Question

Explain the following: (a) \(\mathrm{Be}(\mathrm{OH})_{2}\) is insoluble and \(\mathrm{Ba}(\mathrm{OH})_{2}\) is soluble. (b) Be and Mg do not impart colours to the flame. (c) First IE of \(\mathrm{Mg}\) is more than that of Na but second \(\mathrm{IE}\) of \(\mathrm{Mg}\) is less than that of Na.

Short answer

The solubility of a compound depends on its ionic nature and size of the ions. Due to their small size and high ionization energies, Be and Mg do not impart colors to the flame. The first Ionization energy of Mg is more than Na because of the smaller size of Mg, however, the second Ionization energy of Mg is less than Na.

Step-by-step solution

Explain the Difference in Solubility

The solubility of a compound depends on its ionic nature and size of the ions. \(\mathrm{Be}(\mathrm{OH})_{2}\) is insoluble because the size of Be⁺² ion is small and has high charge density, leading to the formation of a covalent character which makes it insoluble in water. On the other hand, \(\mathrm{Ba}(\mathrm{OH})_{2}\) has a larger Ba⁺² ion with lower charge density. This ion forms an ionic bond with OH- making it highly soluble in water.

Discuss Flame Test

Be and Mg do not impart color to the flame due to their small size and high ionization energies. Their electrons are held tightly to the nucleus and hence, when they are exposed to flame, no visible spectrum is seen because the electrons do not absorb the energy required to jump to the higher energy level and hence do not emit any color on falling back to the ground state.

Identity Ionization Energies

Ionization energy is the energy needed to remove an electron from the outermost shell of an atom. The first ionization energy of Mg is more than that of Na due to the smaller size of Mg. A smaller size means the attraction between the nucleus and outermost electron is stronger, hence more energy is required to remove it. However, the second ionization energy of Mg is less than that of Na because after losing one electron, Mg attains stable noble gas configuration and more energy is needed to remove an electron from this stable configuration. On the other hand, Na, after losing one electron, still has one left in the outer most shell making it relatively easier to remove.

Key concepts

Solubility Trends

Solubility trends in inorganic chemistry are fascinating and can be influenced by several factors like ionic charge, ionic size, and the nature of the compound. In the case of beryllium hydroxide, \( \text{Be(OH)}_2 \), and barium hydroxide, \( \text{Ba(OH)}_2 \), these factors play a crucial role.

The solubility of \( \text{Be(OH)}_2 \) is low because the beryllium ions, \( \mathrm{Be}^{2+} \), are very small and have a high charge density. This leads to a covalent character, making \( \text{Be(OH)}_2 \) poorly soluble in water. Meanwhile, \( \text{Ba(OH)}_2 \) is highly soluble due to the larger size of barium ions, \( \mathrm{Ba}^{2+} \), which lowers their charge density. This results in a more ionic character and stronger attraction to hydroxide ions, \( \text{OH}^- \), which increases its solubility in water. Keeping these solubility trends in mind is essential for predicting the behavior of different hydroxides in water.

Flame Test

The flame test is a simple yet effective method to detect the presence of certain metal ions based on the color they emit when heated in a flame. However, not all metals show color changes during a flame test. This is especially true for beryllium and magnesium.

Both Be and Mg have small atomic sizes and tightly held electrons due to their high ionization energies. This means their electrons do not easily absorb energy to jump to a higher energy state. Consequently, when these elements are subjected to a flame, they do not emit visible light as their electrons do not make significant transitions to emit photons. Thus, flame tests fail to detect Be and Mg due to their unique electron configurations.

Ionization Energy

Ionization energy is a key concept in understanding an element's reactivity, as it measures how much energy is required to remove an electron from an atom. Magnesium, \( \text{Mg} \), and sodium, \( \text{Na} \), offer an interesting comparison in this regard.

The first ionization energy of \( \text{Mg} \) is higher than that of \( \text{Na} \). This is due to the smaller atomic radius of \( \text{Mg} \), which results in a stronger attraction between the nucleus and the valence electrons. Thus, more energy is needed to remove an electron from \( \text{Mg} \).

However, when we consider the second ionization energy, \( \text{Mg} \)'s is lower than that of \( \text{Na} \). After \( \text{Mg} \) loses its first electron, it achieves a stable noble gas configuration, making it harder but not too difficult to remove the second electron. In contrast, removing a second electron from \( \text{Na} \) is tougher because, after losing one, it still has a partially filled outer shell, making the second removal even more challenging due to the energy required to disrupt this partly stable state. Understanding these differences helps clarify why \( \text{Mg} \) and \( \text{Na} \) behave differently in chemical reactions.