Problem 58 - Chemistry The Central Science

Question

(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?

Step-by-Step Solution

Answer

In the series BH4-, CH4, and NH4+, all molecules have a tetrahedral geometry and sp3 hybridization for their central atoms. The bond dipoles will be small due to small electronegativity differences and will cancel out due to the tetrahedral arrangement. The analogous species SiH4 from period 3 also has tetrahedral geometry and sp3 hybridization for its central atom.

Step 1: (a) Molecular Geometry and Hybridization

We will identify the molecular geometry and hybridization of each central atom in BH4-, CH4, and NH4+. 1. BH4-: The central atom B has 3 valence electrons, but it will share an extra electron due to the negative charge, making it 4 electron pairs. From VSEPR theory, 4 electron pairs result in a tetrahedral geometry, and the hybridization will be sp3. 2. CH4: The central atom C has 4 valence electrons, making 4 electron pairs, resulting in a tetrahedral geometry and sp3 hybridization. 3. NH4+: The central atom N has 5 valence electrons, but it will lose 1 electron due to the positive charge, making 4 electron pairs, resulting in a tetrahedral geometry and sp3 hybridization.

Step 2: (b) Bond Dipoles Magnitude and Direction

In BH4-, CH4, and NH4+ molecules, each central atom electrons are not shared evenly because of the varying electronegativity between atoms. Therefore, bond dipoles will be present. The difference in electronegativity will determine the bond dipole's magnitude, while the spatial arrangement of atoms determines the direction. 1. BH4-: In boron, the difference in electronegativity between B and H is small, so the bond dipoles will be small. The bond dipoles will point from H to B, and they will cancel each other out due to tetrahedral geometry. 2. CH4: Carbon and hydrogen have a small difference in electronegativity, so bond dipoles will be small and point from H to C, also canceling out because of the tetrahedral arrangement. 3. NH4+: In nitrogen, the difference in electronegativity between N and H is significant, so the bond dipoles will be larger and point from H to N. Due to the tetrahedral geometry, these bond dipoles will cancel each other out.

Step 3: (c) Formulas for Analogous Species and Hybridization Comparison

For the analogous species of the elements of period 3, SI4-, SiH4, and Pi4+ formulas can be written. The hybridizations of these species can be compared to those of period 2 analogs: 1. SiH4: Si is the central atom, has 4 valence electrons, which form 4 electron pairs. Due to its tetrahedral geometry and VSEPR theory, SiH4 will have sp3 hybridization, the same as CH4. It is important to note that the SiH4 is neutral, so there won't be similar species like period 2 analogs (SI4- and Pi4+). In summary, the given molecules have tetrahedral geometry, and sp3 hybridization while their bond dipoles cancel each other due to their spatial arrangement. The period 3 analogous species SiH4 also has the same tetrahedral geometry and sp3 hybridization.

Key Concepts

Hybridization

Hybridization is a fundamental concept in chemistry that explains the formation of equivalent orbitals for bonding. In the case of \(\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4}, \mathrm{NH}_{4}^{+}\), each molecule undergoes \(\text{sp}^3\) hybridization. This occurs when one \(s\) orbital and three \(p\) orbitals mix to form four equivalent \(\text{sp}^3\) hybrid orbitals. Each of these orbitals contains one unpaired electron, which can pair with an electron from a hydrogen atom to form a covalent bond.
For example:

• \(\mathrm{BH}_{4}^{-}\): The boron atom, initially having 3 valence electrons and gaining one more due to the negative charge, forms 4 covalent bonds using \(\text{sp}^3\) hybrid orbitals.
• \(\mathrm{CH}_{4}\): The carbon atom, with 4 valence electrons, forms 4 covalent bonds in a similar \(\text{sp}^3\) manner.
• \(\mathrm{NH}_{4}^{+}\): The nitrogen atom will have 5 valence electrons, but loses one due to the positive charge, resulting in 4 covalent bonds through \(\text{sp}^3\) hybridization.

Thus, hybridization allows for the formation of stable tetrahedral structures in these molecules.

VSEPR Theory

The Valence Shell Electron Pair Repulsion (VSEPR) theory is crucial for predicting the geometry of molecules. It states that electron pairs around a central atom tend to arrange themselves as far apart as possible to minimize repulsion. This theory helps us understand why molecules like \(\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4}, \mathrm{NH}_{4}^{+}\) exhibit a tetrahedral shape.
In these molecules, the central atom forms 4 equivalent notations of electron pairs (bond pairs) which spread uniformly in three-dimensional space. The VSEPR model predicts the tetrahedral geometry based on this configuration because:

• Each pair of bonding electrons experience mutual repulsion.
• By occupying positions at 109.5 degrees from each other, the repulsions are minimized.

This leads to a stable and symmetrical arrangement, which is vital in determining molecular behavior and interaction.

Tetrahedral Geometry

A tetrahedral geometry is a three-dimensional arrangement where a central atom is bonded to four other atoms at the corners of a tetrahedron. This can be visualized as a pyramid with a triangular base. In molecules like \(\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4}, \mathrm{NH}_{4}^{+}\), each hydrogen atom occupies a corner of this shape.
Here are some key points:

• The bond angles between the hydrogen atoms are approximately 109.5 degrees, providing the maximum separation and minimizing repulsion as per the VSEPR theory.
• This geometry is common in molecules where the central atom is involved in \(\text{sp}^3\) hybridization.

The tetrahedral arrangement is symmetrical, which often leads to non-polar characteristics if all peripheral atoms are identical, as they are in these examples.

Bond Dipoles

Bond dipoles occur due to the difference in electronegativity between the bonded atoms. Each bond behaves like a tiny magnet with a positive and negative end. In \(\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4}, \mathrm{NH}_{4}^{+}\), these dipoles are subtle due to the small differences in electronegativity between hydrogen and the central atoms.
To understand how these dipoles behave:

• In \(\mathrm{BH}_{4}^{-}\) and \(\mathrm{CH}_{4}\), the electronegativity difference is small, leading to weak dipoles pointing from hydrogen to the central atom.
• In \(\mathrm{NH}_{4}^{+}\), the difference is more significant, creating slightly larger dipoles that point from hydrogen to nitrogen.

Despite their presence, in these tetrahedral molecules, the symmetrical geometry means that the dipoles cancel each other out, leading to overall non-polar molecules. Understanding bond dipoles is essential for predicting molecular behavior, especially in interactions with other substances.