Question

Which of the following statement is incorrect. (1) Be dissolves in alkali forming \(\left[\mathrm{Be}(\mathrm{OH})_{4}\right]^{-2}\) (2) LiCl is soluble in pyridine. (3) Alkaline earth metal lon, because of their much larger charge to size ratio exert a much stronger electrostatic attraction on the oxygen of water molecule surrounding them. (4) \(\mathrm{Be} \mathrm{F}_{2}\) form complex ion with \(\mathrm{NaF}\) in which Be goes with cation.

Short answer

The incorrect statement is (2) LiCl is soluble in pyridine.

Step-by-step solution

Analyze Statement 1

Be dissolves in alkali forming \(\text{\[ \text{{[\mathrm{{Be}}(\mathrm{{OH}})_{4}]^{-2}}} \]}\). This is correct because beryllium typically forms the tetrahydroxyberyllate ion in alkali.

Analyze Statement 2

LiCl is soluble in pyridine. This is incorrect because lithium chloride is an ionic compound, typically soluble in polar solvents like water, but not in nonpolar solvents like pyridine.

Analyze Statement 3

Alkaline earth metal ions, because of their much larger charge-to-size ratio, exert a much stronger electrostatic attraction on the oxygen of the water molecule surrounding them. This is correct because alkaline earth metals have higher charge densities compared to alkali metals.

Analyze Statement 4

BeF2 forms complex ions with NaF in which Be goes with the cation. This statement is not correct because in such complexes, Be typically remains as a cation surrounded by fluoride anions.

Key concepts

beryllium chemistry

Beryllium (Be) is an interesting and unique element among the alkaline earth metals. It has several unique properties and forms compounds that are quite different from the others in the same group. One key characteristic of beryllium is its ability to form complex ions. In an alkaline medium, beryllium reacts to form the tetrahydroxyberyllate ion \(\[ \text{{\text{{Be}}(\text{{OH}})_{4}}^{-2}\]\). This is due to its small size and high charge density which allows it to form stable complexes with hydroxide ions.
Generally, beryllium forms covalent bonds due to its high ionization energy and small atomic radius. This makes its chemistry very distinct from other alkaline earth metals like calcium and magnesium. It’s also important to note beryllium does not react with water or steam even at high temperatures and forms a thin oxide layer that protects it from further oxidation.

solubility rules

Solubility rules help predict whether a compound will dissolve in a given solvent. For example, lithium chloride (LiCl) is soluble in water but not in pyridine. Water is a polar solvent, which means it has partial positive and negative charges. Ionic compounds like LiCl dissolve well in polar solvents because the positive and negative ions are surrounded by water molecules.
Pyridine, on the other hand, is a nonpolar solvent. Ionic compounds generally do not dissolve in nonpolar solvents since there are no partial charges to stabilize the ions in solution. Knowing these rules helps chemists understand and predict solubility behaviors of various compounds in different solvents.

alkaline earth metals

Alkaline earth metals comprise Group 2 of the periodic table and include elements such as magnesium (Mg), calcium (Ca), and beryllium (Be). These metals share common properties but also have unique individual characteristics. One prominent feature is their much larger charge-to-size ratio compared to the alkali metals. This high charge density results in a stronger electrostatic attraction to surrounding molecules, such as the oxygen in water molecules.
This explains why alkaline earth metals often form more stable hydration complexes than alkali metals. Their higher melting points, increased hardness, and more reactive nature also differentiate them from alkali metals.

complex ion formation

Complex ion formation is a common theme in chemistry, especially for transition metals and some smaller elements like beryllium (Be). Complex ions consist of a central metal ion bonded to one or more molecules or ions. For instance, when beryllium fluoride (BeF_2) reacts with sodium fluoride (NaF), a complex ion formation occurs where Be typically remains as the central cation.
Such complex formations are influenced by factors like the charge and size of the metal ion, and the availability of donor atoms in the surrounding molecules or ions. These rules help understand the coordination chemistry and reactivity of metals in various chemical environments.