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Despite the expected decrease in atomic size, there is an unexpected drop in the first ionization energy between Groups \(2 \mathrm{~A}(2)\) and \(3 \mathrm{~A}(13)\) in Periods 2 through \(4 .\) Explain this pattern in terms of electron configurations and orbital energies.

Short Answer

Expert verified
The drop in ionization energy from Group 2A to 3A is because 3A elements have a higher energy p-orbital electron, which is easier to remove than the s-orbital electron in 2A elements.

Step by step solution

01

- Understand Ionization Energy

The first ionization energy is the amount of energy required to remove the outermost electron from an atom in its gaseous state.
02

- Review Periodic Trends

Typically, ionization energy increases across a period from left to right due to the increasing nuclear charge which holds the electrons more tightly.
03

- Examine Group 2A Elements

Elements in Group 2A (e.g., Be, Mg, and Ca) have electron configurations ending in the s-orbital configuration (e.g., Be: [He]2s^2, Mg: [Ne]3s^2, Ca: [Ar]4s^2).
04

- Examine Group 3A Elements

Elements in Group 3A (e.g., B, Al, and Ga) have electron configurations ending in the p-orbital configuration (e.g., B: [He]2s^2 2p^1, Al: [Ne]3s^2 3p^1, Ga: [Ar]4s^2 3d^10 4p^1).
05

- Compare Orbital Energies

The energy of an 2s orbital is lower than that of a 2p orbital. Hence, the outermost electron in Group 3A elements is easier to remove compared to Group 2A elements because it is in a higher energy 2p orbital.
06

- Explain the Anomaly

The unexpected drop in ionization energy between Groups 2A and 3A is due to the fact that the electron being removed in Group 3A elements is from a higher energy p-orbital, which is less tightly bound to the nucleus compared to the s-orbital electrons being removed in Group 2A elements.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Periodic Trends
When we talk about the periodic table, we're discussing the arrangement of elements based on their atomic number. One key aspect of this arrangement is the trend in ionization energy, which tends to increase across a period from left to right. This means that as you move from one element to the next across a row, the energy required to remove an electron usually becomes higher. The reason behind this is simple: the number of protons in the nucleus increases, making the nuclear charge stronger. This stronger nuclear charge pulls the electrons closer, making them harder to remove.
However, understanding periodic trends requires more than just looking at the table. One needs to consider how electron configurations and orbital energies play a role in these trends.
Remember, trends are observed to provide a general understanding of chemical behaviors. But exceptions do exist, and ionization energy between Groups 2A and 3A in Periods 2 through 4 is one such anomaly.
Electron Configurations
Each element has its own unique electron configuration, which tells us how the electrons are distributed among the atom's orbitals. For example, elements in Group 2A, like Be, Mg, and Ca, have electron configurations ending in the s-orbital (Be: ot[He]2s^2ot, Mg: ot[Ne]3s^2ot, Ca: ot[Ar]4s^2ot). On the other hand, elements in Group 3A, such as B, Al, and Ga, have configurations ending in the p-orbital (B: ot[He]2s^2 2p^1ot, Al: ot[Ne]3s^2 3p^1ot, Ga: ot[Ar]4s^2 3d^{10} 4p^1ot).
These configurations are important because the location of the outermost electrons—whether in an s-orbital or a p-orbital—affects how tightly they are held by the nucleus. Electrons in higher energy orbitals, like the p-orbital, are easier to remove. This is crucial in explaining the anomaly noticed with ionization energies between Groups 2A and 3A.
Orbital Energies
The energy levels of orbitals dictate how tightly electrons are bound to the nucleus. For instance, the energy of a 2s orbital is lower than that of a 2p orbital. Therefore, an electron in a 2s orbital is more tightly bound compared to an electron in a 2p orbital.
Now, let's link this back to the anomaly in ionization energies. In Group 2A elements, the electron to be removed is from an s-orbital. In contrast, in Group 3A elements, it is from a p-orbital. Since p-orbital electrons are at a higher energy and thus less tightly bound than s-orbital electrons, it takes less energy to remove them.
This explains why, despite an overall trend of increasing ionization energy across a period, there is a noticeable dip between Groups 2A and 3A: the electron being removed in Group 3A elements (p-orbital) requires less energy to ionize than the electron in Group 2A elements (s-orbital). This interplay between electron configurations and orbital energies is key to understanding periodic trends.

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Most popular questions from this chapter

In addition to those in Table \(14.3,\) other less stable nitrogen oxides exist. Draw a Lewis structure for each of the following: (a) \(\mathrm{N}_{2} \mathrm{O}_{2}\), a dimer of nitrogen monoxide with an \(\mathrm{N}-\mathrm{N}\) bond (b) \(\mathrm{N}_{2} \mathrm{O}_{2}\), a dimer of nitrogen monoxide with no \(\mathrm{N}-\mathrm{N}\) bond (c) \(\mathrm{N}_{2} \mathrm{O}_{3}\) with no \(\mathrm{N}-\mathrm{N}\) bond (d) \(\mathrm{NO}^{+}\) and \(\mathrm{NO}_{3}^{-}\), products of the ionization of liquid \(\mathrm{N}_{2} \mathrm{O}_{4}\)

Complete and balance the following: (a) \(\operatorname{As}(s)+\) excess \(\mathrm{O}_{2}(g) \longrightarrow\) (b) \(\mathrm{Bi}(s)+\operatorname{excess} \mathrm{F}_{2}(g) \longrightarrow\) (c) \(\mathrm{Ca}_{3} \mathrm{As}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\)

The oxygen and nitrogen families have some obvious similarities and differences. (a) State two general physical similarities between Group \(5 \mathrm{~A}(15)\) and \(6 \mathrm{~A}(16)\) elements. (b) State two general chemical similarities between Group \(5 \mathrm{~A}(15)\) and \(6 \mathrm{~A}(16)\) elements. (c) State two chemical similarities between \(\mathrm{P}\) and \(\mathrm{S}\). (d) State two physical similarities between \(\mathrm{N}\) and \(\mathrm{O}\). (e) State two chemical differences between \(\mathrm{N}\) and \(\mathrm{O}\).

From its formula, one might expect \(\mathrm{CO}\) to be quite polar, but its dipole moment is actually small \((0.11 \mathrm{D})\). (a) Draw the Lewis structure for CO. (b) Calculate the formal charges. (c) Based on your answers to parts (a) and (b), explain why the dipole moment is so small.

Boron nitride (BN) has a structure similar to graphite but is a white insulator rather than a black conductor. It is synthesized by heating diboron trioxide with ammonia at about \(1000^{\circ} \mathrm{C}\) (a) Write a balanced cquation for the formation of BN; water also forms. (b) Calculate \(\Delta H_{\text {ran }}^{\circ}\) for the production of BN \(\left(\Delta H_{i}^{\circ}\right.\) of \(\mathrm{BN}\) is \(-254 \mathrm{~kJ} / \mathrm{mol})\) (c) Boron is obtained from the mineral borax, \(\mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7} \cdot 10 \mathrm{H}_{2} \mathrm{O}\). How much borax is needed to produce \(1.0 \mathrm{~kg}\) of BN, assuming \(72 \%\) yield?

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