Problem 54 - Chemistry The Central Science

Question

(a) Which geometry and central atom hybridization would you expect in the series \(\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4}, \mathrm{NH}_{4}^{+} ?\) (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

All three molecules, \(\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4}, \mathrm{NH}_{4}^{+}\), exhibit tetrahedral geometry with \(\mathrm{sp^3}\) hybridization. Bond dipoles are weakest in \(\mathrm{BH}_{4}^{-}\) and \(\mathrm{CH}_{4}\), and strongest in \(\mathrm{NH}_{4}^{+}\), pointing from H to the more electronegative central atom. The analogous species for the elements in Period 3 are \(\mathrm{AlH}_{4}^{-}, \mathrm{SiH}_{4}\), and \(\mathrm{PH}_{4}^{+}\), which also exhibit tetrahedral geometry and \(\mathrm{sp^3}\) hybridization.

Step 1: Determine the geometry and central atom hybridization for each molecule#

To determine the geometry and hybridization of the central atom for the given molecules, we first need to find the number of electron pairs surrounding the central atom. Then, we can use VSEPR theory to predict the molecular geometry and hybridization for each molecule. 1. \(\mathrm{BH}_{4}^{-}\): B has 3 valence electrons, and each H has 1 valence electron, plus 1 extra electron from the negative charge. In total, there are 6 valence electrons, and 4 electron pairs surrounding the central B atom. Thus, the geometry is tetrahedral, and the hybridization is \(\mathrm{sp^3}\). 2. \(\mathrm{CH}_{4}\): C has 4 valence electrons, and each H has 1 valence electron. In total, there are 8 valence electrons, and 4 electron pairs surrounding the central C atom. Thus, the geometry is tetrahedral, and the hybridization is \(\mathrm{sp^3}\). 3. \(\mathrm{NH}_{4}^{+}\): N has 5 valence electrons, and each H has 1 valence electron, but 1 electron is lost due to the positive charge. In total, there are 8 valence electrons, and 4 electron pairs surrounding the central N atom. Thus, the geometry is tetrahedral, and the hybridization is \(\mathrm{sp^3}\).

Step 2: Analyze bond dipoles for the series#

Bond dipoles are determined by the difference in electronegativity between the bonded atoms. We'll go through each molecule in the series: 1. \(\mathrm{BH}_{4}^{-}\): The difference in electronegativity between B and H is quite small, so the bond dipoles are relatively weak. 2. \(\mathrm{CH}_{4}\): The difference in electronegativity between C and H is also minimal, so the bond dipoles are quite weak. 3. \(\mathrm{NH}_{4}^{+}\): The difference in electronegativity between N and H is more significant, which results in stronger bond dipoles compared to the other molecules in this series. The direction of the bond dipoles points from the less electronegative atom (H) towards the more electronegative atom (B, C, N).

Step 3: Determine the analogous species for elements in Period 3 and their hybridization#

To find the analogous species for the elements in Period 3, we can simply replace the central atoms (B, C, and N) with their corresponding elements in Period 3: Al, Si, and P. 1. \(\mathrm{AlH}_{4}^{-}\): Analogous to \(\mathrm{BH}_{4}^{-}\), this species has tetrahedral geometry. The central Al atom is surrounded by 4 electron pairs, so the hybridization is \(\mathrm{sp^3}\). 2. \(\mathrm{SiH}_{4}\): Analogous to \(\mathrm{CH}_{4}\), this species has tetrahedral geometry. The central Si atom is surrounded by 4 electron pairs, indicating an \(\mathrm{sp^3}\) hybridization. 3. \(\mathrm{PH}_{4}^{+}\): Analogous to \(\mathrm{NH}_{4}^{+}\), this species has tetrahedral geometry. The central P atom is surrounded by 4 electron pairs, which implies an \(\mathrm{sp^3}\) hybridization. The species formed from Period 3 elements exhibit the same hybridization as their Period 2 counterparts.

Key Concepts

VSEPR Theory

The Valence Shell Electron Pair Repulsion (VSEPR) theory is crucial for determining the molecular geometry of compounds. It relies on the idea that electron pairs around a central atom will arrange themselves as far apart as possible to minimize repulsion. This helps predict the shape of molecules. For example:

• In \( \mathrm{BH}_4^- \), \( \mathrm{CH}_4 \), and \( \mathrm{NH}_4^+ \), each central atom (B, C, and N respectively) is surrounded by four electron pairs.
• According to VSEPR theory, when there are four pairs of electrons, they arrange themselves in a way to maximize distance, leading to a tetrahedral geometry.

Thus, by counting electron pairs around a central atom, and using VSEPR theory, one can easily predict the molecular structure.

Central Atom Hybridization

The concept of hybridization helps explain how atomic orbitals mix to form new, hybrid orbitals in a molecule. These hybrid orbitals dictate the geometry and bonding properties of the molecule. For the series \( \mathrm{BH}_4^- \), \( \mathrm{CH}_4 \), and \( \mathrm{NH}_4^+ \), all three molecules exhibit \( \mathrm{sp^3} \) hybridization.

• Each central atom forms four bonds by mixing one \( s \) and three \( p \) orbitals from its valence shell, creating four equivalent \( \mathrm{sp^3} \) hybrid orbitals.
• This results in a consistent arrangement of electrons that supports the tetrahedral geometry observed in these molecules.

Understanding hybridization not only clarifies the geometry of molecules but also provides insights into bond strength and angles within them.

Bond Dipoles

A bond dipole arises from a difference in electronegativity between two atoms that are bonded together. This means that one side of the bond will be partially negative, and the other side will be partially positive. Let's examine this for our series \( \mathrm{BH}_4^- \), \( \mathrm{CH}_4 \), and \( \mathrm{NH}_4^+ \):

• In \( \mathrm{BH}_4^- \), the electronegativity difference between B and H is slight, so the dipoles are weak.
• In \( \mathrm{CH}_4 \), the electronegativity difference between C and H is similarly minimal, resulting in weak dipoles.
• In \( \mathrm{NH}_4^+ \), there is a greater electronegativity difference between N and H, leading to stronger dipoles.

The direction of each dipole is towards the more electronegative atom. Understanding bond dipoles is essential for predicting molecular polarity and interactions between molecules.