Periodic Trends State and explain the following trends: the answers

State and explain the following trends: 1.Atomic radius across period 3 2.Electrical conductivity from Na  Al 3.Melting/boiling points across period 3 4.First ionisation energy across period 3 5.Atomic radius down group 2 6.ionisation energy trend down group 2

Atomic radius across period 3 Decreases across the period, because the outer electrons are all in the same shell, but the number of protons is increasing so the electrostatic attraction for the outer electrons increases therefore reducing the radius.

Electrical conductivity from Na  Al Increases, as in the metallic lattice Al forms 3+ ions, compared to Na which forms 1+ ions, therefore there are a greater number of delocalised electrons in Al.

Melting/boiling points across period 3 Na  Al: metallic bonding, Na forms 1+ions whereas Al forms 3+ ions. In Al there is a greater electrostatic attraction between the higher number of delocalised electrons and the ionic lattice, therefore, more energy is required to overcome the stronger metallic bond. Si: giant covalent structure (no intermolecular forces) requires a large amount of energy to break the covalent bonds. P  Ar: discreet molecules (or atoms in the case of Ar) held together by weak intermolecular forces, (temporary-induced dipoles).

First ionisation energy across period 3 General trend: increase because atomic number is increasing but the outer electrons are in the same shell, therefore the electrostatic attraction between the nucleus and the outer electrons is greater. Mg  Al: decrease because the outer electrons is now in a p orbital which is at a higher energy and therefore easier to remove. P  S: decrease because P is 3p3 and therefore the electrons are unpaired with parallel spins. S is 3p4 has a pair of electrons in a p orbital so there is a greater mutual repulsion between electrons so they are easier to remove.

Atomic radius down group 2 Increases because there is an increasing number of energy levels/shells. Although the atomic number does increase the electrostatic attraction between the nucleus and the outer electrons decreases due to more shielding by full shells.

ionisation energy trend down group 2 decreases because the number of full shells increases, shielding the nucleus from the outer electrons. The outer electrons are also further from the nucleus. These reduce the electrostatic attraction between the nucleus and the outer electrons, despite the increase in atomic number, making them easier to remove.

reducing power down group 2 Reducing power increases as the elements become more easily oxidised. I.e. they lose electrons more easily – see point 6

Boiling points down group 7 Increase – molecules increase in size, therefore a larger instantaneous dipole can occur leading to stronger van der Waals forces

oxidising power down group 7 Oxidising power decreases as the elements become less easily reduced. I.e. they gain electrons less easily – because the number of full shells increases, shielding the nucleus from the outer electrons. The outer electrons are also further from the nucleus. These reduce the electrostatic attraction between the nucleus and the outer electrons, despite the increase in atomic number, making them harder to gain.

State and explain the following trends: 1.Atomic radius across period 3 2.Electrical conductivity from Na  Al 3.Melting/boiling points across period 3 4.First ionisation energy across period 3 5.Atomic radius down group 2 6.ionisation energy trend down group 2 7.reducing power down group 2 8.Boiling points down group 7 9.oxidising power down group 7