/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Consider the normalized function... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider the normalized functions \\[ \begin{array}{l} G_{1}(x, y, z ; \alpha)=\left(\frac{2048 \alpha^{7}}{9 \pi^{3}}\right)^{1 / 4} x^{2} e^{-\alpha r^{2}} \\ G_{2}(x, y, z ; \alpha)=\left(\frac{2048 \alpha^{7}}{9 \pi^{3}}\right)^{1 / 4} y^{2} e^{-\alpha r^{2}} \\ G_{3}(x, y, z ; \alpha)=\left(\frac{2048 \alpha^{7}}{9 \pi^{3}}\right)^{1 / 4} z^{2} e^{-\alpha r^{2}} \end{array} \\] Which hydrogen atomic orbital corresponds to the linear combination \(G_{1}(x, y, z ; \alpha)-\) \\[ G_{2}(x, y, z ; \alpha) ? \\]

Short Answer

Expert verified
The combination corresponds to the \(d_{x^2-y^2}\) orbital.

Step by step solution

01

Understand the problem

The problem asks us to identify which hydrogen atomic orbital corresponds to the linear combination of two normalized functions, \(G_1(x, y, z; \alpha)\) and \(G_2(x, y, z; \alpha)\).
02

Write the linear combination

We need to consider the linear combination \(G_1(x, y, z; \alpha) - G_2(x, y, z; \alpha)\). This is computed as: \[ \left(\frac{2048 \alpha^7}{9 \pi^3}\right)^{1/4} (x^2 e^{-\alpha r^2} - y^2 e^{-\alpha r^2}) \]
03

Factor out the common terms

Both terms share a common factor \(\left(\frac{2048 \alpha^7}{9 \pi^3}\right)^{1/4} e^{-\alpha r^2}\). Factor these out to simplify the expression: \[ \left(\frac{2048 \alpha^7}{9 \pi^3}\right)^{1/4} e^{-\alpha r^2} (x^2 - y^2) \]
04

Identify the orbital

The combination \(x^2 - y^2\) matches the form of the hydrogen atomic orbital \(d_{x^2-y^2}\). This orbital has a characteristic sign of orthogonality with \(x\) and \(y\), reflecting the patterns in the function \(x^2-y^2\).

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Linear Combination
A linear combination involves adding or subtracting constant multiples of functions. For the given normalized functions, it's the process of combining two specific functions, \(G_1(x, y, z; \alpha)\) and \(G_2(x, y, z; \alpha)\). By subtracting \(G_2\) from \(G_1\), we create a new function. In our specific problem, the linear combination is:
  • \(G_1(x, y, z; \alpha) - G_2(x, y, z; \alpha)\)
This results in: \( \left( \frac{2048 \alpha^7}{9 \pi^3} \right)^{1/4} (x^2 e^{-\alpha r^2} - y^2 e^{-\alpha r^2}) \)Notice how both functions \(x^2 e^{-\alpha r^2}\) and \(y^2 e^{-\alpha r^2}\) are present with similar coefficients before subtraction, emphasizing that linear combinations preserve relationships in similar functions. This combination gives us insight into characteristics or behaviors of new functions, such as identifying particular orbitals.
Normalized Functions
Normalized functions in quantum mechanics are adjusted so that their overall magnitude is consistent and measurable. Simply put, these functions are scaled and adjusted to reflect a total probability of 1. In the context of the problem, the normalization factor is represented now as the common term: \(\left( \frac{2048 \alpha^7}{9 \pi^3} \right)^{1/4} e^{-\alpha r^2} \) This factor ensures the functions \(G_1\), \(G_2\), and \(G_3\) have the same scale when performing operations like linear combination. When working with atomic orbitals, maintaining this normalization helps us accurately depict electron probabilities and spatial distributions. This consistency is essential in fields like quantum chemistry, where predicting electron behavior is key.
Orthogonality
In the context of functions and vectors, orthogonality means being perpendicular or independent. For hydrogen atomic orbitals, orthogonality indicates that combinations of orbitals won't interfere with each other, preserving distinct directional qualities. The function \(x^2 - y^2\) derived from linear combination \( \left( \frac{2048 \alpha^7}{9 \pi^3} \right)^{1/4} e^{-\alpha r^2} (x^2 - y^2)  \)is inherently tied to the concept of orthogonality. This specific form is recognized as matching the orbital \(d_{x^2-y^2}\), known for its orientation along the x and y axes, demonstrating zero overlap with each axis separately.Similarly, understanding orthogonality helps distinguish different functions or orbitals from one another, especially when predicting how they might combine or interact. In chemistry and physics, leveraging orthogonal properties simplifies computation and understanding of multi-component systems.

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

Given the matrices $$ \begin{gathered} C_{3}=\left(\begin{array}{rr} -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{array}\right) \quad \sigma_{\mathrm{v}}=\left(\begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right) \\ \sigma_{v}^{\prime}=\left(\begin{array}{cc} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{array}\right) \quad \sigma_{\mathrm{v}}^{\prime \prime}=\left(\begin{array}{rr} -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{array}\right) \end{gathered} $$ show that $$ \begin{array}{ll} \sigma_{v} C_{3}=\sigma_{v}^{n} & C_{3} \sigma_{v}=\sigma_{v}^{\prime} \\ \sigma_{v}^{\prime \prime} \sigma_{v}^{\prime}=C_{3} & C_{3} \sigma_{v}^{\prime \prime}=\sigma_{v} \end{array} $$ Calculate the determinant associated with each matrix. Calculate the trace of each matrix.

The Gaussian integral \\[ I_{0}=\int_{0}^{\infty} e^{-a x^{2}} d x \\] can be evaluated by a trick. First write \\[ I_{0}^{2}=\int_{0}^{\infty} d x e^{-a x^{2}} \int_{0}^{\infty} d y e^{-a y^{2}}=\int_{0}^{\infty} \int_{0}^{\infty} d x d y e^{-a\left(x^{2}+y^{2}\right)} \\] Now convert the integration variables from Cartesian coordinates to polar coordinates and show that \\[ I_{0}=\frac{1}{2}\left(\frac{\pi}{a}\right)^{1 / 2} \\]

Show that a three-dimensional Gaussian function centered at \(\mathbf{r}_{0}=x_{0} \mathbf{i}+y_{0} \mathbf{j}+z_{0} \mathbf{k}\) is product of three one-dimensional Gaussian functions centered on \(x_{0}, y_{0},\) and \(z_{0}\).

What is meant by the phrase "triple-zeta basis set"?

Show that the Gaussian function \\[ \phi(r)=\left(\frac{2 \alpha}{\pi}\right)^{3 / 4} e^{-\alpha r^{2}} \\] is normalized.

See all solutions

Recommended explanations on Chemistry Textbooks

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.