Consider the following ionization energies for aluminum: ...

Consider the following ionization energies for aluminum: $$ \begin{aligned} \mathrm{Al}(g) & \longrightarrow \mathrm{Al}^{+}(g)+\mathrm{e}^{-} & I_{1} &=580 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{+}(g) & \longrightarrow \mathrm{Al}^{2+}(g)+\mathrm{e}^{-} & I_{2} &=1815 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{2+}(g) & \longrightarrow \mathrm{Al}^{3+}(g)+\mathrm{e}^{-} & I_{3} &=2740 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{3+}(g) & \longrightarrow \mathrm{Al}^{4+}(g)+\mathrm{e}^{-} & I_{4} &=11,600 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ a. Account for the trend in the values of the ionization energies. b. Explain the large increase between $I_{3}$ and $I_{4}$. c. Which one of the four ions has the greatest electron affinity? Explain. d. List the four aluminum ions given in order of increasing size. and explain your ordering. (Hint: Remember that most of the size of an atom or ion is due to its electrons.)

Consider the following ionization energies for aluminum:
$$
\begin{aligned}
\mathrm{Al}(g) & \longrightarrow \mathrm{Al}^{+}(g)+\mathrm{e}^{-} & I_{1} &=580 \mathrm{~kJ} / \mathrm{mol} \\
\mathrm{Al}^{+}(g) & \longrightarrow \mathrm{Al}^{2+}(g)+\mathrm{e}^{-} & I_{2} &=1815 \mathrm{~kJ} / \mathrm{mol} \\
\mathrm{Al}^{2+}(g) & \longrightarrow \mathrm{Al}^{3+}(g)+\mathrm{e}^{-} & I_{3} &=2740 \mathrm{~kJ} / \mathrm{mol} \\
\mathrm{Al}^{3+}(g) & \longrightarrow \mathrm{Al}^{4+}(g)+\mathrm{e}^{-} & I_{4} &=11,600 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$$
a. Account for the trend in the values of the ionization energies.
b. Explain the large increase between $I_{3}$ and $I_{4}$.
c. Which one of the four ions has the greatest electron affinity? Explain.
d. List the four aluminum ions given in order of increasing size. and explain your ordering. (Hint: Remember that most of the size of an atom or ion is due to its electrons.)

Key Concepts

Ionic Size Variation

Ionic size is largely determined by the number of electrons and the balance between electron–electron repulsion and nuclear attraction. With successive ionization, as electrons are removed, the ionic radius decreases because the remaining electrons are pulled closer to the nucleus by the unopposed nuclear charge. Thus, ions with higher positive charges are smaller in size compared to those with fewer removed electrons.

Effective Nuclear Charge and Shielding

Effective nuclear charge is the net positive charge experienced by electrons after accounting for shielding by other electrons. As electrons are removed, the inner electrons experience a stronger pull from the nucleus because there is less shielding. This concept is crucial in understanding the trends in ionization energies and the differences in electron binding strength between valence and core electrons.

Ionization Energy Trends

Ionization energy is the energy required to remove an electron from an atom or ion. In successive ionizations, the remaining electrons experience a greater effective nuclear charge due to reduced electron–electron repulsion and less shielding. This increased electrostatic attraction makes each subsequent removal of an electron require significantly more energy, explaining the overall rising trend in ionization energies.

Core versus Valence Electrons

The large jump in ionization energy, such as between the third and fourth ionizations, occurs when removal shifts from a loosely held valence electron to a more tightly bound core electron. Core electrons experience much less shielding and are closer to the nucleus, which greatly increases the energy required to remove them compared to valence electrons.

Electron Affinity in Ions

Electron affinity generally refers to the energy change when an electron is added to a neutral atom; however, in ions, the concept reflects the tendency or energy gain when adding an electron to a positively charged ion. Ions with a higher positive charge exhibit a stronger attraction for electrons due to a high effective nuclear charge, making them more likely to bind additional electrons tightly.

Transcript

00:01 Okay, so to answer the questions of this problem, i think it is useful to look at the bohr model for aluminum, where we can see that the nucleus of the aluminum atom has 13 protons with a positive charge, and then it is surrounded by those electrons at different energy levels moving away from the nucleus.

00:19 So if we look at the bohr model, we will see that we have three electrons in the outermost energy level.

00:26 These are the electrons with the highest energy.

00:28 Now, if we think of the ionization energy, it means basically the energy required to remove electrons from the atom.

00:38 And the first question is asking us about the trend of values in ionization energy.

00:44 Remember that as we remove electrons, for example, if we remove this first electron, then we're still going to be left with 13 protons, but in this case we're only going to have 12 electrons after removing the first electron.

00:58 Therefore, we're going to have after each one of the ionization energies the same amount of protons, but a lower amount of electrons.

01:09 So this is what remains after removing the first electron, this is what remains after removing the second electron, and so on.

01:18 So in every case, we're going to have the same amount of protons, but less amount of electrons.

01:25 And therefore, the net attractive force over each one of the remaining electrons is going to be higher as we increase in ionization energy.

01:38 That's why ionization energies increase successively for each one of the removed electrons, because we keep the same number of protons, but we have a lower number of electrons.

01:50 Remember that the protons attract the electrons, while the electrons repel each other.

01:55 So if i have less electrons, i have a lower repelling force, but i have the same attractive force from the same amount of protons.

02:04 And therefore, the net attractive force increases as we remove each successive electron.

02:10 That is how we account for the increase in different ionization energies as we remove successive electrons.

02:17 So that would be the answer to the first question.

02:21 Now for the second question, we are asked why is there such a large increase between the third and the fourth ionization energy.

02:31 And for that, we can also look at the bohr model.

02:34 We can see that the first three electrons are on the outermost energy level, whereas the fourth electron that we remove is going to be at an inner energy level.

02:47 Remember that the closer we get to the nucleus, the stronger my attractive force is over those electrons.

02:54 So when removing the first three, i am going to be removing electrons from the outermost energy level...

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