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Which of the following statements is/are not true about the diagonal relationship of \(\mathrm{Be}\) and \(\mathrm{Al} ?\) (1) Both react with \(\mathrm{NaOH}\) to liberate hydrogen (2) Their oxides are basic (3) They are made passive by nitric acid (4) Their carbides give acetylene on treatment with water (a) Only (1) (b) (2) and (3) (c) Only (4) (d) (2) and (4)

Short Answer

Expert verified
The answer is (d) (2) and (4).

Step by step solution

01

Analyze Statement (1)

Beryllium (Be) and Aluminum (Al) both react with \(\text{NaOH} \) to release hydrogen gas. Thus, statement (1) is true.
02

Analyze Statement (2)

Check the nature of the oxides of Be and Al. BeO is amphoteric, meaning it can react with both acids and bases. Similarly, Al\(_2\)O\(_3\) is also amphoteric. Since neither BeO nor Al\(_2\)O\(_3\) are basic, statement (2) is not true.
03

Analyze Statement (3)

Investigate their passivation by nitric acid. Both Be and Al form protective oxide layers when they react with \(\text{HNO}_3\), preventing further reaction, so they undergo passivation. Therefore, statement (3) is true.
04

Analyze Statement (4)

Consider the reaction of their carbides with water. Be\(_2\)C and Al\(_4\)C\(_3\) do not produce acetylene, \(\text{C}_2\text{H}_2\). Instead, Be\(_2\)C forms methane (\(\text{CH}_4\)) and Al\(_4\)C\(_3\) forms methane, not acetylene. Thus, statement (4) is not correct.
05

Select Incorrect Statements

From the analysis, statements (2) and (4) are not true.

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Key Concepts

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

Beryllium (Be) and Aluminum (Al) reactions
Beryllium (Be) and Aluminum (Al) have a fascinating diagonal relationship on the periodic table. This relationship is due to the similar properties they exhibit, despite being in different groups. One of the key similarities is their ability to react with sodium hydroxide (NaOH) to release hydrogen gas. Beryllium behaves differently than most metals, as it does not react with water but will react with NaOH to produce hydrogen. This is similar to aluminum, which also reacts with NaOH to release hydrogen gas and form soluble aluminates. This reaction is a hallmark of their unusual chemistry, making both Be and Al interesting subjects of study. The overall process occurs as metals are introduced to NaOH, forming their respective hydroxides and releasing hydrogen:
Amphoteric oxides
Beryllium oxide (BeO) and aluminum oxide (AlO) are not merely basic or acidic. They are classified as amphoteric, meaning they can react with both acids and bases. This property is rather exceptional and deviates from the typical behavior of many metal oxides. For BeO, it can dissolve in acidic solutions to form beryllium salts, while in basic solutions, it forms beryllates. Similarly, AlO behaves in much the same way, dissolving in acid to form aluminum salts and in bases to produce aluminate ions. In summary, the amphoteric nature of these oxides reflects their versatility in chemical reactions. They do not fit neatly into the category of a purely basic oxide, which differentiates them distinctly from many other metal oxides.
Passivation by Nitric Acid
Both beryllium and aluminum have the ability to undergo passivation when exposed to concentrated nitric acid (HNO). Passivation is a process where a material becomes less affected by environmental factors through the formation of an outer layer. In this case, the exposure to nitric acid results in the formation of an oxide layer that adheres closely to the surface of the metal. This layer acts as a protective barrier, preventing further reaction of the metal with the acid. This protective behavior is important as it helps preserve the metal's integrity in environments that could otherwise cause corrosion or wear, making it a crucial property for practical applications in various industries.
Carbides reactions with water
Carbides are compounds composed of carbon and a metal. Beryllium carbide (BeC) and aluminum carbide (AlC) showcase interesting reactions when combined with water; however, they do not produce acetylene (CH) as might be expected. When BeC reacts with water, it forms methane (CH) instead. Similarly, AlC's reaction with water also results in the production of methane, not acetylene. This outcome is contrary to the behavior of some other metal carbides, which do react with water to produce acetylene. These reactions highlight the importance of understanding specific carbide properties, as they can vary quite significantly among different metals, leading to a range of products upon hydrolysis.

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Most popular questions from this chapter

For the gaseous reaction, \(\mathrm{K}+\mathrm{F} \rightarrow \mathrm{K}^{+}+\mathrm{F}, \Delta \mathrm{H}\) was calculated to be 19 kcal under conditions where the cations and anions were prevented by electrostatic separation from combining with each other. The ionization potential of \(\mathrm{K}\) is \(4.3 \mathrm{eV}\). What is the electron affinity of \(\mathrm{F} ?\) (a) \(3.21\) (b) \(4.28\) (c) \(3.48\) (d) \(1.48\)

The correct statement among the following is: (a) The first ionization potential of \(\mathrm{Al}\) is less than the first ionization potential of \(\mathrm{Mg}\) (b) The second ionization potential of \(\mathrm{Mg}\) is greater than the second ionization potential of Na (c) The first ionization potential of \(\mathrm{Na}\) is less than the first ioniation potential of \(\mathrm{Mg}\) (d) The third ionization potential of \(\mathrm{Mg}\) is greater than the third ionization potential of \(\mathrm{Al}\)

The element with the highest first ionization potential is: (a) Nitrogen (b) Oxygen (c) Boron (d) Carbon

The correct order of ionic radii is: (a) \(\left.\mathrm{I}^{+}>\mathrm{l}^{-}\right\rangle \mathrm{I}\) (b) \(\mathrm{I}>\mathrm{I}^{+}>\mathrm{I}^{-}\) (c) \(\mathrm{I}^{+}>\mathrm{I}>\mathrm{I}^{-}\) (d) \(\mathrm{I}^{-}>\mathrm{I}>\mathrm{I}^{+}\)

Which of the following statements is true about effective nuclear charge? (a) \(Z_{\text {eff }}\) decreases from top to bottom (b) \(Z_{\text {eff }}\) increases from top to bottom (c) \(Z_{\text {eff }}\) increases as we move from left to right in periodic table (d) \(Z_{\text {eff }}=Z \times \sigma\) (here \(\sigma\) is screening constant)

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