Question

Correct order of ionization energy among the elements Be, B, C, N, O is (a) \(\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}\) (b) \(\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{N}<\mathrm{O}\) (c) \(\mathrm{Be}<\mathrm{B}<\mathrm{C}<\mathrm{N}<\mathrm{O}\) (d) \(\mathrm{Be}<\mathrm{B}<\mathrm{O}<\mathrm{N}<\mathrm{C}\)

Short answer

(a) \\(\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}\\) is the correct answer.

Step-by-step solution

Understand Ionization Energy

Ionization energy is the energy required to remove an electron from a gaseous atom or ion. It generally increases across a period as the nuclear charge increases, making it harder to remove an electron.

Analyze Periodic Trend

As you move across a period from left to right in the periodic table, ionization energy typically increases due to greater nuclear charge and decreased atomic radius.

Identify Exceptions

There are specific exceptions in the periodic trend. Notably, within periods 2 and 3, boron has a lower ionization energy than beryllium, and oxygen has a lower ionization energy than nitrogen due to electron-electron repulsion in partially filled p orbitals.

Apply the Concepts to the Given Elements

For simple elements such as Be, B, C, N, O, you need to account for these known exceptions: B < Be due to the onset of p orbital, O < N due to electron pairing repulsion.

Choose the Correct Option

Using the information from prior steps, verify which option matches the correct ionization energy trend, which is B < Be < C < O < N.

Key concepts

Periodic Trends

The periodic table is not only a fundamental tool for chemists but also a map that reveals intriguing trends in element properties. One such property is ionization energy, which increases as you move from left to right across a period. This happens because the number of protons in the nucleus increases, leading to a greater nuclear charge. As the nuclear charge becomes stronger, electrons are held more tightly. Thus, more energy is required to remove an electron.

Another aspect to consider is atomic radius, which decreases as you move across a period. This shrinking radius occurs because electrons are added to the same energy level while the nuclear charge pulls the electron cloud closer to the nucleus. With electrons pulled in tighter, they are more strongly bound, requiring more energy to be released. Consequently, elements on the right side of a period generally have higher ionization energies compared to those on the left. This understanding forms the fundamental basis of predicting ionization energies across periods.

Exceptions in Periodic Trends

Periodic trends in ionization energy are helpful guides, but there are notable exceptions that defy the usual pattern. These occur primarily due to the distribution of electrons across different atomic orbitals. For example, in the second period elements, one specific exception arises between beryllium (Be) and boron (B), where boron has a lower ionization energy despite being to the right of beryllium. This is because boron's additional electron enters the 2p orbital, which is higher in energy than the 2s orbital that marks beryllium's valence electron configuration. As a result, this 2p electron is slightly easier to remove.

Another critical exception is seen between nitrogen (N) and oxygen (O). Although oxygen lies to the right of nitrogen, it has a lower ionization energy. This phenomenon is explained by electron-electron repulsion. In oxygen, the extra electron pairs up in one of the 2p orbitals, increasing repulsive forces and making it easier to remove an electron. Recognizing these exceptions helps in accurately predicting ionization trends among elements and understanding that they don't always match intuitive expectations.

Gaseous Atoms

Understanding ionization energy involves recognizing that the process occurs in gaseous atoms. This condition refers to an atom in the gas phase, where its electrons do not interact with other atoms or interfere with intermolecular forces.

Isolating atoms in this way is essential because it allows the intrinsic properties of an element to be examined without the complexity of additional interactions. The gas phase provides a standardized state where the energy needed to remove an electron purely reflects the element’s nuclear charge and its electron configuration. It contrasts with conditions in solids or liquids where atoms are bound to nearby atoms, which can significantly alter ionization energy measurements.

Electron-Electron Repulsion

In understanding exceptions to periodic trends, electron-electron repulsion plays a crucial role. This concept involves the interactions between electrons that share the same orbital, often leading to increased energy and instability.

In the context of ionization energy, when two electrons occupy the same orbital, their mutual repulsion can cause one of them to be slightly less tightly held by the nucleus. This occurs because as electrons repel each other, they essentially block the attractive pull of the nucleus, making it easier for the outer electron to be removed.

• This repulsion is why oxygen has a lower ionization energy than nitrogen. Oxygen's 2p orbital arrangement leads to increased repulsion compared to nitrogen.
• On the other hand, beryllium benefits from a filled 2s orbital with no repulsion issues, thus explaining its higher ionization energy compared to boron.

These interactions are subtle but significant, illustrating that subatomic forces contribute greatly to the chemical behavior of elements in unexpected ways.