 So in this video we are going to solve a few problems on ionization energy and its trends across the periodic table. So let's look at the first question which says that the first 4 ionization energy values for an element are approximately 300, 400, 650 and 30 to 100 kilojoules per mole. Now what is the number of valence electrons in this element? So before we begin this question, let's have a short recap of what we exactly mean by ionization energy. Ionization energy is the amount of energy required to remove an electron from a neutral atom and the more tightly held the electron is, the greater will be the ionization energy, right? We need to provide a lot more energy to remove this electron as compared to let's say an electron that is present far away from the nucleus and experiences much less attractor first. Now when we look at the ionization values here which are 300 kilojoules per mole, 400 kilojoules per mole, 650 kilojoules per mole and 30 to 100 kilojoules per mole, we can see that there is a significant jump here between the third and the fourth ionization energy and this means that we need to provide a substantial amount of energy to remove this electron, fourth electron because IE1 refers to the energy required to remove the first electron, IE2 would be the second electron, IE3 would be the third electron and IE4 or ionization energy 4 would be the amount of energy required to remove the fourth electron. Now the substantial increase in the ionization energy from the third to the fourth would mean that it is much more closer to the nucleus and experiences a greater attractor force as compared to the other three electrons. Now the first three electrons are easier to remove that means they are more slightly to be the valence electrons whereas this fourth electron is much more harder to remove and that means it's much more likely to be a core electron and not a valence electron. So based on this the number of valence electrons in this element will be 3. So let's look at the second question which asks us what is the correct order of first ionization energy of sodium, magnesium, aluminium and silicon. So how do we tackle this question? You see ionization energy increases across a period. This is because as we move across a period the number of protons in the nucleus increases and this increases the nuclear charge or the force or the pull with which the outer electrons are held close to the nucleus and as a result the ionization energy increases because you know now we need to give more energy to remove this electron. So based on this our order of ionization energy should be sodium less than magnesium less than aluminium less than silicon but wait a minute this really doesn't look like the correct order. Let's look at the electronic configurations to be absolutely sure. Now we know that their atomic numbers are 11, 12, 13, 14. And valence electrons would be 3S1, 3S2, 3S2, 3P1 in case of aluminium and 3S2, 3P2 in case of silicon. Now when you compare magnesium and aluminium, in aluminium the outermost electron is present in the 3P orbital whereas in magnesium 3S2 is a fully filled sub-shell and fully filled sub-shells are much more stable than partially filled sub-shell. And most stability would mean we need to provide more amount of energy to remove the electrons from this 3S orbital. So that is why the ionization energy of magnesium is actually greater than that of aluminium. In fact the first ionization energy of magnesium is about 737 kJ per mole and that of aluminium is about 578 kJ per mole. Although silicon also has its outermost electrons in the 3P orbital, here it has one more proton in the nucleus than aluminium which means the electrons now experience greater attraction from the nucleus and this increases the ionization energy. So based on this the correct order of ionization energy is sodium less than magnesium which is greater than aluminium less than silicon. So let's solve one last problem. So the question asks us which among the following has the highest second ionization enthalpy, neon, oxygen, fluorine and nitrogen. So to go ahead with this question let's first understand what do we actually mean by second ionization enthalpy right? Basically it is the energy required to remove a second electron from an atom after one has already been removed. Now you see second ionization enthalpy is always greater than the first ionization enthalpy and this is because it is much more difficult to remove an electron from a positively charged species as compared to a neutral atom. You see in a positively charged ion you have more number of protons and that attracts the outermost electrons much more strongly as compared to a neutral atom. And this is why second ionization enthalpy is always greater than the first ionization enthalpy. Now to solve this question let's first write down the electronic configurations of the given atoms. Now electronic configuration of neon is 1s2 2s2 2p6, fluorine has an electronic configuration of 1s2 2s2 2p5, oxygen has an electronic configuration of 1s2 2s2 and 2p4 while nitrogen has an electronic configuration of 1s2 2s2 and 2p3. Now for all of these atoms the second ionization removes electrons from the 2p orbitals. But among them neon would have the highest second ionization enthalpy because it has most number of protons as compared to the others. Number of protons in neon is 10 whereas that in fluorine, oxygen and nitrogen is 9, 8 and 7. More the number of protons more tightly the electrons will be held and that means we need to provide much more energy to knock off the electrons. Therefore, atom with the highest second ionization enthalpy would be neon. Now what about the rest? What do you think the correct order of second ionization enthalpy would be for the rest of the atoms? Well you see now after removing one electron i.e. one would give here 2p4 in this case you would get 2p3 and in this case you would get 2p2 and on removing the second electron here you will get 2p3 electronic configuration that means it is a stable half filled electronic configuration and this offers more stability whereas if you look at the oxygen we already have a stable half filled electronic configuration after removing one electron. So in this case the second ionization enthalpy for oxygen would be higher as compared to fluorine because now we are disrupting a relatively stable half filled electronic configuration to achieve a less stable 2p2 configuration. Now in the case of nitrogen the second ionization enthalpy would give you a 2p1 electronic configuration. So based on this you can see that the correct order of the second ionization enthalpy would be neon greater than oxygen greater than fluorine greater than nitrogen.