Periodic Table Consolidation

Question 1

Describe electrical conductivity across Period 3

Answer 1

Across Period 3, electrical conductivity

• increases from sodium to aluminium because the number of delocalised valence electrons increases
• drops sharply from aluminium to silicon(semi – conductor) because the latter is a metalloid and it has poor electrical conductivity
• decreases to zero from phosphorus to argon due to absence of mobile valence electron/charge carriers

Question 2

Across Period 3, the atomic radius of the elements decreases

Answer 2

This is because

• nuclear charge increases as the number of protons increases
• shielding effect remains approximately constant due to the same number of inner electronic shells
• effective nuclear charge increases
• valence electrons are more strongly attracted to the nucleus
• atomic radii decrease

Question 3

Across Period 3, the electronegativity of the elements increases

Answer 3

This is because

• nuclear charge increases as the number of protons increases
• shielding effect remains approximately constant due to the same number of inner electronic shells
• effective nuclear charge increases
• the attraction for the bonding electrons to itself increases
• electronegativity increases

Question 4

Across Period 3, the first ionisation energy of the elements generally increases

Answer 4

This is because

• nuclear charge increases as the number of protons increases
• shielding effect remains approximately constant due to the same number of inner electronic shells
• effective nuclear charge increases
• the amount of energy required to remove the electron increases
• first ionisation energy increases

Question 5

The first ionisation energy of Al is lower than Mg.

not impt

Answer 5

The outermost electron in Mg is in the 3s orbital, while the outermost electron in Al is in the 3p orbital. As the average distance of a 3p orbital from the nucleus is slightly larger than that of a 3s orbital the outermost electron in Al experiences less electrostatic attraction, and hence less energy is required to remove the 3p electron in Al.

Question 6

The first ionisation energy S is lower than P.

not impt?

Answer 6

The electron to be removed in P is one of the three 3p electrons and it is unpaired and the mutual repulsion between these electrons is minimised.

However, in S, the electron to be removed in the 3p-orbital is paired and since these two electrons experience inter-electronic repulsion, it is easier to remove one of them than the p electron in S. Hence less energy is required to remove it.

Question 7

explain how chlorides of aluminium, silicon, and phosphorus exhibit acidic properties when dissolved in water

IMPTTT

Answer 7

AlCl 3 (s) + 6 H2O (l) → [Al(H2O)6 ] 3+ (aq) + Cl– (aq)

[Al(H2O)6 ] 3+ (aq) + H2O (l) ⇌ [Al(H2O) 5(OH– )]2+ (aq) + H3O+ (aq)

AlCl 3 reacts with water to form hydrated ions. Since Al 3+ ions have a high charge density, it has a high polarising power to polarise the electron cloud from the water molecules, weakening the O – H bond, producing H3 O + ions in water, thereby undergoing partial hydrolysis to form an acidic solution of around pH 3.

For SiCl 4 and PCl5 ,

SiCl 4 (l) + 2H2O(l) → SiO 2 (s) + 4HCl(aq)

PCl 5 (s) + 4H2O(l) → H3 PO4 (aq) + 5HCl(aq)

Both chlorides hydrolyse completely in water due to **energetically accessible vacant 3d orbitals ** which can accommodate the lone pairs of electrons from water to form very acidic solutions of around pH 2.