A. Based on the first ionization energy level of the elements provided, you can determine that element 2 is the most metallic. This is because the lower ionization energy level, the more metallic an atom/element is. From the choices, you can see that element 2 has the lowest first ionization level. This means that this elements needs the least amount of energy to remove an electron.
B. Looking at how low the first and second ionization energy of element 3 when you compare it to its third, you can tell that it has 2 valence electrons. Based on the periodic table, the element in period 3 that has only 2 valence electrons, is Magnesium. So element 3 is Magnesium.
C. Now that you know that element 3 is Magnesium, you can write the electronic configuration of that element. The electronic configuration is based on its atomic number, assuming that the element is neutral. In a neutral atom, the number of protons and electrons are equal so there are 12 electrons. So the electronic configuration will be written as:
[tex] 1s^{2} 2s^{2} 2p^{6} 3s^{2} [/tex]
D. Look at the ionization energy levels of element 3. You can see that there is a big gap between the first and second ionization energy. This means that the amount of energy for that element to lose a second electron is very high and it would be difficult. But when you look at the first ionization energy level, only a small amount energy is needed. So you can now expect that oxidation state of the most common ion of element 2 will be 1+.
E. Looking again at the ionization energy of the elements, you can see that element 2 has a lower ionization energy than element 3. According to the periodic trends, the ionization increases from left to right. Since you know that element 3 is magnesium, you can assume that the element to the left of it should be lower. And since element 2 is lower, just think of the atom which is found to the left of Magnesium in period three and that would be Sodium. So the chemical symbol for element 2 is Na.
F. Atomic radius according to periodic trends increases from right to left, while ionization energy increases from left to right. So in short they are directly proportional. You can tell which one has the bigger atomic radius by comparing their ionization energy level. Because element 2 has the lowest ionization energy, you can assume that it has the biggest atomic radius. So the element that has the highest ionization energy is Element 1 and we can now assume that Element 1 has the smallest atomic radius.
