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(a) Which geometry and central atom hybridization would you expect in the series \(\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4}, \mathrm{NH}_{4}^{+} ?(\mathbf{b})\) What would you expect for the magnitude and direction of the bond dipoles in this series? (c) Write the formulas for the analogous species of the elements of period 3 ; would you expect them to have the same hybridization at the central atom?

Short Answer

Expert verified
(a) Tetrahedral geometry; \( sp^3 \) hybridization for all. (b) Nonpolar due to symmetry. (c) Analogous species: \( \mathrm{AlH}_{4}^{-}, \mathrm{SiH}_{4}, \mathrm{PH}_{4}^{+} \); same hybridization.

Step by step solution

01

Determine Hybridization and Geometry

For each molecule in the series \( \mathrm{BH}_{4}^{-}, \mathrm{CH}_{4}, \mathrm{NH}_{4}^{+} \), determine the hybridization of the central atom and the molecular geometry. Using VSEPR theory, each molecule can be described as tetrahedral. The hybridization for each central atom (B, C, and N) is \( sp^3 \) since they form four sigma bonds with hydrogen alongside lacking any lone pairs on the central atom.
02

Analyze Bond Dipoles

For each molecule, consider the bonding and the difference in electronegativity between the central atom and hydrogen. In \( \mathrm{CH}_{4} \), the bond dipoles cancel each other out due to symmetry, resulting in a nonpolar molecule. \( \mathrm{BH}_{4}^{-} \) is also nonpolar due to symmetry, but \( \mathrm{NH}_{4}^{+} \) has bonds slightly more polar toward N, still cancelling due to symmetry, thus nonpolar as well.
03

Write Period 3 Analogous Species

For period 3, analogous species are \( \mathrm{AlH}_{4}^{-}, \mathrm{SiH}_{4}, \mathrm{PH}_{4}^{+} \). These have the same tetrahedral geometry and \( sp^3 \) hybridization for the central atom. Each central atom in these molecules forms 4 bonds, like their period 2 counterparts.

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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 provides insight into how atomic orbitals mix to form new orbitals, which can form bonds in molecules. In the series \( \text{BH}_4^- \), \( \text{CH}_4 \), and \( \text{NH}_4^+ \), each central atom (Boron, Carbon, Nitrogen) displays a specific type of hybridization known as \( sp^3 \) hybridization. This occurs when one \( s \) orbital and three \( p \) orbitals merge to form four equivalent \( sp^3 \) hybrid orbitals. Each of these new orbitals can form a sigma bond with a hydrogen atom.
By understanding hybridization, we can predict the bonding and geometry around a central atom. This is crucial in VSEPR Theory, which helps to determine the 3D arrangement of atoms in a molecule. In these molecules, the \( sp^3 \) hybridization indicates a tetrahedral shape as each orbital is oriented towards the corners of a tetrahedron, keeping the electron pairs as far apart as possible.
Molecular Geometry
The molecular geometry of a molecule is the three-dimensional arrangement of the atoms that constitute it. For the molecules in the series \( \text{BH}_4^- \), \( \text{CH}_4 \), and \( \text{NH}_4^+ \), VSEPR theory predicts a tetrahedral geometry. This is due to the fact that these molecules have four regions of electron density around the central atom, which are all bond pairs with no lone pairs.
  • In a tetrahedral molecule, the bond angles are typically around \(109.5^\circ\).
  • The central atom is positioned at the center of a symmetrical layout, which allows for the minimization of electron pair repulsion.
This geometrical shape is crucial in defining the directional properties of these molecules. A tetrahedral geometry ensures that the bond dipoles, which we will discuss further in the electronegativity differences section, can potentially cancel each other out resulting in nonpolar molecules.
Electronegativity Differences
Electronegativity differences between atoms in a bond influence the distribution of electron density and the molecule's polarity. In \( \text{BH}_4^- \), \( \text{CH}_4 \), and \( \text{NH}_4^+ \), the central atom's tendency to attract electrons compared to hydrogen varies.
Electronegativity is a measure of an atom's ability to attract bonding electrons, and differences in electronegativity between two atoms can lead to polar bonds. However, in these examples:
  • In \( \text{CH}_4 \), the carbon and hydrogen electronegativity difference is minimal, leading to nearly nonpolar bonds, and due to the symmetry of the molecule, the dipoles cancel, resulting in a nonpolar molecule.
  • In \( \text{BH}_4^- \), the boron-hydrogen bonds are also symmetrically arranged, canceling out any dipoles produced by the electronegativity difference, making the molecule nonpolar.
  • For \( \text{NH}_4^+ \), nitrogen is slightly more electronegative than hydrogen, creating polar bonds; however, the symmetrical tetrahedral shape ensures that these dipoles cancel, leaving a nonpolar molecule overall.
As you study these molecules, consider how the arrangement and bond polarity affect the molecule's overall polarity. This understanding is foundational when predicting how a molecule will behave in different chemical environments.

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Most popular questions from this chapter

How does a trigonal pyramid differ from a tetrahedron so far as molecular geometry is concerned?

The phosphorus trihalides \(\left(\mathrm{PX}_{3}\right)\) show the following variation in the bond angle \(\mathrm{X}-\mathrm{P}-\mathrm{X}: \mathrm{PF}_{3}, 96.3^{\circ} ; \mathrm{PCl}_{3}, 100.3^{\circ}\); \(\mathrm{PBr}_{3}, 101.0^{\circ} ; \mathrm{PI}_{3}, 102.0^{\circ} .\) The trend is generally attributed to the change in the electronegativity of the halogen. (a) Assuming that all electron domains are the same size, what value of the \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle is predicted by the VSEPR model? (b) What is the general trend in the \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle as the halide electronegativity increases? (c) Using the VSEPR model, explain the observed trend in \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle as the electronegativity of \(X\) changes. (d) Based on your answer to part (c), predict the structure of \(\mathrm{PBrCl}_{4}\).

(a) Predict the electron-domain geometry around the central S atom in \(\mathrm{SF}_{2}, \mathrm{SF}_{4}\), and \(\mathrm{SF}_{6}\). (b) The anion \(\mathrm{IO}_{4}^{-}\) has a tetrahedral structure: three oxygen atoms form double bonds with the central iodine atom and one oxygen atom which carries a negative charge forms a single bond. Predict the molecular geometry of \(\mathrm{IO}_{6}{ }^{5-}\).

Would you expect the nonbonding electron-pair domain in \(\mathrm{NCl}_{3}\) to be greater or smaller in size than the corresponding one in \(\mathrm{PCl}_{3}\)?

How would you expect the extent of overlap of the bonding atomic orbitals to vary in the series IF, ICl, IBr, and \(\mathrm{I}_{2}\) ? Explain your answer.

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