Period 3

Question 1

Trend in atomic radius as you go across period 3 and why?

Answer 1

• Atomic radius decreases.
• Increase in nuclear charge/ shielding remains similar so the outer electrons/ shells are more strongly attracted to the nucleus (drawing the electron shells in, decreasing distance between positive nucleus and outer electrons.)

() - extra info for clarification.

Question 2

Why is it that when you go across periods (from group 2 to group 3 horiontally), there is a small dip in the ionisation energy? (Ie. when you go from Mg to Al.)

Also in “ionisation energy” deck.

Answer 2

• (Outer) Electron in higher-energy 3p sub -orbital.
• Electron further away from the nucleus.
• Little more electronic shielding (from previous 2s sub-orbital.)
• Electrons are more easily lost from the higher- energy sub-orbital.

Question 3

Why is there a slight dip in ionisation energy when going across periods from group 5 to group 6?

Also in “ionisation energy” deck

Answer 3

• At this point, 2 electrons need to be paired up in the p orbital.
• So the electrons in the same sub-orbital will repel each other (so the electron is more easily lost - even though the nuclear charge increases.)
• Shielding is THE EXACT SAME (because we HAVE NOT gone into another orbital!!)

Question 4

DESCRIBE trends in melting points of period 3

Answer 4

• Na to Al: melting point increases.
• Si: very high melting point relative to rest of the elements.
• Melting point increases betwee P and S and then decreases to Cl and decreases even more to Ar.

Refer to graph on gautum notes, file name: “Gautum period 3”

Question 5

EXPLAIN why melting point increases from Na to Al in period 3.

Answer 5

• Ionic charge increases and ionic radius decreases
• Stronger metallic bonds will be formed.
• More energy required to break the metallic bonds as you go from Na to Al.

Question 6

EXPLAIN why Si has a very high melting point relative to the rest of elements in period 3.

Answer 6

• Giant covalent structure.
• Large amount of energy required to break strong, covalent bonds in Si.

Question 7

EXPLAIN why melting point increases between phosphorous to sulphur and then decreases again back to chlorine.

Answer 7

• Non-polar molecules (as all are elements), only have Van der Waals forces between their molecules.
• Size of molecule/ Mr increases from Cl₂ to P₄ to S₈, there is larger surface area for VDW forces to act over, therefore more energy required to break the VDW forces.

Question 8

EXPLAIN why Ar has the lowest melting point out of all the elements in period 3.

Answer 8

• Ar exists as individual atoms with weak VDW forces between its atoms, relatively small amounts of energy are required to break these weak VDW forces.

Question 9

Trend in ionic radius across period 3.

Answer 9

• Na⁺, Mg²⁺, Al³⁺, ionic radius decreases because no of protons increases and no of electrons stays same.
• P³⁻, S²⁻, Cl⁻, these 3 ions are larger than Na⁺, Mg²⁺, Al³⁺ - due to an extra electron shell but ionic radius will decrease from P3- to Cl- here because no of protons increases and no of electrons stay same.
  (same principle as atmic radius in terms of ionic radius decreasing due to electron shells being drawn inwards!!)

() - extra info for clarification.