Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 10

(a) For a CASSCF calculation of the ground-state wavefunction of diatomic \(\mathrm{C}_{2},\) describe a reasonable choice for the distribution of \(\sigma\) and \(\pi\) molecular orbitals into active, inactive and virtual orbitals. (b) How many inactive and active clectrons arc there in the calculation? (c) In an RASSCF calculation, how might the set of active orbitals be further divided?

Short Answer

Expert verified
In a CASSCF calculation for the ground-state wavefunction of \( \mathrm{C}_{2} \), the most reasonable distribution of the \( \sigma \) and \( \pi \) molecular orbitals is to treat the two \( \sigma \) orbitals as inactive, the four \( \pi \) orbitals as active and the remaining orbitals as virtual. In the process, there are 4 inactive and 4 active electrons. For the RASSCF calculation, the active \( \pi \) orbitals are divided into two spaces - RAS1 and RAS2 - where \( \pi \) bonding orbitals can placed into RAS1 and \( \pi \) antibonding orbitals can be placed into RAS2}

Step by step solution

01

CASSCF Calculation for \(\mathrm{C}_{2}\)

In a diatomic \( \mathrm{C}_{2} \) molecule, the \( \sigma \) molecular orbitals are the bonding orbitals while the \( \pi\) molecular orbitals are the antibonding orbitals. For the CASSCF calculation of the ground state wavefunction, the most reasonable distribution of these orbitals would be: \n- two \( \sigma \) orbitals are regarded as inactive because they are fully occupied and represent the core orbitals, \n- four \( \pi\) orbitals are considered active as they are involved in electron correlation, \n- remaining \( \sigma \) and \( \pi\) orbitals are treated as virtual as they are unoccupied in the ground state.
02

Calculation of inactive and active electrons

In the ground state each carbon atom contributes 4 valence electrons. So, in a \( \mathrm{C}_{2} \) molecule, there are 8 electrons in total. Two \( \sigma \) orbitals are inactive and each can accommodate two electrons, which implies there are 4 inactive electrons. The remaining 4 electrons are in the four active \( \pi\) orbitals, hence there are 4 active electrons.
03

Division of active orbitals in RASSCF calculation

In a Restricted Active Space Self-Consistent Field (RASSCF) calculation, the active orbitals are further divided into two spaces - RAS1 and RAS2. The \( \pi \) bonding orbitals could be placed into RAS1, meaning they can be doubly occupied, singly occupied, or unoccupied. The \( \pi \) antibonding orbitals are placed into RAS2, are they are not allowed to be fully occupied.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use Moller-Plesset perturbation theory to obtain an expression for the ground- state wavefunction corrected to first order in the perturbation.

Consider the two-clectron integrals over the basis functions defined in eqn 9.17 . If the basis functions are taken to be real, a number of the integrals are equivalent; for example, \((a b | c d)=(a d | c b)\). Find the other integrals that are equal to \((a b | c d)\)

Show that the product of an s-type Gaussian centred at \(R_{\text {A with exponent } \alpha_{\text {A and an s-type Gaussian centred at }} R_{\text {nd }}}\) with cxponcnt \(\alpha_{\mathrm{B}}\) can be writtcn in terms of a single s-typc Gaussian centred between \(R_{\lambda}\) and \(R_{B}\)

Prove Brillouin's theorem; that is, show that hamiltonian matrix elements between the HF wavefunction \(\Phi_{0}\) and singly excited determinants are identically zero. Hint. Use the Condon-Slater rules.

Consider two Slater determinants \(\Phi_{1}\) and \(\Phi_{2}\) that differ by only one spinorbital; that is, $$\begin{array}{l} \Phi_{1}=\left\|\cdots \varphi_{m} \varphi_{i} \cdots\right\| \\ \Phi_{2}=\left\|\cdots \varphi_{p} \varphi_{i} \cdots\right\| \end{array}$$ Derive the following Condon-Slater rule: $$\begin{aligned} \left\langle\Phi_{1}|H| \Phi_{2}\right\rangle=&\left\langle\varphi_{m}(1)\left|h_{1}\right| \varphi_{p}(1)\right\rangle+\sum_{i}\left\\{\left[\varphi_{m} \varphi_{i} | \varphi_{p} \varphi_{i}\right]\right.\\\ &\left.-\left[\varphi_{m} \varphi_{i} | \varphi, \varphi_{p}\right]\right\\} \end{aligned}$$ where we have used the notation $$\left[\varphi_{4} \varphi_{b} | \varphi_{c} \varphi_{d}\right]=j_{0} \int \varphi_{a}^{*}(1) \varphi_{b}^{*}(2)\left(\frac{1}{r_{12}}\right) \varphi_{c}(1) \varphi_{d}(2) \mathrm{d} r_{1} \mathrm{d} r_{2}$$
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free