Problem 42 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following statement is true? (a) Hybridization of the central atom in \(\mathrm{NH}_{3}\) and \(\mathrm{CH}_{4}\) is \(\mathrm{sp}^{2}\) (b) \(\mathrm{BeCl}_{2}\) has " \(\mathrm{V}\) " shape while \(\mathrm{SO}_{2}\) is linear (c) \(\mathrm{SF}_{6}\) is octahedral and \(\mathrm{F}-\mathrm{S}-\mathrm{F}\) bond angle is \(90^{\circ}\) (d) \(\mathrm{CO}_{2}\) has dipole moment

Step-by-Step Solution

Answer

Option (c) is true: \(\text{SF}_6\) is octahedral with \(90^{\circ}\) bond angles.

Step 1: Determine the Hybridization

Let's analyze the hybridization of the central atoms in \(\text{NH}_3\) and \(\text{CH}_4\).\[\text{NH}_3\]: Nitrogen has 5 valence electrons, and it forms 3 bonds with hydrogen atoms and has one lone pair. It is \(\text{sp}^3\) hybridized.\[\text{CH}_4\]: Carbon forms 4 bonds with hydrogen, and it is also \(\text{sp}^3\) hybridized.\ Therefore, option (a) is incorrect.

Step 2: Assess Molecular Shapes

Examine the shapes and geometry of \(\text{BeCl}_2\) and \(\text{SO}_2\).\[\text{BeCl}_2\]: Beryllium is \(\text{sp}\) hybridized with linear geometry.\[\text{SO}_2\]: Sulfur is \(\text{sp}^2\) hybridized with a bent molecular geometry forming a V-shape.\ Thus, option (b) is incorrect.

Step 3: Analyze Geometry and Bond Angles

Consider \(\text{SF}_6\) geometry and bond angles. Sulfur hexafluoride (\(\text{SF}_6\)) is \(\text{sp}^3d^2\) hybridized forming an octahedral shape. The \(\text{F-S-F}\) bond angles are exactly \(90^{\circ}\).\ Thus, option (c) is correct.

Step 4: Check for Dipole Moment

Test whether \(\text{CO}_2\) has a dipole moment.\(\text{CO}_2\) has a linear shape with symmetric distribution of charge, therefore it has no net dipole moment.\ Thus, option (d) is incorrect.

Key Concepts

Hybridization

Chemical bonding often involves hybridization—a concept that explains how atomic orbitals mix together to form new hybrid orbitals. These hybrid orbitals are used by atoms to form covalent bonds. Take \( ext{NH}_3\) and \( ext{CH}_4\) as examples. In ammonia (\( ext{NH}_3\)), nitrogen uses its five valence electrons to form three bonds with hydrogen atoms and has one lone electron pair. This requires \( ext{sp}^3\) hybridization, which involves mixing one s and three p orbitals to form four equivalent orbitals. Similarly, in methane (\( ext{CH}_4\)), carbon forms four bonds with hydrogen atoms, also utilizing \( ext{sp}^3\) hybridization. This makes its bond geometries efficient and symmetrical. Understanding hybridization is crucial because it explains many molecular structures and bonding patterns. Without this concept, predicting how atoms combine to form molecules would be difficult. So whenever you encounter a molecule, consider the type of bonds it forms and the hybridization state of its atoms.

Molecular Geometry

Molecular geometry describes the three-dimensional arrangement of atoms in a molecule. It directly influences a molecule's properties and behavior. For instance, \( ext{BeCl}_2\) and \( ext{SO}_2\) have different geometries due to their hybridization and molecular makeup.

• \( ext{BeCl}_2\): Beryllium chloride features a linear geometry. This is due to beryllium's \( ext{sp}\) hybridization, where one s and one p orbital mix, creating two linearly arranged orbitals.
• \( ext{SO}_2\): Sulfur dioxide showcases a bent or V-shaped structure. Here, sulfur undergoes \( ext{sp}^2\) hybridization, involving one s and two p orbitals, leading to a geometry where two bonded pairs and a lone pair form a less symmetric shape.

These different shapes impact how molecules interact with each other and other substances. They can define reaction capabilities and physical properties such as boiling and melting points.

Dipole Moment

A dipole moment occurs when there's a separation of charge within a molecule, leading to a molecule with a positive and a negative end. Not all molecules exhibit dipole moments, as it largely depends on the molecular geometry and the distribution of the electronegative atoms. Take carbon dioxide (\( ext{CO}_2\)), for instance. Although oxygen is more electronegative than carbon, \( ext{CO}_2\) does not have a dipole moment due to its linear geometry. The two oxygen atoms are symmetrically arranged around the central carbon atom, and their polar bond vectors point in exactly opposite directions, canceling each other out. Dipole moments are crucial in chemistry:

• They influence molecule interactions.
• They affect solubility in polar or nonpolar solvents.
• They are critical in determining the physical properties of substances.

Understanding whether a molecule has a dipole moment helps predict its behavior in different environments.

Bond Angles

Bond angles are the angles between adjacent bonds in a molecule and are crucial in determining molecular shape and potential interactions. These angles are influenced by hybridization and the spatial arrangement of atoms. For instance, consider sulfur hexafluoride (\( ext{SF}_6\)). With its \( ext{sp}^3d^2\) hybridization, \( ext{SF}_6\) forms an octahedral shape, leading to bond angles of exactly \ 90^\circ\. Such standardized angles are typical in symmetrical geometries. In molecules like water and sulfur dioxide, lone pairs force a distortion, often leading to smaller bond angles than expected when only counting atoms. This especially happens in geometries like the bent shape of \( ext{SO}_2\), where lone pairs exert repulsion, decreasing bond angles from the ideal. Understanding bond angles provides insight into molecular stresses and reactivity patterns, helping us predict how molecules might behave in chemical reactions.