Periodic table templates

Atomic radii decreases across period 3. Explain why.

• Number of protons increases, increasing nuclear charge

• Number of electrons increases but they are all added to the Valence electron shell. Shielding effect remains relatively constant

• Effective nuclear charge increases, causing stronger electrostatic forces of attraction between nucleus and Valence electrons

• Hence Valence electrons are closer to the nucleus, causing atomic radii to decrease

Describe and explain the trend of ionic radius across period 3 elements.

• Anions are larger than cations

• All of these anions have 10 electrons

• Number of occupied electron shells remain the same

• Shielding effect remains constant

• As charge increases, the electrostatic forces of attraction between the nucleus and the Valence electrons increase

• Hence Ionic radius of anions decreases

• There is a large increase in ionic radius from Si4+ to P3-

• P3- has an additional occupied electron shell

• Shielding effect is greater in P3- than Si4+

• Electrostatic forces of attraction between nucleus and Valence electron is lower in P3- than in Si4+

• Cations are larger

Describe and explain the trend of the 1st ionisation energy across period 3 elements.

• 1st I.E. generally increases across period 3

• As the number of protons increases, the nuclear charge increases

• However, the number of quantum shells remain the same so shielding effect remains constant

• Effective nuclear charge increases

• Leads to stronger electrostatic forces of attraction between nucleus and Valence electrons

• 1st I.E. decreases from Mg to Al.

• Valence electron in Al is in the higher energy 3p subshell

• Valence electron in Mg is in the lower energy 3s subshell

• Less energy needed to remove one Valence electron from Al than from Mg.

• 1st I.E. decreases from P to S

• Inter electron repulsion present in the 3p sushell in S

• Less energy needed to remove a Valence electron from S than from Mg

Describe and explain trend of electronegativity across period 3

• Electronegativity increases across period 3

• As number of protons increases, nuclear charge increases

• Number of quantum shells remain constant so shielding effect is constant

• Effective nuclear charge increases

• Stronger electrostatic forces of attraction between Valence electrons and nucleus

Describe and explain the trend of melting points of oxides across period 3.

• Na2O, MgO and Al2O3 all have giant ionic lattice structures.

• There are strong electrostatic forces of attraction between the oppositely charged ions.

• Large amounts of energy is needed to overcome the electrostatic forces between the ions

• Melting point of MgO is higher than Na2O

• Mg2+ cation has higher ionic charge and smaller atomic radius than Na+ cation

• Mg2+ has greater charge density than Na+

• Stronger electrostatic forces of attraction between oppositely charged ions in MgO compared to Na2O

• More energy needed to overcome stronger ionic bonding in MgO than in Na2O

• Melting point of Al2O3 is lower than MgO

• Al3+ cation has the highest charge density and is able to polarised O2-

• This gives Al2O3 some covalent character which weakens Ionic bond strength

Describe and explain the trend of melting points of oxides across period 3.

• NaCl and MgCl2 have giant ionic lattice structure with strong electrostatic forces of attraction between oppositely charged ions

• Large amount of energy required to overcome the electrostatic forces of attraction

• Mg2+ has a relatively high charge density compared to Na+

• Mg2+ polarises the electron cloud of the Cl- anion

• There is a partial covalent character within the ionic bonding

• AlCl3 has a simple molecular structure with weak intermolecular id id forces of attraction

• Little energy required to overcome the id id forces of attraction

• The Al3+ ion has the highest charge density of all metal captions

• Al3+ ions have the greatest polarising power so it is able to polarise the Cl- anion to the greatest extent

• Al-Cl bond is covalent so the compound is a simple molecular structure

• Hence AlCl3 has a low melting point

• SiCl4 and PCl5 both have simple molecular structures with weak intermolecular id id forces of attraction

• Little energy required to overcome id id forces of attraction

• Hence SiCl4 and PCl5 both have low melting points

• Size of electron cloud in SiCl4 is less than in PCl5

• Extent of intermolecular id id forces of attraction is lower in SiCl4 than in PCl5

• Less energy needed to overcome id id forces of attraction in SiCl4 than in PCl5

• Hence SiCl4 has a lower melting point than PCl5