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Chapter 4: Problem 21

The \(\nu_{3}\) vibrational wavenumber for \(\left[\mathrm{BF}_{4}\right]^{-}\) comes at \(1070 \mathrm{cm}^{-1},\) whereas the corresponding band for \(\left[\mathrm{BCl}_{4}\right]^{-},\left[\mathrm{BBr}_{4}\right]^{-}\) and \(\left[\mathrm{BI}_{4}\right]^{-}\) comes at 722,620 and \(533 \mathrm{cm}^{-1},\) respectively. Rationalize this trend.

Short Answer

Expert verified
Heavier and weaker B-X bonds lead to lower vibrational wavenumbers in \( [\mathrm{BX}_{4}]^{-} \) molecules.

Step by step solution

01

Understanding Vibrational Wavenumbers

The vibrational wavenumber is related to the frequency of the vibrations of the atoms within a molecule. Higher wavenumbers indicate higher vibrational frequencies. In tetrahedral molecules like \( [\mathrm{BX}_{4}]^{-} \), these vibrations refer to the symmetrical stretching of the B-X bonds where X can be F, Cl, Br or I.
02

Analyzing Molecular Mass

The vibrational frequency is inversely related to the mass of the atom involved in the bond (based on the reduced mass in the vibration formula \( u = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}} \), where \(k\) is the force constant and \(\mu\) the reduced mass). In our molecules, as we move from F to Cl to Br to I, the mass of X increases, and thus the reduced mass \(\mu\) of the B-X bond increases, lowering the vibrational frequency.
03

Evaluating Force Constants

The force constant \(k\) reflects the bond strength. Generally, lighter atoms like fluorine (F) form stronger bonds with boron (B) compared to heavier halogens like Cl, Br, or I, resulting in a higher force constant. A higher \(k\) leads to a higher vibrational wavenumber.
04

Synthesizing the Information

Combining these factors, \( [\mathrm{BF}_{4}]^{-} \) has the highest wavenumber due to the low mass of F and a strong B-F bond. As we replace F with heavier halogens (Cl, Br, I), the mass increases and the bond strength decreases, resulting in decreased wavenumbers. Thus, we observe \( [\mathrm{BF}_{4}]^{-} \), \( [\mathrm{BCl}_{4}]^{-} \), \( [\mathrm{BBr}_{4}]^{-} \), and \( [\mathrm{BI}_{4}]^{-} \) in diminishing order of vibrational wavenumbers.

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Key Concepts

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

Tetrahedral molecules
In the world of chemistry, the shape of a molecule can greatly influence its properties. One common shape is the tetrahedral structure. Imagine a central atom bonded to four other atoms, almost like a three-dimensional pyramid.
In the case of the molecules discussed 鈥� \( [\mathrm{BF}_{4}]^{-} \), \( [\mathrm{BCl}_{4}]^{-} \), \( [\mathrm{BBr}_{4}]^{-} \), and \( [\mathrm{BI}_{4}]^{-} \) 鈥� boron is at the center of the pyramid, and each halogen forms the pyramid's corners.
This shape allows for the symmetrical stretching of bonds which is important in vibrations.
  • The tetrahedral geometry ensures that each B-X bond angles are equal, making the vibration more predictable.
  • Tetrahedral structures are relatively compact, contributing to higher vibrational frequencies.
Understanding the shape helps us make sense of the vibrational properties observed in these molecules.
Symmetrical stretching
Symmetrical stretching refers to a specific type of molecular vibration where bonds expand and contract in unison. Picture a tetrahedral molecule where all bonds are pulsing in and out together.
This type of vibration is common in molecules like \( [\mathrm{BF}_{4}]^{-} \).
The characteristics of symmetrical stretching include:
  • All changes in bond length happen at the same rate, maintaining the symmetry around the central atom.
  • Symmetrical stretching occurs at distinct frequencies that can be detected via vibrational spectroscopy techniques like IR spectroscopy.
An interesting aspect is how this stretching frequency shifts depending on the mass of the atoms involved and bond strength. In our analysis, it helps rationalize why wavenumbers differ between the molecules.
Reduced mass in vibration
Reduced mass is a concept that helps to better understand how two bonded atoms move with respect to each other. It is especially important when considering vibrational frequencies.
The formula used is \( \mu = \frac{m_{1}m_{2}}{m_{1}+m_{2}} \), where \( m_{1} \) and \( m_{2} \) are the masses of the two atoms involved in the bond.
Here's why reduced mass matters for vibrations:
  • It influences the vibrational frequency: The greater the reduced mass, the lower the vibrational frequency, because heavier atoms are harder to accelerate.
  • In the context of \( [\mathrm{BX}_{4}]^{-} \) molecules, as we move from F to I, the mass of X increases, which influences the reduced mass and subsequently the vibrational frequency.
Understanding reduced mass helps to explain why lighter molecules like \( [\mathrm{BF}_{4}]^{-} \) have higher wavenumbers compared to heavier ones like \( [\mathrm{BI}_{4}]^{-} \).
Force constant in bonds
The force constant, often denoted as \( k \), measures the stiffness of a bond and its ability to resist stretching or compressing. In vibrational spectroscopy, it directly affects the vibrational wavenumber.
The force constant is higher in bonds where atoms are tightly bound, like in \( B-F \) bonds. Here are some key ideas:
  • Lighter atoms, such as fluorine, form stronger bonds with boron, resulting in a larger force constant \( k \).
  • Conversely, as we move down the halogen group to iodine, the B-I bond becomes weaker and thus has a lower force constant.
  • Higher force constants translate to more energy required to stretch the bond, hence a higher vibrational frequency or wavenumber.
By considering the force constant, it's clear why \( [\mathrm{BF}_{4}]^{-} \) has the highest vibrational wavenumber among the series of analyzed molecules.

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