Problem 89 - A Complete Resource Book in Chemistry for JEE Main

Question

Select incorrect statement about hydrides of group 15 elements (a) the central atom in the hydride is sp \(^{2}\) hybridized (b) \(\mathrm{NH}_{3}\) readily form \(\mathrm{NH}_{4}^{+}\)salts with \(\mathrm{H}^{+} ; \mathrm{PH}_{4}^{+}\)salts are formed with \(\mathrm{H}^{+}\)under anhydrous condition (c) the tetrahedron is distorted due to repulsion between the lone pair of electrons and the bond pairs (d) the bond energy of the \(\mathrm{M}\)-H bond decreases from \(\mathrm{NH}_{3}\) to \(\mathrm{BiH}_{3}\) because of increase in the size of the element.

Step-by-Step Solution

Answer

Option (a) is incorrect; the central atom is sp³ hybridized, not sp².

Step 1: Understanding Hybridization

In group 15 hydrides like NH₃, the central atom (e.g., N in NH₃) is actually sp³ hybridized, not sp², because it forms three sigma bonds with hydrogen atoms and has one lone pair. Therefore, option (a) is incorrectly stating the hybridization as sp².

Step 2: Salt Formation with Protons

Ammonia ( NH₃ ) readily forms ammonium ( NH₄^{+}}) salts when reacting with protons ( H^{+} ), while phosphine ( PH₃ ) forms PH₄^{+} salts only under anhydrous conditions. This statement (b) is correct based on typical chemical reactions.

Step 3: Geometry and Electron Repulsion

The tetrahedral shape in these hydrides is indeed distorted. The distortion is due to lone pair-bond pair repulsion being greater than bond pair-bond pair repulsion, affecting the geometry. Therefore, statement (c) accurately describes this aspect.

Step 4: Bond Energy Trend Analysis

As you move down the group from nitrogen to bismuth, the M-H bond energy decreases due to an increase in atomic size, which weakens the overlap between orbitals and reduces bond strength. Statement (d) correctly explains this trend.

Key Concepts

Hybridization

Hybridization is an important concept when discussing the structure of molecules, especially those involving central atoms surrounded by other atoms or lone pairs. In group 15 hydrides, like ammonia (\(\mathrm{NH}_3\)), the nitrogen atom undergoes \(\text{sp}^3\) hybridization. This means that one "s" orbital and three "p" orbitals mix to form four equivalent sp³ hybrid orbitals.
These sp³ orbitals arrange themselves in a tetrahedral shape to minimize repulsion, but remember that one of these orbitals contains a lone pair, which affects the actual shape and bonding.
The lone pair occupies more space and has a greater repulsion effect than a bonding pair, which can lead to a slight distortion in the tetrahedral shape.

Salt formation reaction

The ability of group 15 hydrides to form salts is an intriguing property. \(\mathrm{NH}_3 \) can readily react with \( \mathrm{H}^{+} \) ions to form the ammonium ion \( \mathrm{NH}_4^{+} \). This forms the basis of many ammonium salts, which are commonly found in nature and used in various applications.
Phosphine (\( \mathrm{PH_3} \)), on the other hand, is less eager to react with protons in aqueous conditions compared to ammonia. It can form \( \mathrm{PH_4^{+}} \) salts, but this usually requires special anhydrous (water-free) conditions.
This difference in reactivity is due to the varying electronegativities and orbital hybridizations of nitrogen and phosphorus, where nitrogen is more electronegative and forms stronger hydrogen bonds.

Lone pair-bond pair repulsion

The concept of lone pair-bond pair repulsion is key to understanding the molecular geometry of group 15 hydrides. Lone pairs of electrons need more space than bonded pairs because they aren’t shared by multiple atoms. This results in greater repulsion over bond pairs, leading to changes in bond angles.
In molecules like ammonia (\(\mathrm{NH}_3\)), this repulsion causes a compression of the bond angles between hydrogen atoms, making it slightly less than the ideal tetrahedral angle of \(109.5^\circ\). Instead, the angles are around \(107^\circ\).
This pattern is found across other hydrides of group 15, with lone pair repulsions contributing significantly to their three-dimensional structure.

Bond energy trend

A clear trend is observed in the bond energies of group 15 hydrides as you descend the group from nitrogen to bismuth. As you move down the group, the central atom increases in size, which leads to weaker overlap between the hydrogen's 1s orbital and the central atom's orbitals.
Consequently, the \( \text{M–H} \) bond energy decreases. For example, the N-H bonds in ammonia are quite strong due to effective orbital overlap, but as we look to heavier elements like bismuth in \(\mathrm{BiH}_3\), the bonds are weaker, because the increased atomic size reduces the quality of overlap.
This decrease in bond strength correlates with lower bond energies, which can influence the stability and reactivity of these hydrides.