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Beryllium and aluminium exhibit many properties which are similar. But the two elements differ in [2004] (a) forming polymeric hydrides (b) forming covalent halides (c) exhibiting maximum covalency in compounds (d) exhibiting amphoteric nature in their oxides.

Short Answer

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Beryllium and aluminum differ in exhibiting maximum covalency in compounds.

Step by step solution

01

Identify the Elements and Their Properties

Beryllium (Be) and aluminum (Al) are both metals that can exhibit similarities in properties due to their positions in the periodic table. Beryllium is in Group 2, while aluminum is in Group 13. It is important to identify how each element typically behaves and what properties they share or differ in.
02

Understanding Polymer Formation

Polymeric hydrides are larger molecules made up of smaller units. Aluminum is known to form polymeric structures due to its ability to extend bonding through d-orbitals. Beryllium, however, does not typically form polymeric hydrides due to its limited ability to expand its coordination.
03

Analyzing Covalent Character

Both beryllium and aluminum form covalent halides. Due to their relatively small size and high charge density, they tend to form covalent bonds. Hence, they are similar in this regard.
04

Considering Maximum Covalency

Maximum covalency in compounds is related to the number of covalent bonds an element can form. Beryllium typically forms only a limited number of covalent bonds. Aluminum can expand its covalency due to available empty p-orbitals and sometimes d-orbitals, reaching covalency of 6 or more in some cases.
05

Examining Amphoteric Oxides

Amphoteric substances can behave both as acids and as bases. Both beryllium oxide (BeO) and aluminum oxide (Al2O3) are amphoteric, meaning they can react with both acids and bases. Thus, they share this property.
06

Conclusion: Key Differentiating Property

Analyzing these properties shows that beryllium and aluminum are different in terms of maximum covalency. While beryllium has limited covalency, aluminum can reach higher covalency due to more available orbitals.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Beryllium and Aluminum properties
Beryllium and aluminum, despite being in different groups on the periodic table, exhibit some similar chemical properties.
Both elements are metals with relatively small ionic radii and high charge densities.
These similarities arise because they both have the ability to form covalently bonded compounds.

However, this is where a key difference lies.
Beryllium, belonging to Group 2, typically does not expand its covalency due to the absence of available d-orbitals.
In contrast, aluminum, from Group 13, can utilize its empty p-orbitals and sometimes even its d-orbitals for bonding.
Thus, aluminum can achieve a higher state of covalency compared to beryllium.
In summary:
  • Beryllium is generally limited in covalent bond formation.
  • Aluminum can form more complex structures due to available orbital space.
This distinction is crucial in understanding their broader chemical behavior.
Polymeric hydrides
Polymeric hydrides refer to structures where hydrogen atoms are part of extensive networks of more complex materials.
Aluminum, due to its ability to utilize additional orbitals, is capable of forming polymeric structures as seen in compounds like aluminum hydride ( ext{(AlH}_3 ext{)}_n ext{).
Beryllium, on the other hand, with its limited orbital availability, typically does not form such extended networks.

This is largely because beryllium does not have accessible d-orbitals to engage in extended bonding.
Thus, beryllium hydrides remain simple in structure without forming polymers.
  • Aluminum hydrides tend to be polymeric.
  • Beryllium hydrides are not polymeric; they do not extend into large network structures.
This property highlights the structural versatility of aluminum compared to beryllium.
Covalent bonding
Covalent bonds are formed when atoms share electron pairs.
Beryllium and aluminum are both known for forming covalent halides, largely due to their small atomic sizes and high charge densities.
In these compounds, electrons are shared between the metal atom and the halogen atoms.

The bond formation arises because the electronegativity difference isn't as high as seen in typical ionic compounds.
For example, beryllium chloride ( ext{BeCl}_2 ext{)} and aluminum chloride ( ext{AlCl}_3 ext{)} both exhibit covalent characteristics.
  • Beryllium and aluminum can both form covalent bonds with halides.
  • The covalent nature is due to their electronegativity and size.
This characteristic is significant in defining their chemical reactivity and the types of compounds they form.
Amphoteric oxides
Amphoteric oxides have the unique ability to act as both acids and bases depending on the reacting substances.
Both beryllium oxide ( ext{BeO} ext{)} and aluminum oxide ( ext{Al}_2 ext{O}_3 ext{)} exemplify amphoteric behavior.

This means they can react with both acids to form salts and water and with bases to form complex ions.
This duality makes them versatile in chemical reactions and applications.
  • Beryllium oxide reacts with acids and bases.
  • Aluminum oxide also shows amphoteric properties.
Understanding this property is critical in industrial and laboratory contexts where these compounds are frequently utilized.

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Most popular questions from this chapter

Out of the following metals, those that cannot be obtained by electrolysis of the aqueous solution of their salts are 1\. \(\mathrm{Ag}\) 2\. \(\mathrm{Mg}\) 3\. Cu 4\. Al (a) 2,4 (b) 2,3 (c) 1,4 (d) all

Which of the following is not correct? (a) Al reacts with \(\mathrm{NaOH}\) and liberates \(\mathrm{H}_{2}\) (b) \(\mathrm{AlCl}_{3}\) is a Lewis acid (c) Al is used in the manufacture of electrical cables (d) \(\mathrm{NaOH}\) is used during Halls process of purification of bauxite

A metal (A) heating in nitrogen gas gives (B). (B) on treatment with \(\mathrm{H}_{2} \mathrm{O}\) gives a colourless gas which on passing through \(\mathrm{CuSO}_{4}\) solution turns it deep blue. Here (B) is (a) \(\mathrm{NH}_{3}\) (b) \(\mathrm{MgO}\) (c) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) (d) \(\mathrm{Mg}_{3} \mathrm{~N}_{2}\)

Match the following: List-I List-II 1\. Beryl (i) \(\mathrm{KCl} \cdot \mathrm{MgCl}_{2} .6 \mathrm{H}_{2} \mathrm{O}\) 2\. Carnalite (ii) \(\mathrm{MgCO}_{3}\) 3\. Asbestos (iii) \(3 \mathrm{BeO} \cdot \mathrm{Al}_{2} \mathrm{O}_{3} \cdot 6 \mathrm{SiO}_{2}\) 4\. Magnesite (iv) \(\mathrm{Ca}_{2} \mathrm{Mg}_{5} \mathrm{Si}_{8} \mathrm{O}_{22}(\mathrm{OH})_{2}\) (v) \(\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) The correct matching is: \(\begin{array}{lll}1 & 2 & 3\end{array}\) 4 (a) (iii) (iv) (i) (ii) (b) (iv) (iii) (ii) (i) (c) (iii) (i) (iv) (ii) (d) (iii) (i) (ii) (iv)

\(\mathrm{MgSO}_{4}\) on reaction with \(\mathrm{NH}_{4} \mathrm{OH}\) and \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) forms a white crystalline precipitate. What is its formula? (a) \(\mathrm{Mg}\left(\mathrm{NH}_{4}\right) \mathrm{PO}_{4}\) (b) \(\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) (c) \(\mathrm{MgCl}_{2} \cdot \mathrm{MgSO}_{4}\) (d) \(\mathrm{MgSO}_{4}\)

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