Question

The order of first ionization energies of the elements \(\mathrm{Li}, \mathrm{Be}, \mathrm{B}, \mathrm{Na}\) is: (a) \(\mathrm{Be}>\mathrm{Li}>\mathrm{B}>\mathrm{Na}\) (b) \(\mathrm{B}>\mathrm{Be}>\mathrm{Li}>\mathrm{Na}\) (c) \(\mathrm{Na}>\mathrm{Li}>\mathrm{B}>\mathrm{Be}\) (d) \(\mathrm{Be}>\mathrm{B}>\mathrm{Li}>\mathrm{Na}\)

Short answer

(d) Be > B > Li > Na

Step-by-step solution

Understand Ionization Energy

Ionization energy is defined as the amount of energy required to remove the outermost electron from a gaseous atom. It is influenced by several factors including atomic size, nuclear charge, and electron shielding.

Analyze Periodic Trends

In a period, ionization energy generally increases from left to right due to increased nuclear charge and decreased atomic radius. In a group, it decreases downwards as atomic size increases.

Element Position Overview

- Lithium (Li) and Beryllium (Be) are in the second period, with Be to the right of Li. - Boron (B) is also in the second period, further to the right of Be. - Sodium (Na) is in the third period, directly below Li.

Predict Ionization Energies

Based on the position: - Be should have the highest ionization energy in the group due to its position to the right of Li. - B should have slightly less than Be, as it has a filled subshell which provides some shielding. - Li, being to the left of both Be and B, will have lower ionization energy than both. - Na, with a new electron shell, will have the lowest ionization energy.

Compare with Given Options

Now, compare different options to find which one corresponds to the predicted order of ionization energies: - (a) Be > Li > B > Na : This option is incorrect because B should have a higher ionization energy than Li. - (b) B > Be > Li > Na : This is also incorrect since Be has a higher ionization energy than B due to its filled 2s subshell. - (c) Na > Li > B > Be : Incorrect as Na should have the lowest ionization energy. - (d) Be > B > Li > Na : This correctly follows the trend we established.

Key concepts

Periodic Table Trends

Understanding **Periodic Table Trends** is essential when predicting ionization energy. As you move across a period from left to right, the ionization energy generally increases. This is mainly due to the increase in nuclear charge, which causes the outer electrons to be held more tightly to the nucleus.
Moving down a group, the ionization energy tends to decrease as the atomic size increases. This increase in size reduces the force of attraction between the electrons and the nucleus, making it easier to remove the outermost electron. These trends help in comparing the ionization energies of the elements like Lithium (Li), Beryllium (Be), Boron (B), and Sodium (Na).

Atomic Size

**Atomic Size** plays a crucial role in determining ionization energy. As we move across a period, atomic size decreases. This is because an increased nuclear charge pulls the electron cloud closer to the nucleus.
Conversely, as we move down a group, atomic size increases due to the addition of electron shells. Larger atomic size means that the outermost electrons are further from the nucleus, resulting in a decrease in ionization energy.
For instance, Beryllium (Be) has a smaller atomic size than Lithium (Li) and thereby a greater ionization energy. Sodium (Na), being in the next period, has a larger atomic size compared to all three, resulting in the lowest ionization energy amongst them.

Nuclear Charge

The **Nuclear Charge** is a key factor in determining ionization energy. It refers to the total charge of the nucleus, i.e., the number of protons. A higher nuclear charge results in a stronger electrostatic force of attraction between the nucleus and the outer electrons.
Moving across a period, the nuclear charge increases, enhancing the ionization energy. For example, as you move from Lithium (Li) to Beryllium (Be) and then to Boron (B), the nuclear charge increases, resulting in higher ionization energies.
However, the increasing nuclear charge can be counteracted by electron shielding, influencing the actual ionization energy observed.

Electron Shielding

**Electron Shielding** affects ionization energy by reducing the effective nuclear charge experienced by the outermost electrons. When inner-shell electrons partially shield the outer electrons from the full attractive force of the nucleus, it becomes easier to remove these outer electrons.
As you move across a period, this shielding effect is minimal, which allows the ionization energy to increase as the nuclear charge increases. But moving down a group, additional electron shells significantly increase the shielding effect, decreasing the ionization energy.
Beryllium (Be) has a full 2s subshell, which provides effective shielding to its outer electrons, contributing to its higher ionization energy. Boron (B), even though it follows Be, experiences slightly more shielding due to its extra electron in the 2p subshell.