Problem 149 - A Complete Resource Book in Chemistry for JEE Main

Question

Isostructural group of molecule is (a) \(\mathrm{XeO}_{4}, \mathrm{NH}_{4}^{+}, \mathrm{CH}_{4}\) (b) \(\mathrm{CH}_{3}^{-}, \mathrm{NH}_{3}, \mathrm{NF}_{3}\) (c) \(\mathrm{NH}_{3}, \mathrm{NF}_{3}, \mathrm{BF}_{3}\) (d) \(\mathrm{NO}_{3}, \mathrm{NO}_{2}, \mathrm{SF}_{4}\)

Step-by-Step Solution

Answer

Option (a) is correct: \(\mathrm{XeO}_{4}\), \(\mathrm{NH}_{4}^{+}\), and \(\mathrm{CH}_{4}\) are isostructural.

Step 1: Understand Isostructural Molecules

Isostructural molecules have similar shapes and bond angles, often having the same number of atoms attached to the central atom and occupying the same positions in space.

Step 2: Analyze Option A

The molecules are \(\mathrm{XeO}_{4}\), \(\mathrm{NH}_{4}^{+}\), and \(\mathrm{CH}_{4}\). \(\mathrm{XeO}_{4}\) and \(\mathrm{CH}_{4}\) both have a tetrahedral shape. \(\mathrm{NH}_{4}^{+}\) is also tetrahedral due to sp3 hybridization. Therefore, all three are isostructural.

Step 3: Analyze Option B

The molecules are \(\mathrm{CH}_{3}^{-}\), \(\mathrm{NH}_{3}\), and \(\mathrm{NF}_{3}\). \(\mathrm{NH}_{3}\) and \(\mathrm{NF}_{3}\) both have a trigonal pyramidal shape due to sp3 hybridization, while \(\mathrm{CH}_{3}^{-}\) is planar, so they are not all isostructural.

Step 4: Analyze Option C

The molecules are \(\mathrm{NH}_{3}\), \(\mathrm{NF}_{3}\), and \(\mathrm{BF}_{3}\). \(\mathrm{NH}_{3}\) and \(\mathrm{NF}_{3}\) are trigonal pyramidal, whereas \(\mathrm{BF}_{3}\) is trigonal planar due to sp2 hybridization, so they are not isostructural.

Step 5: Analyze Option D

The molecules are \(\mathrm{NO}_{3}\), \(\mathrm{NO}_{2}\), and \(\mathrm{SF}_{4}\). \(\mathrm{NO}_{3}\) is trigonal planar, \(\mathrm{NO}_{2}\) is bent, and \(\mathrm{SF}_{4}\) is see-saw shaped, so they are not isostructural.

Step 6: Determine the Correct Answer

Recognizing that isostructural molecules must have the same shape, only the molecules in option (a): \(\mathrm{XeO}_{4}\), \(\mathrm{NH}_{4}^{+}\), and \(\mathrm{CH}_{4}\) are all tetrahedral and isostructural.

Key Concepts

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. This spatial configuration is pivotal in determining a molecule's properties, such as polarity and reactivity.
When exploring molecular geometry, we rely on the notion that the arrangement of atoms is influenced by the repulsive forces between electron pairs surrounding the central atom. This is often visualized using the VSEPR theory (Valence Shell Electron Pair Repulsion theory).
For instance, in our exercise,

• \(\text{CH}_4\) exhibits a tetrahedral molecular geometry, where carbon is the central atom surrounded evenly by four hydrogen atoms.
• Similarly, \(\text{NH}_4^+\) maintains a tetrahedral shape due to the symmetric distribution around the nitrogen atom despite having a positive charge.

Molecular geometry helps us understand the "isostructural" nature of these molecules, as they share similar shapes and bond angles.

Hybridization

Hybridization offers insight into the type and number of bonds a molecule's central atom can form. It describes the mixing of atomic orbitals to generate new hybrid orbitals.
This process helps achieve a molecule's ideal geometry by using hybrid orbitals for bonding.

• For example, in \(\text{CH}_4\), the carbon atom undergoes \(\text{sp}^3\) hybridization. This means one \(\text{s}\) and three \(\text{p}\) orbitals combine to create four equivalent hybrid orbitals, fostering a tetrahedral shape.
• In \(\text{NH}_4^+\), the nitrogen atom similarly exhibits \(\text{sp}^3\) hybridization, allowing it to form bonds in a tetrahedral layout.

Recognizing the hybridization in a molecule aids in predicting its geometry and understanding intermolecular interactions.

Tetrahedral Structure

Tetrahedral structures are a common theme in chemistry, representing a molecular shape with four bonds emanating from a central atom symmetrically.
Such a configuration is characterized by bond angles of approximately 109.5 degrees, creating space-filling geometries that minimize repulsion among bond pairs.
In our specific example:

• Both \(\text{XeO}_4\) and \(\text{NH}_4^+\) depict tetrahedral geometries, resulting in identical molecular frameworks.
• This geometry arises from the atoms being evenly spaced, offering stability and symmetry, as visible in tetrahedral isostructural molecules.

Understanding this shape is crucial for students as it is pivotal in elucidating chemical bonding and molecular properties.

Chemical Bonding

Chemical bonding outlines how atoms attach within a molecule, profoundly impacting their shapes and functions.
There are several types of chemical bonds, but in the context of isostructural molecules, we are primarily concerned with covalent bonds.

• For tetrahedral molecules, prevalent bonding involves sigma bonds formed between the central atom's hybrid orbitals and other's atomic orbitals.
• In \(\text{CH}_4\), carbon forms four covalent bonds with hydrogen through these sigma interactions.

By understanding these bonds' nature, students can predict molecular geometry and how molecules may interact or react chemically in various environments. Emphasizing chemical bonding clarifies the structure-functional relationship seen in isostructural groups.