Video Transcript
An element, X, from the third period of the periodic table has a second ionization
energy that is higher than that of the neighboring two elements. What is the electronic configuration of element X? (A) 1s2 2s2 2p6 3s2, (B) 1s2 2s2 2p6 3s2 3p1, (C) 1s2 2s2 2p6 3s2 3p2, (D) 1s2 2s2
2p6 3s2 3p3, or (E) 1s2 2s2 2p6 3s1 3p5.
This question involves second ionization energy, which can be defined as the energy
required to remove an electron from a gaseous ion that has a one-plus charge
state. Let’s have a look at an example of this, using a helium atom. Generally speaking, ionization energy measures the energy required to remove an
electron. The energy required to remove the most loosely bound electron from a gaseous isolated
atom is the first ionization energy. This process forms a cation with a one-plus charge state. For the second ionization energy event, generally more energy is required to overcome
the electrostatic attraction between the next electron and the nucleus.
We are told that element X is located in the third period of the periodic table. The elements in period three are shown in the diagram. And thus, element X must have an electronic configuration that corresponds to one of
these elements. As we go across a period from left to right, generally, the ionization energy of an
element increases. As the atomic number increases, there are more protons and electrons. The increasing electrostatic attraction across a period draws the electrons closer to
the nucleus, decreasing the atomic radius and making the electrons harder to
remove.
However, there is an exception to this trend. It is energetically favorable for an atom to have a full subshell. So removing the first electron from a full subshell requires more energy than the
trend would predict. We are looking for an element whose second ionization energy is unusually high. In other words, element X is an element whose second ionization energy event involves
removing the first electron from a full subshell.
Let’s have a look at the given electronic configurations and transform them into
their one-plus cations.
The one-plus cation of answer choice (A) has the outermost electron in the 3s, which
is not a full subshell. The second ionization energy would not be related to removing an electron from a full
subshell. So it is not the answer to this question.
Answer choices (C), (D), and (E) similarly would not have a second ionization energy
event that removes an electron from a full subshell. We can see this by the one-plus cation electronic configurations, which show that the
next electron removed in all three of these answers is not from a full subshell.
Answer choice (B) would have an unusually high second ionization energy value. The electronic configuration of the one-plus cation of this element shows that the
second ionization energy event would involve removing an electron from the 3s
subshell. Since this is a full subshell and is energetically favorable, removing this electron
would require more energy than is expected. Therefore, this electronic configuration fits the description for element X, which
means element X is likely aluminum.
So the electronic configuration of X is answer choice (B): 1s2 2s2 2p6 3s2 3p1.
