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 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

Statement (a) is incorrect.

Step 1: Identify Statements

Carefully read each given statement about the hydrides of group 15 elements and evaluate what each statement is implying about these compounds.

Step 2: Analyze Group 15 Hydrides

Understand and recall the properties of group 15 hydrides ( ext{NH}_3, ext{PH}_3, ext{AsH}_3, ext{SbH}_3, ext{BiH}_3). Note their common properties such as hybridization, ability to form salts, molecular geometry, and trends in bond energy.

Step 3: Evaluate Statement (a)

Statement (a) claims that the central atom in the hydride is sp extsuperscript{2} hybridized. However, group 15 hydrides have sp extsuperscript{3} hybridization due to the presence of three sigma bonds and one lone pair.

Step 4: Evaluate Statement (b)

Statement (b) is true. ext{NH}_3 readily forms ext{NH}_4^+ salts with ext{H}^+, while ext{PH}_3 salts like ext{PH}_4^+ can be formed under anhydrous conditions. This is well-documented for group 15 hydrides.

Step 5: Evaluate Statement (c)

Statement (c) is true. The presence of lone pairs causes distortion in the tetrahedral molecular geometry of group 15 hydrides due to increased lone pair-bond pair repulsion.

Step 6: Evaluate Statement (d)

Statement (d) is correct. The bond energy of the M-H bond generally decreases from ext{NH}_3 to ext{BiH}_3, which aligns with the increase in atomic size as you move down the group from nitrogen to bismuth.

Step 7: Determine Incorrect Statement

Since statement (a) is incorrect, as the hydrides are sp extsuperscript{3} hybridized, it is the statement that does not correctly describe a property of group 15 hydrides.

Key Concepts

Hydrides

Hydrides of group 15 elements include ammonia (\(\text{NH}_3\)), phosphine (\(\text{PH}_3\)), arsine (\(\text{AsH}_3\)), stibine (\(\text{SbH}_3\)), and bismuthine (\(\text{BiH}_3\)).These compounds have a central atom bonded to hydrogen atoms.The central atom is from group 15 of the periodic table, so these hydrides share some common properties.

• Ammonia is the most well-known and stable among these hydrides due to nitrogen's high electronegativity.
• The stability generally decreases as you move from \(\text{NH}_3\) to \(\text{BiH}_3\).
• Group 15 hydrides tend to exhibit reducing properties due to their ability to donate lone pair electrons.

Understanding these basic characteristics can help in grasping why the improper hybridization statement in the exercise was incorrect.

Hybridization

Hybridization is a concept that explains the mixing of atomic orbitals to form new hybrid orbitals. This is crucial for understanding the geometry and reactivity of a molecule. In the case of group 15 hydrides, each central atom undergoes \(sp^3\) hybridization. This is because:

• Each central atom, such as nitrogen in \(\text{NH}_3\), forms 3 sigma bonds with hydrogen atoms.
• One of the available p-orbitals contains a pair of non-bonding (lone) electrons.

Thus, because of the presence of three bonding domains and one lone pair, the central atom is \(sp^3\) hybridized, explaining its tetrahedral arrangement, although distorted.

Bond Energy

Bond energy is an important aspect that influences the chemical behavior and stability of molecules.In group 15, the bond energy of M-H bonds decreases as you move from \(\text{NH}_3\) to \(\text{BiH}_3\).This trend can be explained by:

• An increase in atomic size as you go down the group leading to longer and weaker M-H bonds.
• Lower bond energy implies less stable hydrides, making compounds like \(\text{BiH}_3\) less favorable to form compared to \(\text{NH}_3\).

The bond energy works inversely with atomic size and is an essential factor in predicting molecule reactivity and compound formation.

Molecular Geometry

The molecular geometry of the group 15 hydrides predominantly features a tetrahedral shape.However, the presence of a lone pair slightly distorts this geometry, leading to:

• A trigonal pyramidal structure, rather than a perfect tetrahedral one, due to lone pair-bond pair repulsion.
• This lone pair exerts more repulsion than bond pairs, compressing the bond angles slightly.
• For instance, the bond angle in \(\text{NH}_3\) is about \(107^ ext{o}\), which is less than the tetrahedral angle of \(109.5^ ext{o}\).This compression is larger in hydrides lower in the group.

The understanding of molecular geometry is vital because it affects physical and chemical properties including boiling points, solubility, and interactions with other molecules.