Chemistry Paper 2: Blind Spot Part 3

I'll skip nucleophile and electrophile with inorganic chemistry first because of the higher demand of the latter chapter.

For convenience, I will use point form in this section.


Since the Periodic Table tool is malfunctioning, we will use the one on top of this section.

The Basic

• Basically (this word is getting redundant), all of this chapter is about the electrons and protons. That's it!
• This is what I always assume. Imagine an element which possess 7 protons, 7 neutrons and 7 electrons. Then let's say I magically add 1 proton, 1 neutron and 1 electron (nothing less). This changes will affect their chemical and physical characteristics, hence its uses and chemical reactions . But we know that this kind of thing is virtually impossible. Just to give an example on how electrons and protons (and neutrons) is the crucial point in inorganic chemistry.
• In this part, we need to understand first the effect of the electron, especially their position at the outermost shell. As we all know, all chemical reactions depend on the transfer of electron.
  
• Like we all know, as the proton number increases, the number of electron increases.
• As the number of electron increases, hence the number of shell and orbital will also increase, as the electron need a space to move right?
• So as the number of shell increases, wouldn't the size of the atom will increase?
• So as the size of the atom increases, wouldn't the atomic radius increases?

Okay, grasp this concept first before we continue.

The Important Point

• Get back to the important point. So when atomic radius increases, the distance of its valence electrons from its nucleus will increase.
• Hence its attraction will decrease as the distance gets larger, right? (Important point)
  
• When the distance gets larger, the ionization energy will decrease as there is less attraction so less energy is required to remove them, right?
• Like we discussed before, as the proton number increases, so does electron and their shell. Down the group, they all have more proton numbers so their electron will increase too and so their shells and their atomic radius.
• Factors that affects ionization energy is not only their atomic radius. Size of their nucleus (their positive charge) and the shielding effect of electron will also play.
  
• So for group I-III, they will become more reactive down the group because it is easier for them to remove their outermost electrons down the group. However, the non metal which belongs to group V-VII accepts electron rather than donating them. To accept electron, they rely on their attractive force of their nucleus. This is another important point. Down the group, their reactivity will decrease as the distance between the nucleus and the outermost electron will become larger and hence their attraction will decrease.
• Another note for ionic radii. A lot of people often have misconception about this. You may notice in your textbook that, unlike atomic radius, ionic radius of elements have uneven trend. This is the reason. For positive ions, they will remove their outermost electron to form a positive ion. When we remove the electron, we will also remove its shell/orbital hence we will also reduce its size and atomic radius. The opposite things happen for negative ions. As they receive electrons, they will need to create shell/orbital to place their newfound electron. When this happen, the size of the atom will get larger and ionic radius will become larger. (Assuming that all elements will form cation, their ionic radii will decrease throughout the period).
• Electronegativity of elements decrease down the group. The reason is because of their attraction gets weaker as atom gets larger.
• Ionization energy is uneven across a period. This is due the fact that electron occupies different orbital and there is repulsion effect of electron when they occupy the same orbital.
  
• Take period 3 as an example. Sodium electron configuration is [Ne] 3s1, while magnesium's is [Ne] 3s2. Aluminium electronic configuration is [Ne] 3s2 3p1. The p orbital is located farther than the s orbital and it is at higher energy level (less energy required to remove electron at higher energy level). Use this logic to understand why first ionisation energy of aluminium is lower than magnesium, the second ionisation energy of aluminium is higher than magnesium and the third ionisation energy of magnesium is higher than aluminium.
  
• Again we will use period 3 as an example. Phosphorus electronic configuration is [Ne] 3s2 3p3, while sulphur's is [Ne] 3s2 3p4. Refer back to "Chemistry Paper 2: Blind Spot Part 2". The extra electron in sulphur is located at the px orbital. Inside the px orbital in sulphur, there are two electrons whilst in phosphorus there are only one electron. The two electrons will create a repulsion effect and it will decreases the ionisation energy. Hence phosphorus ionisation energy is higher than sulphur.

Group II

• Group II sulphates had its solubility decreases down the group. Oxide, on the other hand, becomes more soluble down the group.
  
• All of Group II nitrates are soluble and all of its carbonate are insoluble.
• Group II thermal stabilities increases down the group which means down the group, the metal carbonates and nitrates will be more stable towards heating down the group. Memorize all the three of them fully.
• Yes I know that Group II oxides are able to dissociate. But only at high temperature. What do I mean by high is 3000 C to 4500C.
  

Group VII

• There is a lot to remember. Firstly, down the group, Group VII reactivity decreases and its colour becomes darker. Their extent and ease of reaction will also decrease means that when it reacts with something, the yield will not be maximum (if it is supposed to form hydrogen halide, the result will be a mixture of hydrogen halide and the halogen)
• Hydrogen halides bond energy will decrease down the group. Hence it will become thermally less stable down the group and their reactivity will increases. Be careful on this part. Hydrides and elements are two different things. Many people (including myself) made mistakes when the questions ask about their states of matter (Iodine is solid, bromine is liquid at r.t.p) result after reaction (when you put a hot iron metal inside compound, example) and their reactivity with acids or salt (NaX).
• Each halides ion can be oxidised by the halogen above it. Example, I-(aq) can be oxidised to Br2 or Cl2.
• About oxidation, down the group it is easier to oxidise halide ions to halide element.
  

Thank you for reading this post. I hope that it will be useful for you guys.