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Chapter 4: Problem 42

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 "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

Short Answer

Expert verified
Option (c) is true: \(\mathrm{SF}_{6}\) is octahedral with bond angles of \(90^{\circ}\).

Step by step solution

01

Understanding the Options

The task is to identify the true statement among the given options. Each statement is about the molecular structure or properties of the compounds. We will analyze each statement for accuracy based on known chemical principles.
02

Analyze Option (a)

In option (a), we need to determine the hybridization of the central atom in \(\mathrm{NH}_{3}\) and \(\mathrm{CH}_{4}\). \(\mathrm{NH}_{3}\) has a central nitrogen atom with three hydrogen bonds and a lone pair, which leads to \(\mathrm{sp}^3\) hybridization. \(\mathrm{CH}_{4}\) has a central carbon atom with four hydrogen bonds, also leading to \(\mathrm{sp}^3\) hybridization. Thus, the statement that both have \(\mathrm{sp}^2\) hybridization is false.
03

Analyze Option (b)

For option (b), we check the molecular shapes. \(\mathrm{BeCl}_{2}\) in its gaseous form is linear because beryllium forms two bonds with chlorine atoms without any lone pairs, leading to a linear shape. \(\mathrm{SO}_{2}\) has a bent or "V" shape due to the presence of lone pairs on sulfur. Therefore, this option is also false.
04

Analyze Option (c)

Option (c) involves \(\mathrm{SF}_{6}\), which has an octahedral geometry. In an octahedral arrangement, the bond angle between \(\mathrm{F}-\mathrm{S}-\mathrm{F}\) is indeed \(90^{\circ}\). Therefore, this statement is true.
05

Analyze Option (d)

In option (d), \(\mathrm{CO}_{2}\) is a linear molecule with two polar carbon-oxygen bonds. However, due to the linearity, the dipoles cancel each other out, resulting in no net dipole moment. Thus, this statement is false.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hybridization
Hybridization is a concept that explains the mixing of atomic orbitals to form new hybrid orbitals. These new orbitals have different energies and shapes than the original orbitals and are necessary for the formation of bonds in molecules. The type of hybridization is determined by the number of electron domains (bonding or lone pairs) around the central atom.
  • For example, in ammonia (\(\mathrm{NH}_3\)), the central nitrogen atom has three bonding pairs and one lone pair, resulting in \(\mathrm{sp}^3\) hybridization.
  • This means four orbitals mix to form four equivalent hybrids that explain the molecule's 3D shape.
  • Similarly, methane (\(\mathrm{CH}_4\)) has four bonding pairs and no lone pairs, yielding \(\mathrm{sp}^3\) hybridization, where four equivalent \(\mathrm{sp}^3\) orbitals align in a tetrahedral shape.
Understanding hybridization helps clarify the geometry and bonding capabilities of molecules. This is crucial for predicting molecular properties and reactivity.
Molecular Shape
The molecular shape or geometry of a molecule is determined by the spatial arrangement of its atoms. It is a direct consequence of molecular hybridization and the VSEPR (Valence Shell Electron Pair Repulsion) theory.
  • For instance, beryllium chloride (\(\mathrm{BeCl}_2\)) is linear because beryllium forms only two bonds, with no lone pairs on the central atom, allowing a 180-degree separation.
  • Sulfur dioxide (\(\mathrm{SO}_2\)), in contrast, exhibits a "V" shape or bent geometry due to the presence of lone pairs on sulfur that repel bonding pairs, compressing the bond angle.
  • Molecular shapes such as linear, bent, tetrahedral, trigonal planar, and octahedral are among the basics you'll encounter when exploring molecular geometry.
Understanding the molecular shape is vital for predicting physical and chemical properties, such as solubility, polarity, and reaction pathways.
Dipole Moment
Dipole moment is a measure of the separation of positive and negative charges in a molecule. It determines the molecule's overall polarity. A polar molecule has a net dipole moment due to unequal distribution of electrons between atoms with different electronegativities.
  • For example, carbon dioxide (\(\mathrm{CO}_2\)) has polar bonds; however, because of its linear shape, the dipoles from the two \(\mathrm{C=O}\) bonds cancel each other out, resulting in no overall dipole moment.
  • This is why \(\mathrm{CO}_2\) is considered a non-polar molecule despite having polar bonds.
The direction and magnitude of dipole moments affect interaction with other molecules. Thus, they influence boiling points, solubility, and the molecule's behavior in electric fields.
Bond Angles
Bond angles are the angles formed between adjacent lines representing bonds. They are closely related to a molecule's hybridization and shape. Understanding bond angles is essential for visualizing the 3-dimensional structure of molecules.
  • In sulfur hexafluoride (\(\mathrm{SF}_6\)), the octahedral geometry creates bond angles of \(90^{\circ}\) between the \(\mathrm{F}-\mathrm{S}-\mathrm{F}\) bonds.
  • This arrangement ensures the six fluorine atoms are as far apart as possible, minimizing repulsion.
  • On the other hand, a \(\mathrm{CH}_4\) molecule with \(\mathrm{sp}^3\) hybridization has bond angles of \(109.5^{\circ}\).
Recognizing bond angles helps in determining optimal electron pair arrangements, which is key in predicting a molecule鈥檚 stability and interactions.

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Most popular questions from this chapter

\(\mathrm{H}_{2} \mathrm{O}\) is dipolar, whereas \(\mathrm{BeF}_{2}\) is not. It is because: (a) The electronegativity of \(\mathrm{F}\) is greater than that of \(\mathrm{O}\) (b) \(\mathrm{H}_{2} \mathrm{O}\) involves hydrogen bonding whereas \(\mathrm{BeF}_{2}\) is a discrete molecule (c) \(\mathrm{H}_{2} \mathrm{O}\) is linear and \(\mathrm{BeF}_{2}\) is angular (d) \(\mathrm{H}_{2} \mathrm{O}\) is angular and \(\mathrm{BeF}_{2}\) is linear

What is the hybridization state of the central atom in the conjugate base of \(\mathrm{NH}_{4}^{+}\) ion? (a) sp (b) \(\mathrm{sp}^{3}\) (c) \(\mathrm{sp}^{3}\) (d) \(\mathrm{dsp}^{2}\)

The correct order of bond angles (smallest first) in \(\mathrm{H}_{2} \mathrm{~S}, \mathrm{NH}_{3}, \mathrm{BF}_{3}\) and \(\mathrm{SiH}_{4}\) is: (a) \(\mathrm{H}_{2} \mathrm{~S}<\mathrm{SiH}_{4}<\mathrm{NH}_{3}<\mathrm{BF}_{3}\) (b) \(\mathrm{NH}_{3}<\mathrm{H}_{2} \mathrm{~S}<\mathrm{SiH}_{4}<\mathrm{BF}_{3}\) (c) \(\mathrm{H}_{2} \mathrm{~S}<\mathrm{NH}_{3}<\mathrm{SiH}_{4}<\mathrm{BF}_{3}\) (d) \(\mathrm{H}_{2} \mathrm{~S}<\mathrm{NH}_{3}<\mathrm{BF}_{3}<\mathrm{SiH}_{4}\)

The sequence that correctly describes the relative bond strength pertaining to oxygen molecule and its cation or anion is: (a) \(\mathrm{O}_{2}^{2-}>\mathrm{O}_{2}^{-}>\mathrm{O}_{2}>\mathrm{O}_{2}^{+}\) (b) \(\mathrm{O}_{2}>\mathrm{O}_{2}^{+}>\mathrm{O}_{2}^{-}>\mathrm{O}_{2}^{2}\) (c) \(\mathrm{O}_{2}^{2}>\mathrm{O}_{2}>\mathrm{O}_{2}^{2-}>\mathrm{O}_{2}^{2}\) (d) \(\mathrm{O}_{2}^{+}>\mathrm{O}_{2}>\mathrm{O}_{2}^{-}>\mathrm{O}_{2}^{2}\)

In \(\mathrm{X}-\mathrm{Hi}-\mathrm{Y}\), where both \(\mathrm{X}\) and \(\mathrm{Y}\) are electronegative elements: (a) Electron density of \(\mathrm{X}\) will increase and the eletron density on \(\mathrm{H}\) will decrease (b) On both species electron density will increase (c) On both species electron density will decrease (d) On \(\mathrm{X}\) the electron density will decrease and on \(\mathrm{H}\) it will increases
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