Trends

Question 1

Period 3:

• Atomic Radius
• 1st Ionisation Energy
• Electronegativity

Answer 1

• general decrease; same no. shells, more protons w/ attractive force on outer e-, so more inward attraction.
• general increase; more protons, larger nuclear charge, stronger nuclear attraction to outer e-, harder to lose.
• general increase; more protons, more nuclear attraction to outer e-, smaller atomic radius,no more nuclear shielding, outer e- closer to nucleus, more attraction.

Question 2

Period 3:

• Electrical Conductivity
• Melting and Boiling Points

Answer 2

• increase then decrease; Al more delocalised e- than Na/Mg so more e- to carry charge. Si,PS no delocalised e- so no charge carried.
• B.P: Na-Al increase, more delocalised e-, stronger metallic bonds between atoms. Al-Si decrease as only weak i-m forces in Si (as liquid) but metallic bonds in Al. Remainder general decrease; simple molecular structures so weak i-m forces, in order of decreasing no. e-. (S>P>Cl>Ar)
  M.P: Na-Al increase, more delocalised e-, stronger metallic bonding. Increases Si as giant covalent structure, strong covalent bonds. Remainder decreases as simpler molecular structures so weak i-m forces, in order of decreasing no. e-. (S>P>Cl>Ar)

Question 3

Group 2:

• Atomic Radius
• 1st Ionisation Energy
• Melting Points
• Reactivity

Answer 3

• general increase; increasing nuclear shielding as more e-.
• general decrease; atomic radius increases, more nuclear shielding, weaker nuclear attraction to outer e-.
• general decrease; all metallic bonding, size of ions increases while charge remains 2+, larger ions less able to attract outer delocalised e- so weaker metallic bonding.
• general increase; all group 2 trying to lose 2 outer electrons- for full outer shell. more nuclear shielding so less nuclear attraction t outer e- so can be lost easier.

Question 4

Group 7:

• Atomic Radius
• Electronegativity
• Melting and Boiling Points

Answer 4

• general increase; more nuclear shielding as more e-.
• general decrease; atomic radius increases, weaker nuclear attraction to outer e- as more nuclear shielding.
• general increase, more e- so stronger van der waals forces.

Question 5

Group 7:

• 1st Ionisation energy
• Oxidising power
• Reducing power

Answer 5

• general decrease; larger atomic radius, more nuclear shielding, weaker nuclear attraction to outer e-
• general decrease; chlorine strongest as smallest ionic radius and least nuclear shielding, so strongest nuclear attraction to outer e-.
• general increase; iodine strongest as largest ionic radius and most nuclear shielding, so weakest nuclear attraction to outer e-.