Week 1

Question 1

What factors effect shielding?

Answer 1

• Electrons in orbitals with higher penetration (like s orbitals) are closer to the nucleus and thus can shield outer electrons more effectively
• principal quantum number
  n is significant in determining the distance of electrons from the nucleus and their energy levels

Question 2

What is the principle of Slater’s rules?
What is the flaw?

Answer 2

• the actual charge felt by an electron is equal to what you’d expect the charge to be from a certain number of protons, but minus a certain amount of charge from other electrons
• assumes all shielding is equivalent

Question 3

What is the trend of effecive nuclear charge?

Answer 3

LEFT TO RIGHT INCREASE
nuclear charge (no. protons) increases and the electrons are added to the same principle quantum numberthe additional shielding provided by the same-shell electrons is not sufficient to completely counteract the increase in nuclear charge. As a result, Zeff experienced by the outermost electrons rises
DOWN A GROUP DECREASE
Moving down a group, the electron is added to a higher principle quantum number level further from the nucleus, experiencing significant shielding and so lowering the nuclear charge felt

Question 4

What is the trend of atomic/ covalent radii?

Answer 4

LEFT TO RIGHT DECREASE
As more protons are added to the nucleus, the Zeff felt by the electrons increases - the electrons are pulled closer to the nucleus, resulting in a smaller atomic radius.
DOWN A GROUP INCREASE
each successive element down a group has an additional principal energy level which means the outermost electrons are further from the nucleus and so are pulled less tightly and have a larger radius

Question 5

Define: ionisation enthalpies

Answer 5

the energy required to remove an electron from a gaseous atom or ion

Question 6

Describe the general trend of ionisation enthalpies

Answer 6

LEFT TO RIGHT INCREASE
electrons are added to the same principal energy level, so the effective nuclear charge Zeff felt by the electrons increases. As a result, electrons are held more tightly to the nucleus, requiring more energy to remove an electron
DOWN A GROUP DECREASE
As you move down a group, electrons are added to higher principal energy levels, which are farther from the nucleus and so electrons are held less tightly and require less energy to remove

Question 7

What factors effect ionisation enthalpies?

Answer 7

• Nuclear Charge: Higher nuclear charge increases ionization enthalpy because the electrons are more strongly attracted to the nucleus.

-Atomic Radius: Larger atomic radius decreases ionization enthalpy because the outer electrons are farther from the nucleus and less tightly bound.

-Shielding Effect: Increased shielding from inner electron shells decreases ionization enthalpy because the effective nuclear charge felt by the outer electrons is reduced.

-Electron Configuration: Atoms with stable electron configurations (such as noble gases) have very high ionization enthalpies because removing an electron would disrupt their stability.

Question 8

What is the reason for the N O ionisation enthalpy kink?

Answer 8

Nitrogen (N): 1s² 2s² 2p³
The 2p orbital is half-filled, which is a relatively stable configuration (symmetrical distribution minimises repulsion)
Oxygen (O): 1s² 2s² 2p⁴
The 2p orbital has one more electron than a half-filled state, leading to increased electron-electron repulsion

Expected trend: Increase from N to O
Actual: slight decrease from N to O
Reason: N configuration is stabilised compared to O so requires more energy to move an electron

Question 9

Why does Mg have a higher IE than Al?

Answer 9

Mg: 1s² 2s² 2p⁶ 3s²
Al: 1s² 2s² 2p⁶ 3s² 3p¹
3p electron is less tightly bound due to increased shielding and higher energy level so the electron is easier to remove

Question 10

What is exchange energy?
Compare the exchange energies of N and O.

Answer 10

The stabilization energy that arises from the exchange interaction between electrons of the same spin in degenerate orbitals
e.g N: The 2p orbitals are half-filled with parallel spins, resulting in high exchange energy and increased stability
O: The additional electron in one of the 2p orbitals leads to a pairing of spins, which decreases the exchange energy compared to nitrogen, reducing its relative stability

Question 11

What is electron attachment energy?

Answer 11

The energy change (typically released) when an electron is added to a neutral atom in the gas phase to form a negatively charged ion. It is usually expressed in units of kJ/mol

Question 12

What is the trend of electron affinity?

Answer 12

It is more exothermic if the added electron stabilises the configuration e.g reduces Columbic repulsion or completes a subshell
GENERAL: increase across a period and decrease down a group

Question 13

Describe the trend of covalent bond strength in s block

Answer 13

Group 1 - generally weak
Alkali metals (Li, Na, K, Rb, Cs, Fr) have a single valence electron in an s orbital. These atoms have large atomic radii, especially as you move down the group, which means the valence electron is far from the nucleus and less tightly held
Li - strongest; Cs - weakest

Group 2 - stronger than group 1 but still weak compared to p block
atoms are smaller than their Group 1 counterparts and can form stronger covalent bonds due to a higher effective nuclear charge and smaller atomic radii
Be - strongest, Ba - weakest

Question 14

Why is Ga harder to ionise than Al?

Answer 14

Ga: [Ar] 3d¹⁰ 4s² 4p¹
Al: [Ne] 3s² 3p¹
- Ga has a smaller covalent radius due to the post-transition effect
-3d orbitals (2 angular nodes) are poor at shielding 4s and 4p electrons. 4s and 4p orbitals have higher electron probabilities near the nucleus than 3d, so penetrate better
4p orbital of Ga is at similar energy to 3p of aluminium => similar IE1

Question 15

Why does 6th period deviate from group trends?

Answer 15

-inclusion of filled f-orbitals for the first time, 4f orbitals shield poorly (3 angular nodes), both 6s and 6p orbitals experience increase in Zeff (6s more so as more penetrating)
- Relativistic effects, for heavy elements the mass of electrons increase due to travelling at speeds ca. 50% of speed of light. This effects electrons in 6s orbitals most
MAJOR EFFECT: 6s orbital stabilised/contracted and energy lower compared to that expected by the group trend

Question 16

Most common oxidation states:
Tl+1, Pb+2, Bi+3 Why?

Answer 16

Inert pair effect - the 6s electrons are inert and don’t contribute
e.g Tl is grp 13 but 2 electrons don’t contribute so end up with +1

Question 17

PbCl4 ->
TlCl3 ->

Answer 17

PbCl4 -> PbCl2 + Cl2
TlCl3 -> TlCl + Cl2

Question 18

How do ionic radii vary with oxidation state? E.g. Which will possess the larger radius Pb2+ or Pb4+ ?

Answer 18

r(Pb2+) > r(Pb4+) : the more positive the oxidation state the higher Zeff is on the remaining electrons, and therefore the smaller the size.

Question 19

How might coordination number effect ionic radius? Eg which will have an apparent smaller ionic radius Si in SiF4 or in [SiF6]2- ?

Answer 19

The 4 coordinate Si will appear smaller, as four fluorines can be placed closer to Si than six, so SiF4 has shorter Si-F bonds and therefore appears to be the “smaller” Si.

Question 20

How would χ vary with oxidation state? E.g. Tl(III) vs Tl(I) ?

Answer 20

As oxidation state increases the orbital energy will become more negative (more
stabilised) as Z remains the same, but the number of electrons decreases. Therefore χ increases with oxidation state. E.g. Tl(III) = 2.04, Tl(I) = 1.6.

Question 21

Suggest reasons to explain why gallium and mercury are low melting point metals. [Ga m.p.=30°C, Al>660°C, Hg m.p.=-39°C, Cd>320°C.]

Answer 21

Due to insertion of d-block and f-block elements orbitals of the elements that follow are contracted so less good overlap of the valence orbitals hence atoms are held less tightly, so lowering the melting point.

Question 22

Explain whether Fr will have a higher or lower (IE1) compared with Cs.
Element IE1 (KJ mol-1)
Li 513
Na 495
K 419
Rb 403
Cs 376
Fr ?

Answer 22

Fr 392
These ionisations are removal of an electron from an s-orbital, which will be effected by the relativistic effects that arise as elements become more massive. So we expect the 7s orbital to be contacted and stabilised, therefore higher IE1 (although not by very much).