Problem 172 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following statement is wrong? (a) Single \(\mathrm{N}-\mathrm{N}\) bond is weaker than the single \(\mathrm{P}-\mathrm{P}\) bond (b) Nitrogen cannot form \(\mathrm{d} \pi-\mathrm{p} \pi\) bond (c) The stability of hydrides increases from \(\mathrm{NH}_{3}\) to \(\mathrm{BiH}_{3}\) in group 15 of the periodic table. (d) \(\mathrm{N}_{2} \mathrm{O}_{4}\) has two resonance structures.

Step-by-Step Solution

Answer

Statement (c) is wrong.

Step 1: Understanding Each Statement

Let's first understand what each statement is addressing. (a) compares bond strengths between \ \( \mathrm{N}-\mathrm{N} \) and \ \( \mathrm{P}-\mathrm{P} \); (b) discusses nitrogen's ability to form certain types of bonds; (c) compares the stability of hydrides in group 15; (d) involves the resonance of \ \( \mathrm{N}_{2} \mathrm{O}_{4} \).

Step 2: Analyzing Statement (a)

Single \ \( \mathrm{N}-\mathrm{N} \) bonds are generally weaker compared to \ \( \mathrm{P}-\mathrm{P} \) bonds due to \ \( \sigma \)-bond strength differences and ferocious repulsion in small Nitrogen atoms.

Step 3: Analyzing Statement (b)

Nitrogen does not have vacant \ \( d \)-orbitals, thus cannot form a \ \( \mathrm{d} \pi-\mathrm{p} \pi \) bond. This statement is correct.

Step 4: Analyzing Statement (c)

From \ \( \mathrm{NH}_3 \) to \ \( \mathrm{BiH}_3 \), the stability of hydrides actually decreases, hence the statement that it increases is incorrect.

Step 5: Analyzing Statement (d)

\ \( \mathrm{N}_{2} \mathrm{O}_{4} \) indeed has resonance, featuring two structures due to electron delocalization.

Step 6: Conclusion

Based on the analysis, statement (c) is incorrect because the stability of hydrides decreases down the group from \ \( \mathrm{NH}_3 \) to \ \mathrm{BiH}_3 \.

Key Concepts

Nitrogen Bonding

Nitrogen is one of the most fascinating elements in the periodic table, especially when it comes to bonding. Unlike its heavier counterparts in Group 15, nitrogen lacks vacant d-orbitals.

This absence makes it impossible for nitrogen to form \(d\pi-p\pi\) bonds, a characteristic bond type involving d-orbitals and p-orbitals. Because of this limitation, nitrogen relies chiefly on \(s\) and \(p\) orbitals to make strong covalent bonds.

In the world of bonds, nitrogen prefers to form triple bonds, as seen in \(N_2\), or involve itself in strong hydrogen bonds, such as those found in ammonia (\(NH_3\)). The triple bond in \(N_2\) is particularly notable for its strength, resulting from three shared electron pairs between two nitrogen atoms.

• The bond energy of a single \(N-N\) bond is relatively low compared to the energy of a \(P-P\) bond.
• This is due to factors like smaller atomic radius and high mutual electron repulsion between nitrogen atoms.

The bonding capabilities of nitrogen make it an excellent bridge between simple structures and more complex chemical environments.

Hydride Stability

Hydrides in Group 15 of the periodic table, particularly those involving nitrogen, phosphorus, arsenic, antimony, and bismuth exhibit varying stability.

Understanding hydride stability comes down to looking at a few key aspects, including bond strength and molecular interactions. Among the Group 15 hydrides, \(NH_3\) or ammonia is the most stable. This is primarily because of the strong \(N-H\) bonds and hydrogen bonding interactions. As you move down the group, the stability of hydrides like \(PH_3\), \(AsH_3\), \(SbH_3\), and \(BiH_3\) decreases.

This downward trend in stability is attributed to:

• The increasing atomic size which leads to weaker \(E-H\) (element-hydrogen) bonds.
• The heavier elements in Group 15 form weaker bonds with hydrogen because of the lower bond strength and energy required to break these bonds.

Furthermore, the weaker intermolecular forces observed in heavier hydrides contribute to the overall lower stability as you move from \(NH_3\) to \(BiH_3\).

Resonance Structures

Resonance is an insightful concept that unveils the potential of electron delocalization in molecules. In the case of \(N_2O_4\), the presence of resonance structures significantly affects its properties.

Resonance structures describe the same molecule being shown in different forms to reflect the distribution of electrons. For \(N_2O_4\), it implies that electrons are shared across multiple atoms, rather than being confined to a single bond.

Visualizing \(N_2O_4\):

• The dinitrogen tetroxide molecule can be represented through two major resonance forms.
• These structures typically show differences like various double bond placements between the nitrogen and oxygen atoms.

The importance of resonance lies in its ability to provide molecules like \(N_2O_4\) with stability. The delocalization of electrons across atoms in different resonance structures means that the electron cloud is more spread out, lowering the energy and making the molecule more stable. This concept beautifully explains why resonance is central to understanding molecular stability and reactivity.