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Determine whether the following transitions are allowed or forbidden: a. \(\sqrt[3]{\Pi}_{u} \rightarrow^{3} \Sigma_{g}^{-}\) \(\mathbf{b} .^{1} \Sigma_{g}^{+} \rightarrow^{1} \Pi_{g}\) \(\mathbf{c} .^{3} \Sigma_{g}^{-} \rightarrow^{3} \Pi_{g}\) d. \(^{1} \Pi_{g} \rightarrow^{1} \Delta_{u}\)

Short Answer

Expert verified
Transitions a and d are allowed (as all selection rules are satisfied), while transitions b and c are forbidden (as the parity selection rule is violated).

Step by step solution

01

Transition a

For the transition \(\sqrt[3]{\Pi}_{u} \rightarrow^{3} \Sigma_{g}^{-}\), we need to check the selection rules: 1. Change in electronic spin: Initial state: S=1, Final state: S=1 ∆S = 1 - 1 = 0, which satisfies the selection rule. 2. Change in total orbital angular momentum: Initial state: Λ=1 (π), Final state: Λ=0 (Σ) ∆Λ = 1 - 0 = 1, which satisfies the selection rule. 3. Parity: Initial state: u (ungerade), Final state: g (gerade) The parity has changed, which satisfies the selection rule. Since all the selection rules are satisfied, the transition a is allowed.
02

Transition b

For the transition \(^{1}\Sigma_{g}^{+} \rightarrow^{1}\Pi_{g}\), we need to check the selection rules: 1. Change in electronic spin: Initial state: S=0, Final state: S=0 ∆S = 0 - 0 = 0, which satisfies the selection rule. 2. Change in total orbital angular momentum: Initial state: Λ=0 (Σ), Final state: Λ=1 (π) ∆Λ = 1 - 0 = 1, which satisfies the selection rule. 3. Parity: Initial state: g (gerade), Final state: g (gerade) The parity has not changed, which violates the selection rule. Since the parity selection rule is violated, the transition b is forbidden.
03

Transition c

For the transition \(^{3}\Sigma_{g}^{-} \rightarrow^{3}\Pi_{g}\), we need to check the selection rules: 1. Change in electronic spin: Initial state: S=1, Final state: S=1 ∆S = 1 - 1 = 0, which satisfies the selection rule. 2. Change in total orbital angular momentum: Initial state: Λ=0 (Σ), Final state: Λ=1 (π) ∆Λ = 1 - 0 = 1, which satisfies the selection rule. 3. Parity: Initial state: g (gerade), Final state: g (gerade) The parity has not changed, which violates the selection rule. Since the parity selection rule is violated, the transition c is forbidden.
04

Transition d

For the transition \(^{1}\Pi_{g} \rightarrow^{1}\Delta_{u}\), we need to check the selection rules: 1. Change in electronic spin: Initial state: S=0, Final state: S=0 ∆S = 0 - 0 = 0, which satisfies the selection rule. 2. Change in total orbital angular momentum: Initial state: Λ=1 (π), Final state: Λ=2 (Δ) ∆Λ = 2 - 1 = 1, which satisfies the selection rule. 3. Parity: Initial state: g (gerade), Final state: u (ungerade) The parity has changed, which satisfies the selection rule. Since all the selection rules are satisfied, the transition d is allowed. In conclusion, transitions a and d are allowed, while transitions b and c are forbidden.

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Key Concepts

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

Selection Rules
In the world of molecular spectroscopy, selection rules serve as vital guidelines to determine whether a particular electronic transition is allowed or forbidden. Fundamentally, these rules are based on the symmetries and properties of an electron's quantum states and the interaction with electromagnetic radiation.

For instance, when examining transitions like \(\sqrt[3]{\Pi}_{u} \rightarrow^{3} \Sigma_{g}^{-}\), we apply selection rules concerning electronic spin, orbital angular momentum, and parity. If all the rules are satisfied, the transition is deemed 'allowed'; otherwise, it's 'forbidden'. This concept is instrumental in predicting the likelihood of a transition occurring, which in turn affects the absorption or emission spectrum of a molecule.
Parity Selection Rule
The parity selection rule is crucial in molecular electronic transitions. It dictates that for an electronic transition to be allowed, the parity must change from even (gerade, 'g') to odd (ungerade, 'u') or vice versa. This is rooted in the fact that electromagnetic radiation (photons) have odd parity.

When we look at a transition like \(^{1}\Sigma_{g}^{+} \rightarrow^{1}\Pi_{g}\), we can quickly determine it violates the parity selection rule as the transition is from a gerade to another gerade state. Hence, conservation of parity in this context results in a forbidden transition. Understanding this rule helps predict the observed spectra and the exact conditions under which an electronic transition may occur.
Orbital Angular Momentum
Another critical concept in molecular transitions is orbital angular momentum, which is determined by the orbital angular momentum quantum number, represented by \(\Lambda\). In electronic transitions, the selection rule states that \(\Delta\Lambda = \pm 1\) is allowed, meaning the quantum number can change by one unit.

For instance, when considering a transition from \(\Pi\) to \(\Sigma\), where \(\Pi\) has \(\Lambda=1\) and \(\Sigma\) has \(\Lambda=0\), such a change satisfies the orbital angular momentum selection rule, signaling an allowed transition. A grasp of this concept is essential for discerning the types of possible transitions within a molecule's electron configuration.
Electronic Spin
Finally, electronic spin plays a determining role in the selection rules for molecular transitions. An electron has a spin quantum number (S), which can impact transitions due to the 'spin selection rule'. This rule specifies that for an electronic transition to be allowed, there should be no change in the multiplicity of the electronic state, which implies \(\Delta S = 0\).

Transitions such as \(\sqrt[3]{\Pi}_{u} \rightarrow^{3} \Sigma_{g}^{-}\) conform to this rule since the initial and final states both have a spin multiplicity of three (\(S=1\)), resulting in \(\Delta S = 0\). Understanding electronic spin and its selection rules is indispensable for predicting the possible electronic transitions and interpreting spectroscopic patterns.

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Most popular questions from this chapter

The ground electronic state of \(\mathrm{O}_{2}\) is \(^{3} \Sigma_{g}^{-},\) and the next two highest energy states are \(^{1} \Delta_{g}\left(7918 \mathrm{cm}^{-\mathrm{f}}\right),\) and \(^{1} \Sigma_{g}^{+}\) \(\left(13195 \mathrm{cm}^{-1}\right)\) where the value in parentheses is the energy of the state relative to the ground state. a. Determine the excitation wavelength required for a transition between the ground state and the first two excited states. b. Are these transitions allowed? Why or why not?

Structural changes in proteins have been measured using FRET with the amino acid tryptophan as the donor and dansyl as the acceptor where dansyl is attached to the protein through addition to amino acids with aliphatic amine groups such as lysine. For this pair \(R_{0}=2.1 \mathrm{nm},\) and the excited state lifetime of tryptophan is \(\sim 1.0\) ns. Determine the rate of energy transfer for \(r=0.50,1.0,2.0,3.0,\) and \(5.0 \mathrm{nm}\)

Ozone \(\left(\mathrm{O}_{3}\right)\) has an absorptivity at \(300 .\) nm of 0.00500 torr \(^{-1} \mathrm{cm}^{-1} .\) In atmospheric chemistry the amount of ozone in the atmosphere is quantified using the Dobson unit (DU), where 1 DU is equivalent to a \(10^{-2} \mathrm{mm}\) thick layer of ozone at 1 atm and \(273.15 \mathrm{K}\) a. Calculate the absorbance of the ozone layer at \(300 .\) nm for a typical coverage of \(300 .\) DU. b. Seasonal stratospheric ozone depletion results in a decrease in ozone coverage to values as low as \(120 .\) DU. Calculate the absorbance of the ozone layer at this reduced coverage. In each part, also calculate the transmission from the absorbance using Beer's Law.

Consider a diatomic molecule for which the bond force constant in the ground and excited electronic states is the same, but the equilibrium bond length is shifted by an amount \(\delta\) in the excited state relative to the ground state. For this case the vibrational wavefunctions for the \(n=0\) state are \\[ \psi_{g, 0}=\left(\frac{\alpha}{\pi}\right)^{1 / 4} e^{-\frac{1}{2} a r^{2}}, \psi_{e, 0}=\left(\frac{\alpha}{\pi}\right)^{1 / 4} e^{-\frac{1}{2} a(r-\delta)^{2}}, \alpha=\sqrt{\frac{k \mu}{\hbar^{2}}} \\] Calculate the Franck-Condon factor for the 0 - 0 transition for this molecule by evaluating the following expression: \\[ \left|\int_{-\infty}^{\infty} \psi_{g, 0}^{*} d e_{e, 0} d r\right|^{2} \\] In evaluating this expression, the following integral will be useful: \\[ \int_{-\infty}^{\infty} e^{-a x^{2}-b x} d x=\left(\frac{\pi}{a}\right)^{1 / 2} e^{b^{2} / 4 a}(a>0) \\]

One method for determining Franck-Condon factors between the \(n=0\) vibrational state of the ground electronic state and the \(n_{\mathrm{th}}\) vibrational level of an electronic excited state is: \\[ F C_{0-n}=\frac{1}{n !}\left(\frac{\delta^{2}}{2}\right)^{n} \exp \left(-\frac{\delta^{2}}{2}\right) \\] Where \(\delta\) is the dimensionless displacement of the excited state relative to the ground state and can be related to atomic displacements through \\[ \delta=\left(\frac{\mu \omega}{\hbar}\right)^{1 / 2}\left(r_{e}-r_{g}\right) \\] a. Determine the Franck-Condon factors for \(n=0\) to \(n=5\) when \(\delta=0.20\) corresponding to the excited-state potential surface being slightly displaced from that of the ground state. b. How would you expect the Franck-Condon factors to change if the excited- state displacement increases to \(\delta=2.0 ?\) Verify your expectation by calculating the FranckCondon factors from \(n=0\) to \(n=5\) for this displacement.

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