Periodicity

Question 1

State and explain the trend in atomic radius across a period

Answer 1

Atomic radius decreases due similar shielding present, yet an increase in the electrostatic force of attraction between nucleus and outer electron shell due to an increase in nuclear charge since proton number is increasing

Question 2

Describe the trend in melting point across period 3

Answer 2

Between Na and Al:
General increase as metal ions have an increasing positive charge, increasing number of delocalised electrons and smaller atomic radius. This means an increasing strength of the metallic bond’
Si:
Highest M.P as it is giant covalent, lots of energy needed to overcome the many strong covalent bonds between atoms
P:
Much lower M.P than Si as it is simple molecular. Little energy is required to overcome the weak VDW forces between molecules
S to Ar
S has a slightly higher MP than P as it is a slightly bigger molecule, therefore stronger VDW forces
Trend from S to Ar is a decrease due to weaker VDW forces (Cl2 is a smaller molecule than P4 or S8, and Ar only exists as individual atoms)

Question 3

Why does Ar have the lowest melting point across period 3

Answer 3

It only exists as individual atoms and so it has much smaller VDW forces and thus, a lower melting point

Question 4

What is ionisation energy

Answer 4

The enthalpy change required to remove one mole of electrons from one mole of gaseous atoms

Question 5

What 3 factors affect ionisation energy

Answer 5

Shielding (increased shielding means lower ionisation energy)
Nuclear charge (Increase in nuclear charge means higher ionisation energy
Atomic radius (Increase in atomic radius means a lower ionisation energy

Question 6

What is periodicity

Answer 6

The repeating pattern of physical or
chemical properties going across the periods

Question 7

What is the trend in successive ionisation energies, and where do we get the biggest jumps

Answer 7

Trend is an increase in ionisation energy successively due to electrons being removed from a more positive ion when comparing electrons being removed from the same orbital

When comparing electrons from different orbitals, this is where we get the biggest jumps as removing electrons from an orbital closer to the nucleus requires much more energy due to an increase in the electrostatic force of attraction between positive nucleus and outer electron

Question 8

Explain the trend in ionisation energy down a group

Answer 8

Ionisation energy generally decreases
Increase in atomic radius
Increase in shielding
Weaker electrostatic force of attraction between nucleus and outer electron

Question 9

What does decreasing ionisation energy down a group prove

Answer 9

Proves that electrons are in shells

Question 10

State and explain the general trend in ionisation energy across period 3, what are the exceptions

Answer 10

General increase:

Increase in the number of protons, shielding is similar and atomic radius marginally decreases therefore more energy is required to remove an outer electron

Mg and S are the exceptions

Question 11

Explain the exceptions to the trend in ionisation energies across period 3

Answer 11

Al - The outer electron in Al sits in a higher energy 3p subshell which is slightly further from the nucleus and therefore has a lower first ionisation energy compared to Mg, whose outer electron sits in the 3s subshell

S - The outer electron in S sits in an orbital with 2 electrons in it. Thus, as opposite charges repel, less energy is required to remove an electron from an orbital with 2 electrons compared to an orbital with 1 electron such as in P