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 \(\mathrm{d} \pi-\mathrm{p}\) z 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 because hydride stability decreases down the group.

Step 1: Analyze statement (a)

Statement (a) claims that the single \(\mathrm{N}-\mathrm{N}\) bond is weaker than the \(\mathrm{P}-\mathrm{P}\) bond. This statement is generally correct, as nitrogen tends to form stable triple bonds, such as in \(\mathrm{N}_2\), due to its small size and high electronegativity. This means that single bonds involving nitrogen atoms can be relatively weak compared to those of heavier elements like phosphorus, making statement (a) true.

Step 2: Analyze statement (b)

Statement (b) says nitrogen cannot form \(\mathrm{d} \pi-\mathrm{p} z\) bonds. Nitrogen, being in the second period, does not have accessible \(d\) orbitals needed for \(\mathrm{d} \pi-\mathrm{p} z\) bonding. Hence, this statement is also correct.

Step 3: Analyze statement (c)

Statement (c) suggests that the stability of hydrides increases from \(\mathrm{NH}_{3}\) to \(\mathrm{BiH}_{3}\) within group 15. Actually, the stability decreases as we proceed down the group from \(\mathrm{NH}_{3}\) to \(\mathrm{BiH}_{3}\) due to decreasing bond strengths. Thus, this statement is incorrect.

Step 4: Analyze statement (d)

Statement (d) says \(\mathrm{N}_{2} \mathrm{O}_{4}\) has two resonance structures. \(\mathrm{N}_{2} \mathrm{O}_{4}\) indeed has multiple resonance structures as the electrons can be delocalized over the structure. Hence, statement (d) is correct.

Key Concepts

Nitrogen Bonding

Nitrogen is a fascinating element when it comes to bonding, primarily due to its position in the periodic table and its electronic configuration. It is located in Group 15 and has the electron configuration of [He] 2s² 2p³. This configuration allows nitrogen to exhibit some unique bonding behaviors.

• Nitrogen typically forms triple bonds, such as in the diatomic molecule \(\mathrm{N}_2\), which is exceptionally strong and stable due to the overlap of three pairs of electrons. This triple bond is due to the effective overlap of 2p orbitals.
• The smaller size and high electronegativity of nitrogen make \(\mathrm{N}-\mathrm{N}\) single bonds weaker compared to the \(\mathrm{P}-\mathrm{P}\) bonds. Phosphorus, being larger, forms more stable single bonds than nitrogen due to the decreased repulsion between the non-bonding electron pairs in phosphorus.

Such features of nitrogen greatly impact its chemical behavior and stability.

Hydride Stability

Hydride stability within Group 15 elements reveals an interesting trend. Hydrides are compounds in which hydrogen is bonded to a more electropositive element, and the hydrides of Group 15 elements include \(\mathrm{NH}_3\), \(\mathrm{PH}_3\), \(\mathrm{AsH}_3\), \(\mathrm{SbH}_3\), and \(\mathrm{BiH}_3\).

• As one moves down the group from nitrogen (\