/*! 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 170 In the following questions, two ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In the following questions, two statements (Assertion) \(\mathrm{A}\) and Reason (R) are given. Mark a. If \(\mathrm{A}\) and \(\mathrm{R}\) both are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\) b. If \(\mathrm{A}\) and \(\mathrm{R}\) both are correct but \(\mathrm{R}\) is not the correct explanation of A c. A is true but \(\mathrm{R}\) is false d. A is false but \(\mathrm{R}\) is true e. \(\mathrm{A}\) and \(\mathrm{R}\) both are false (A): Shapes of the orbitals are represented by boundary surface diagrams of contrast probability density. (R): Boundary surface diagram helps in interpreting and visualizing an atomic orbital.

Short Answer

Expert verified
Option a: Both A and R are correct, and R is the correct explanation of A.

Step by step solution

01

Understanding Assertion and Reason

The assertion (A) states that shapes of orbitals are represented by boundary surface diagrams of contrast probability density. This refers to the conventional method used in quantum chemistry to understand the shapes of atomic orbitals by depicting regions with a high probability of finding an electron. This statement is correct.
02

Evaluating the Reason Statement

The reason (R) states that boundary surface diagrams help in interpreting and visualizing an atomic orbital. This is also correct, as boundary surface diagrams indeed provide a visual representation of the spatial distribution of electrons in an orbital.
03

Checking Explanation Validity

Now, we must determine if (R) correctly explains (A). The diagrams described in (R) are indeed the visual tools (A) mentions, as they help represent the probabilistic regions where electrons are most likely to be found. Thus, (R) is a correct explanation of (A).

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.

Orbital Shapes
In quantum chemistry, understanding orbital shapes is fundamental. Each orbital, which is a region around an atom's nucleus, corresponds to a specific energy level filled by electrons. These shapes aren't arbitrary; they are determined by solutions to the Schrödinger equation for electrons in atoms.

Electrons in an atom do not travel in fixed orbits like planets around the sun. Instead, they are found in regions called orbitals. These orbitals can have various shapes:
  • s-orbitals: These are spherical in shape. Only one s-orbital exists at each principal energy level.
  • p-orbitals: Shaped like dumbbells, these orbitals exist in groups of three, perpendicular to each other, along the x, y, and z axes.
  • d-orbitals: These have more complex shapes, typically appearing as clovers or double dumbbells.
  • f-orbitals: Even more complicated, these orbitals feature intricate shapes important in rare and complex elements.
These shapes aren't just for visual satisfaction; they reflect areas where electrons are most likely to exist at any given time.
Probability Density
At the heart of orbital shapes lies the concept of probability density. When discussing atomic orbitals, probability density refers to the likelihood of finding an electron within a particular volume of space surrounding the nucleus.

This concept stems from quantum mechanics, where an electron doesn’t have a precise location. Instead, we talk about where it is most likely to be located. Mathematically, probability density is derived from the wave function— a fundamental solution to the Schrödinger equation—which, when squared, gives the probability of finding an electron at a certain point:
\[ P(x, y, z) = |(x, y, z)|^2\]
This equation shows how essential probability is in quantum mechanics. The probability density helps reveal where the electron is most dense, hence showing regions like the maxima in p-orbitals or nodes where the probability equates to zero. Understanding this concept provides a clearer picture of how electrons occupy space in an atom.
Boundary Surface Diagrams
Boundary surface diagrams offer a visual representation of atomic orbitals, depicting regions with high electron probability density. These diagrams create a 3D image of the space around an atom's nucleus where electrons are likely to be found.

These diagrams are crucial because:
  • They simplify atomic interactions by visually showing where electron density is high.
  • They help predict bonding behavior as certain shapes will overlap more easily between atoms, aiding in bond formation.
When interpreting these diagrams, it's important to note that they only represent areas where there is a 90-95% chance of locating an electron. Beyond this boundary, while electrons might still exist, their probability is low.

This visualization guides chemists in predicting how atoms might interact within molecules, especially in more complex substances where simple Lewis models fall short. Understanding boundary surface diagrams empowers students to move beyond abstract numbers into palpable shapes in molecular chemistry.

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

Which of the following represent electronic configurations that are allowed but do not represent ground-state configurations? I. \([\mathrm{Ne}] 3 \mathrm{~s}^{1} 3 \mathrm{p}^{5}\) II. \([\mathrm{Kr}] 4 \mathrm{~d}^{12} 5 \mathrm{~s}^{2} 5 \mathrm{p}^{3}\) III. \([\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^{2} 4 \mathrm{p}^{2}\) a. I and II b. II and III c. only \(\mathbb{I}\) d. only I

Energy of H-atom in the ground state is \(-13.6 \mathrm{eV}\), hence energy in the second excited state is a. \(-6.8 \mathrm{eV}\) b. \(-3.4 \mathrm{eV}\) c. \(-1.51 \mathrm{eV}\) d. \(-4.53 \mathrm{eV}\)

Match the following: Column I (Principle) A. Exclusion principle B. Multiplicity rule C. Uncertainty principle D. Quantum theory Column II (Discoverer) (p) Hund (q) Heisenberg (r) Einstein (s) Planck (t) Pauli

In the Bohr's model of the hydrogen atom, which of the following statements is correct? a. The transition \(\mathrm{n}=1 \rightarrow \mathrm{n}=3\) represents absorption of energy b. The transition \(\mathrm{n}=2 \rightarrow \mathrm{n}=4\) represents emission of energy c. When \(\mathrm{n}=\infty\), the electron is in its ground state d. When \(\mathrm{n}=1\), the electron is in an excited state

The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1 , would be (Rydberg constant \(=1.097\) \(\left.\times 10^{7} \mathrm{~m}^{-1}\right)\) a. \(91 \mathrm{~nm}\) b. \(192 \mathrm{~nm}\) c. \(406 \mathrm{~nm}\) d. \(9.1 \times 10^{-3} \mathrm{~nm}\)

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.