C8 Flashcards
Trend in atomic radius along a period?
Decreases because as you go along the period, each element will have more protons, therefore a greater nuclear charge, however will have the same amount of inner shells, therefore will have a similar shielding effect. Therefore as you go along the period, this greater nuclear charge and similar shielding will cause the valence shell to be pulled in closer to the nucleus, decreasing the size of the atom
Trend in atomic radius down a group?
Increases because as you go down a group, each atom will have an extra electron shell, therefore the shielding effect each atom will have down the group will increase, as a result, as you go down the group it becomes harder for the nucleus to pull the valence electrons closer to itself therefore the atomic radius will increase
Trend in ionisation energy across a period?
Across a period, ionisation energy increases because across the period, each element has more protons, therefore nuclear charge increases, while each atom will have the same amount of inner shells, therefore the shielding effect will stay similar, As a result, as the nuclear charge increases and shielding stays similar, the efa of the nucleus on the most outer electron will increase across the period, therefore will require more energy to knock it off.
Trend in ionisation energy down a group?
Decrease, this is because as you go down a group, each atom will have more electron shells, therefore will experience more shielding. As a result, the efa between the outer most electron and nucleus will decrease down the group, meaning that it will be easier to knock off and will consequently will be knocked off easier.
Explain the trend in melting points in period 3?
In period 3, out of Na, Mg and Al, the melting points go: Na–>Mg–>Al. This is because each metal will release more electrons to gain an increasing ionic charge (+1, +2, +3), therefore as the charge on the ion increases, more electrons will be donated per ion into the sea of delocalised electrons. Overall, all of this will increase the efa between each positive metal ion and its sea of delocalised electrons. Si has the greatest melting point as it has a macromolecular structure. This means it has lots of strong covalent bonds that require lots of energy to break that run all through out its crystalline structure. All of P4, S8 and CL2 are simple molecular, therefore have weak vdw forces that are easy to break. S8 has the highest out of these 3 molecules as one S8 molecule has the most electrons, therefore has the greatest vdw forces. Ar has the lowest boiling point because it is a noble gas and has a full electron configuration therefore is monatomic and the vdw between each atom is very very weak.
What 2 elements in period 2 deviate from the pattern?
Boron because the outer electron is in the 2p subshell, which has a higher energy than the 2s subshell, therefore will require less energy to remove the 2p valence electron than the 2s valence electron from boron.
Oxygen because in its 2p subshell, it has 4 electrons: 2 paired, and 2 unpaired. Therefore, since electrons repel, it will require less energy to remove the paired electrons in its subshell than any of the unpaired 2p electrons in Nitrogen.
What 2 elements in period 3 deviate from the pattern and why?
Aluminium because its valence electron lies in the 3p subshell which is of a higher energy level than the 3s subshell, therefore will require less energy to knock off than Magnesium’s valence electron that is the lower 3s subshell
Sulfur because in the 3p subshell, it has 4 total electrons: 2 unpaired and 2 paired. Therefore because electrons repel each other, it will take less energy to ionise the paired 3p electron than any of the unpaired 3p electrons in Phosphorus.
