Animated Solution for Chemistry - s and p-Block Elements: Which of the following statements is wrong?

The beauty of chemistry often lies not just in the predictable periodic trends, but in the fascinating exceptions and anomalies that arise from the fundamental properties of atoms. In this problem, we are tasked with identifying the incorrect statement among four claims about Group 15 elements. Let's embark on a journey to dissect each statement and uncover the underlying chemical principles.

Let's begin with the first statement, which claims that the thermal stability of hydrides increases from ${\text{NH}}_3$ to ${\text{BiH}}_3$. To evaluate this, we must look at what happens as we descend Group 15 (Nitrogen, Phosphorus, Arsenic, Antimony, Bismuth).

As we move down the group, the principal quantum number increases, leading to a larger atomic radius for the central atom. When these larger atoms bond with hydrogen, the resulting $\text{M-H}$ bond length naturally increases.

According to the principles of chemical bonding, a longer bond is generally a weaker bond. Because the orbital overlap between the large, diffuse orbitals of heavier elements (like Bismuth) and the tiny $1s$ orbital of Hydrogen is highly ineffective, the bond dissociation energy drops drastically.

Therefore, the thermal stability actually decreases down the group:

Since the statement claims the exact opposite, we have already found our incorrect statement! However, a true chemist never stops at the first clue. Let's verify the remaining options.

The second statement asserts that Nitrogen cannot form $d\pi -p\pi$ bonds. Is this true?

Nitrogen is a second-period element. Its electronic configuration is $1s^{2}2s^{2}2p^3$. The valence shell is $n=2$, which only accommodates $s$ and $p$ subshells. The $d$-orbitals do not appear until the third principal energy level ($n=3$).

Because Nitrogen completely lacks $d$-orbitals in its valence shell, it is physically impossible for it to participate in $d\pi -p\pi$ bonding. This is a classic restriction for second-period elements, making the statement absolutely correct.

The third statement compares the strength of the $\text{N-N}$ single bond to the $\text{P-P}$ single bond, claiming the former is weaker. This might seem counterintuitive at first—shouldn't smaller atoms form stronger bonds due to better overlap?

While true for multiple bonds (like $\text{N}\equiv \text{N}$), single bonds tell a different story. The Nitrogen atom is exceptionally small. When two Nitrogen atoms form a single bond, they are forced into close proximity. Each Nitrogen atom carries a lone pair of electrons.

Because the atoms are so close, these lone pairs experience severe interelectronic repulsion. This intense repulsion acts like a compressed spring, pushing the atoms apart and significantly weakening the $\text{N-N}$ single bond. Phosphorus, being a larger atom, has a longer bond length, which keeps its lone pairs far enough apart to minimize this repulsion. Thus, the $\text{P-P}$ single bond is indeed stronger than the $\text{N-N}$ single bond. The statement is correct.

Finally, the fourth statement claims that ${\text{N}}_2{\text{O}}_4$ has two resonance structures.

Dinitrogen tetroxide (${\text{N}}_2{\text{O}}_4$) is a dimer of nitrogen dioxide (${\text{NO}}_2$). It is a planar molecule featuring a direct $\text{N-N}$ bond. The $\pi$ electrons in the molecule are delocalized over the oxygen atoms. This delocalization can be perfectly represented by two equivalent resonance structures, where the double bonds to the oxygen atoms shift symmetrically.

This resonance stabilization is a key feature of the molecule, confirming that the fourth statement is also correct.

After a thorough analysis of atomic sizes, orbital availability, interelectronic repulsions, and resonance, we can confidently conclude that the only incorrect statement is the first one regarding the thermal stability of hydrides.