Chapter 11: Problem 1

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

Expert verified

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.

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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

Step by step solution

Explain the Difference in Solubility

Discuss Flame Test

Identity Ionization Energies

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