The following data lists the ionization energies for a given atom: IE1 = 738Kj/mol; IE2 = 1451Kj/mol; IE3 = 7733Kj/mol. Predict the valence electron configuration for this atom, and explain your reasoning.
i do not understand this question but i will give it a try, please check if correct and if incorrect help me understand the concept.
Ans: since it has three ionization energies,it can be an element from the group 15, since the first orbit which is closest to the nucleus is filled, the second orbit is filled as well, and the third one will have 5 filled.
Your answer is partially correct. Let me explain the concept and help you arrive at the correct valence electron configuration.
Ionization energy is the amount of energy required to remove an electron from an atom in its gaseous state. The given ionization energies IE1 = 738Kj/mol, IE2 = 1451Kj/mol, IE3 = 7733Kj/mol represent the energy required to remove the first, second, and third electrons from the atom, respectively.
To predict the valence electron configuration, we need to look at the pattern of these ionization energies.
A couple of observations can be made:
1. The first ionization energy (IE1) is relatively low (738Kj/mol), indicating that it does not require a large amount of energy to remove the outermost electron. This suggests that the element is not in group 18 (noble gases), as noble gases have completely filled valence shells and have very high ionization energies.
2. The second ionization energy (IE2) is significantly higher (1451Kj/mol) than the first ionization energy. This indicates that it is more difficult to remove the second electron, suggesting that the valence shell is partially filled.
From these observations, we can conclude that the atom is most likely from the group 13 (Boron Group) or group 15 (Nitrogen Group) of the periodic table. Both groups have partially filled valence shells.
To further narrow down the possibilities, we need to examine the third ionization energy (IE3). The third ionization energy is exceptionally high (7733Kj/mol), indicating that it is very difficult to remove a third electron. This suggests that the element is most likely from the group 15 (Nitrogen Group), as elements in this group tend to have high ionization energies for removing multiple electrons.
Based on this information, we can predict that the valence electron configuration for this atom is likely [valence shell]5 (e.g., 5 valence electrons), which corresponds to group 15 elements.