for Slater’s rules?
The groupings of principle quantum numbers for Slater’s rules are:
(1s), (2s,2p), (3s,3p), (3d), (4s,4p), (4d), (4f), (5s,5p)
What is the formula for calculating effective nuclear charge (Z*/Zeff)?
The formula for calculating effective nuclear charge (Z) is: Z = Z - s (Z=atomic number, s=shielding constant).
What are the rules for estimating the shielding constant, s?
The rules are estimating the shielding constant are:
1. Write out the full electron configuration and group the orbitals by principle quantum number as follows: (1s), (2s,2p), (3s,3p), (3d), (4s,4p), (4d), (4f), (5s,5p).
2. For ns/np electrons:
a) Each electron in the same grouping contributes s=0.35.
b) Each electron in the n-1 shell contribute s=0.85.
c) Each of the electrons in the n-2 or lower shells contribute s=1.00
3. For nf/nd electrons:
a) Each of the other electrons in the nd nf group contributes s=0.35
b) Each of the electrons in a lower group than the one being considered contributes s=1.00
What are the trends in size of covalent radii in the p-block?
In the p-block radii decrease from right to left because there is increasing nuclear charge means electrons are more tightly held which means the orbital size (and energy) decrease so radius contracts.
But radii increase from top to bottom. This is because orbitals of the next principal quantum shell are being populated and these are further from the nucleus.
What are the trends in size of ionic radii in the p-block?
In the p-block size of the ions depends on their charge Anions are larger because of greater inter electron repulsion from more electrons. Cations are smaller because of greater Z*.
Radii will also vary: as oxidation state changes because a higher oxidation = state higher Z** and
depending on coordination number, because for example ligands around a metal centre would mean more electron density at the metal and so an increased apparent size.
How does the difference in electronegativity between F & O affect the bonding in oxides and flourides?
The increase in electronegativity from Oxygen up to Fluorine vs O) shifts the boundary between covalent and ionic compounds. Atoms in fluoride compounds have higher Covalency parameter (which is the average electronegativity).
What are the rules for oxidation states in the p-block?
The rules for oxidation states are:
- similar electron configuration give similar oxidation state,
- The first row must statisfy an octet and there are only s and p orbitals.
- To the left of the block are generally +ve oxidation states whereas the right has -ve oxidation states.
What is the inert pair and effect and why does it occur?
The inter pair effect is the observation that as a group is descended, the n-2 oxidation state becomes more
favoured (where n = group oxidation state).
This is because the strength of covalent bonds decreases down a group due to poor orbital overlap. This means bond enthalpy does not offset the rehybridisation energy cost. Also, relativistic effects stabilise the 6s orbital for Tl and Pb.
What is the trend with multiple bonding in the p-block?
Generally the trend with multiple bonding in the p-block is that there are weaker bonds as group descended. This is because orbitals are more diffuse, and these are longer bonds. First row elements commonly form multiple bonds as there is good overlap with a small covalent radius of the atom.
What are The Triels?
The Triels are the Group 13 elements.
Why are The Triel’s compounds Lewis Acids?
Compounds of The Triels are Lewis Acids because they have fewer valence electrons than the number of valence orbitals. Meaning the orbitals are electron deficient.
Define a Lewis Acid and a Lewis Base?
A Lewis Acid is an electron pair acceptor. A Lewis Base is an electron pair donor.
Explain the bodning in BF3 as an example of halide bonding.
BF3 is a planar molecule and electrons from the full P-orbital on F are donated into vacant P-orbital on B. The pi bonding is not strong as there large electronegativity difference between B and F. Although it still has a major influence on theproperties of BF3:
Lewis acid strengths BI3 > BBr3 > BCl3 > BF3
What is the trend in Lewis Acid strength of boron halides and why is this seen?
Lewis Acid strength increases as you go down the halides i.e. boron triiodide is the most strong. This is because the valence electrons of I are larger, there is a much poorer match. Meaning pi contribution is weaker, so the B is least electron rich in triiodide.
Why can’t BH3 be formed?
BF3 cannot be formed because B cannot pi-bond with H as there are no lone pairs. Instead an electron deficient dimer (B2H6) is formed.
Why does valence bond theory fail for B2H6?
Valence bond theory fails for B2H6 because it cannot account for this bonding as there are not enough valence electrons.
What are the steps to assigning Wade’s Rules?
The steps to assigning Wade’s rules are as follows:
1. Count the number of BH units. Each gives 2e-
2. Each H atom gives 1e-
3. Include the overall charge (2- charge means add in 2e-)
4. This should get an EVEN NUMBER of polyhedral skeletal electron pairs (PSEPS), n.
5. n-1 = the number of vertices.
6. Use the number of BH units, if:
- No. of BH units = no. Vertices –> CLOSO - place each unit at one vertix.
- No. of BH = 1 fewer than no. of vertices –> NIDO. Place BH unit on all but one of the vertices.
- No. BH = 2 fewer than no. vertices –> ANARCHNO (COBWEB) –> fill all but two vertices. The unfilled vertices should be adjacent.
What name is given to group 14?
Group 14 is called The Tetrels.
What happens the stability of hydrides down group 14 and why?
Hydride stability decreases down group 14. This is because the bond between the Tetrel element and H get weaker due to larger, more diffuse orbitals. As well as more polar bonds, the metal not being shielded, and low lying LUMOs.
What property do the tetrels’ halides have and what does this mean?
The Tetrel’s halides are all hydrolytically stable meaning they don’t react with water. Meaning they’re susceptible to hydrolysis. This is because there are more polar bonds, low lying LUMOs, and electrophilic metal(loid).
Why is the Si-O bond so strong?
The Silicon-Oxygen bond is strong because of p-pi-d-pi bonding. The filled 2p orbital on the O can donate electron density into the empty 3d silicon orbital.
