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 \(\mathrm{NH}_{4}{\underline{\phantom{xx}}}^{+}\)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 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

The incorrect statement is (a).

Step 1: Analyze Statement (a)

Statement (a) suggests that the central atom in the hydride is \(sp^2\) hybridized. In the group 15 hydrides, all central atoms are actually \(sp^3\) hybridized due to the presence of three bonding pairs and one lone pair of electrons. Therefore, statement (a) is incorrect.

Step 2: Analyze Statement (b)

Statement (b) correctly states that \(\mathrm{NH}_3\) readily forms \(\mathrm{NH}_4^+\) salts with \(\mathrm{H}^+\) ions due to the lone pair on nitrogen easily accepting a proton. On the other hand, \(\mathrm{PH}_4^+\) salts form under more controlled, anhydrous conditions. Thus, statement (b) is correct.

Step 3: Analyze Statement (c)

Statement (c) mentions the distortion of the tetrahedron in hydrides due to the lone pair-bond pair repulsion. This is accurate, as the lone pair causes a greater repulsion compared to bond pairs, leading to a distorted tetrahedral shape. Hence, statement (c) is correct.

Step 4: Analyze Statement (d)

Statement (d) claims that the bond energy of the M-H bond decreases from \(\mathrm{NH}_3\) to \(\mathrm{BiH}_3\) due to the increase in the size of the element. This is true as larger atomic sizes result in weaker M-H bonds due to less effective overlap. Thus, statement (d) is correct.

Key Concepts

Hybridization in Hydrides

In the world of chemistry, understanding the hybridization of molecular structures is key. For Group 15 hydrides such as \( \mathrm{NH}_3 \), \( \mathrm{PH}_3 \), and others, hybridization plays a crucial role in determining their shape and properties. These hydrides are composed of a central atom linked to hydrogen atoms through chemical bonds. The central atom in these hydrides is actually \( sp^3 \) hybridized, not \( sp^2 \). This is due to the presence of three bonding pairs with hydrogen atoms and one lone pair of electrons, making a total of four regions of electron density.

• The \( sp^3 \) hybridization supports the formation of a tetrahedral geometry.
• However, the lone pair causes distortion, leading to shapes such as pyramidal in \( \mathrm{NH}_3 \).

Interestingly, this configuration helps in understanding why Group 15 hydrides adopt their unique shapes and properties.

Bond Energy Trends

Bond energy is an important concept when discussing the strength of chemical bonds. For the hydrides of Group 15, you'll notice a pattern. As you move down the group—from \( \mathrm{NH}_3 \) to \( \mathrm{BiH}_3 \)—the bond energy of the M-H bond decreases. Why does this happen? As the central atom gets larger in atomic size, the bond it forms with hydrogen becomes longer and weaker.

• Larger elements have more easily polarizable electron clouds.
• This results in weaker overlap of orbitals between the central atom and hydrogen.

As a result, the bond energy decreases due to less effective overlap, making these bonds easier to break.

Lone Pair-Bond Pair Repulsion

In Group 15 hydrides, lone pair-bond pair repulsion significantly impacts the shape and stability of these molecules. The presence of lone pairs on the central atom creates an interesting scenario in molecular geometry. The lone pairs exert greater repulsive forces compared to bond pairs because they are held more closely to the central atom. This results in a distorted tetrahedral shape:

• This repulsion reduces the bond angles.
• In \( \mathrm{NH}_3 \), for instance, the bond angle is about \( 107.3^\circ \) instead of the ideal \( 109.5^\circ \).

These adjustments give each molecule a unique shape that influences its chemical reactivity and interactions.

Protonation of Hydrides

Protonation is a fascinating aspect of hydrides, particularly with ammonia (\( \mathrm{NH}_3 \)). Due to the presence of a lone pair on nitrogen, \( \mathrm{NH}_3 \) can easily accept a proton (\( \mathrm{H}^+ \)), forming \( \mathrm{NH}_4^+ \). This feature underpins the ability of ammonia to act as a base:

• Ammonia readily forms ammonium salts in reaction with acids.
• This property is less pronounced in phosphine (\( \mathrm{PH}_3 \)), where \( \mathrm{PH}_4^+ \) formation occurs under specific, often anhydrous conditions.

Protonation reflects the basicity of these compounds and highlights differences among Group 15 hydrides. It shows how structural differences influence chemical behavior.