Chapter 15: Problem 37
\(\mathrm{Ti}\left(\mathrm{NH}_{3}\right)_{6}{ }^{3+}\) has a d-orbital electron transition at \(399 \mathrm{~nm}\). Find \(\Delta_{o}\) at this wavelength.
Short Answer
Expert verified
Answer: The energy difference between the d orbitals involved in the transition is \(\Delta_{o} = 3.102 \mathrm{eV}\).
Step by step solution
01
Recall the relationship between energy and wavelength
The energy of a photon is related to its wavelength as follows: $$E = \frac{hc}{\lambda}$$ where \(E\) is the energy, \(h\) is Planck's constant (\(6.626 \times 10^{-34} \mathrm{Js}\)), \(c\) is the speed of light (\(3.0 \times 10^8 \mathrm{ms}^{-1}\)), and \(\lambda\) is the wavelength of the transition.
02
Convert the wavelength to meters
Given the wavelength is \(399 \mathrm{~nm}\), we need to convert it into meters. $$\lambda = 399 \mathrm{~nm} \times \frac{1 \mathrm{~m}}{10^9 \mathrm{~nm}} = 3.99 \times 10^{-7}\mathrm{~m}$$
03
Calculate the energy of the transition
Now that we have the wavelength in meters, we can calculate the energy of the transition using the equation from Step 1. $$E = \frac{6.626 \times 10^{-34} \mathrm{Js} \cdot 3.0 \times 10^8 \mathrm{ms}^{-1}}{3.99 \times 10^{-7} \mathrm{m}} = 4.978 \times 10^{-19} \mathrm{J}$$
04
Convert the energy to electron volts
Since the energy difference between d orbitals is usually expressed in electron volts (eV), we need to convert the energy we just calculated to eV. We can do that using the conversion factor: \(1 \mathrm{eV} = 1.602 \times 10^{-19} \mathrm{J}\). $$\Delta_{o} = 4.978 \times 10^{-19} \mathrm{J} \times \frac{1 \mathrm{eV}}{1.602 \times 10^{-19} \mathrm{J}} = 3.102 \mathrm{eV}$$
05
Final Answer
The energy difference between the d orbitals involved in the transition is \(\Delta_{o} = 3.102 \mathrm{eV}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
d-orbital electron transition
When we talk about d-orbital electron transition, we’re dealing with the movement of electrons between d-orbitals within a metal ion. These transitions are often observed in complexes like \(\mathrm{Ti}\left(\mathrm{NH}_3\right)_6^{3+}\), where electrons are excited from a lower energy d-orbital to a higher one.
This movement is crucial for understanding the color and properties of the complex, as the light absorbed corresponds to specific wavelengths. These are identified by the accompanying change in energy, which is often calculated in chemistry to understand the molecular structure and energy distribution.
The specific wavelength at which this transition occurs tells us about the energy difference \(\Delta_o\) between these d-orbitals.
This movement is crucial for understanding the color and properties of the complex, as the light absorbed corresponds to specific wavelengths. These are identified by the accompanying change in energy, which is often calculated in chemistry to understand the molecular structure and energy distribution.
The specific wavelength at which this transition occurs tells us about the energy difference \(\Delta_o\) between these d-orbitals.
energy calculation in chemistry
Energy calculation in chemistry involves determining the energy changes associated with reactions or transitions, like the d-orbital transitions we discussed. The formula \(E = \frac{hc}{\lambda}\) is pivotal here.
- \(h\) is Planck's constant, \(6.626 \times 10^{-34} \mathrm{Js}\).
- \(c\) is the speed of light, \(3.0 \times 10^8 \mathrm{ms}^{-1}\).
- \(\lambda\) is the wavelength. This formula allows us to find the energy change resulting from electron transitions, helping predict the behavior of molecules.
In practical terms, converting nanometers to meters is necessary to use this formula accurately.
- \(h\) is Planck's constant, \(6.626 \times 10^{-34} \mathrm{Js}\).
- \(c\) is the speed of light, \(3.0 \times 10^8 \mathrm{ms}^{-1}\).
- \(\lambda\) is the wavelength. This formula allows us to find the energy change resulting from electron transitions, helping predict the behavior of molecules.
In practical terms, converting nanometers to meters is necessary to use this formula accurately.
wavelength to energy conversion
Converting wavelength to energy is an essential step to determine how much energy is involved in electronic transitions.
The wavelength, given in nanometers (nm), must be converted to meters since physics equations generally use the metric system.
For example, \(399 \mathrm{~nm}\) becomes \(3.99 \times 10^{-7} \mathrm{~m}\). This conversion enables us to use formulas like \(E = \frac{hc}{\lambda}\) correctly.
The wavelength, given in nanometers (nm), must be converted to meters since physics equations generally use the metric system.
For example, \(399 \mathrm{~nm}\) becomes \(3.99 \times 10^{-7} \mathrm{~m}\). This conversion enables us to use formulas like \(E = \frac{hc}{\lambda}\) correctly.
- Measure light energy in joules (J) initially.
- The process bridges the gap from observed wavelength to calculated energy.
electron volts
An electron volt (eV) is a unit of energy used widely in physics and chemistry. It's often more convenient than the joule for small-scale interactions, such as electron transitions in molecules.
For example, converting from joules to electron volts involves using the conversion factor \(1 \mathrm{eV} = 1.602 \times 10^{-19} \mathrm{J}\).
So, if we have \(4.978 \times 10^{-19} \mathrm{J}\), converting it gives \(3.102 \mathrm{eV}\). This allows chemists to discuss energy changes in simpler, more relatable terms.
For example, converting from joules to electron volts involves using the conversion factor \(1 \mathrm{eV} = 1.602 \times 10^{-19} \mathrm{J}\).
So, if we have \(4.978 \times 10^{-19} \mathrm{J}\), converting it gives \(3.102 \mathrm{eV}\). This allows chemists to discuss energy changes in simpler, more relatable terms.
- Great for small-scale processes.
- Makes complex transition energies understandable.
crystal field splitting energy
Crystal field splitting energy \(\Delta_o\) is a concept in chemistry that describes the separation of d-orbitals in a crystal field, which happens due to the influence of ligands. As seen in complexes like \(\mathrm{Ti}\left(\mathrm{NH}_3\right)_6^{3+}\), ligands cause a difference in energy levels among d-orbitals.
This energy difference \(\Delta_o\) can be calculated using the energy of the absorbed light, as seen in the equation \(\Delta_o = 3.102 \mathrm{eV}\).
This explains the color and electronic properties of the complex,
as each wavelength of absorbed light equates to a specific energy difference. Understanding \(\Delta_o\) is essential for predicting the behavior and characteristics of metal complexes.
This energy difference \(\Delta_o\) can be calculated using the energy of the absorbed light, as seen in the equation \(\Delta_o = 3.102 \mathrm{eV}\).
This explains the color and electronic properties of the complex,
as each wavelength of absorbed light equates to a specific energy difference. Understanding \(\Delta_o\) is essential for predicting the behavior and characteristics of metal complexes.