/*! 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 95 In the ground state of mercury, ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 95

In the ground state of mercury, Hg, a. how many electrons occupy atomic orbitals with \(n=3 ?\) b. how many electrons occupy \(d\) atomic orbitals? c. how many electrons occupy \(p_{z}\) atomic orbitals? d. how many electrons have spin "up" \(\left(m_{s}=+\frac{1}{2}\right) ?\)

Short Answer

Expert verified
There are \(18\) electrons occupying orbitals with n=3, \(30\) electrons occupying d orbitals, \(8\) electrons occupying p_z atomic orbitals, and \(40\) electrons with spin "up" (m_s = +1/2) in the ground state of mercury (Hg).

Step by step solution

01

a. Electrons occupying orbitals with n=3

To find the number of electrons in orbitals with n=3, we need to count all the electrons in the 3s, 3p, and 3d orbitals. From the electronic configuration, we can see that: 3s: 2 electrons, 3p: 6 electrons, 3d: 10 electrons. Hence, there are 2 + 6 + 10 = \(18\) electrons occupying orbitals with n=3.
02

b. Electrons occupying d orbitals

To determine the number of electrons in d orbitals, we can examine the electronic configuration and sum up the electrons in the 3d, 4d, and 5d orbitals: 3d: 10 electrons, 4d: 10 electrons, 5d: 10 electrons. The total number of electrons in d orbitals is 10 + 10 + 10 = \(30\) electrons.
03

c. Electrons occupying p_z orbitals

The p_z orbitals are a subset of the p orbitals, and there is one p_z orbital in each p subshell. Since each orbital can hold a maximum of 2 electrons, there will be 2 electrons in the 2p_z, 3p_z, 4p_z, and 5p_z orbitals. Therefore, there are \(4\) sets of 2 electrons for a total of \(8\) electrons occupying p_z atomic orbitals.
04

d. Electrons with spin "up" (m_s = +1/2)

Finally, to determine the number of electrons with spin "up" (m_s = +1/2), we can assume that, in each orbital, one electron has spin "up" and the other has spin "down". Since there are 80 electrons in mercury, half (40) will have spin "up" and half (40) will have spin "down". Thus, there are \(40\) electrons with spin "up" (m_s = +1/2).

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 91影视!

Key Concepts

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

Atomic Orbitals
Atomic orbitals are regions in an atom where there is a high probability of finding electrons. They are defined by the quantum numbers, which determine their size, shape, and orientation in space. The different types of atomic orbitals are designated as s, p, d, and f. These labels come from the words sharp, principal, diffuse, and fundamental, respectively.

Each type of orbital has a different shape and capacity for holding electrons:
  • s orbitals: Spherical in shape and can hold up to 2 electrons.
  • p orbitals: Dumbbell-shaped and come in sets of three (p_x, p_y, p_z), each capable of holding 2 electrons, for a total of 6 electrons per p subshell.
  • d orbitals: More complex in shape and come in sets of five, each holding 2 electrons, totaling 10 electrons per d subshell.
  • f orbitals: Even more complex, comprising of seven orbitals, and can hold a total of 14 electrons.
These orbitals play a crucial role in the arrangement of electrons in an atom, influencing its chemical properties and reactivity.
Electron Spin
Electron spin is an intrinsic form of angular momentum carried by electrons, described by the spin quantum number, \(m_s\). Electrons can have a spin of either +\(\frac{1}{2}\) or -\(\frac{1}{2}\). These can be thought of as 'spin up' and 'spin down' states.

Having opposite spins allows two electrons to occupy the same orbital without repelling each other, due to the Pauli Exclusion Principle. This is an essential concept as it helps maximize electron occupancy in orbitals while maintaining stability within an atom.

When filling in electrons in an atom's orbitals, one electron will occupy an orbital with a 'spin up' configuration, and when paired, the second will be 'spin down'. This balanced arrangement minimizes energy and stabilizes the electron configuration.
Quantum Numbers
Quantum numbers are sets of numerical values that provide solutions to the Schr枚dinger equation, describing the state of an electron in an atom. These numbers are crucial for defining the electron configuration and behavior in an atom.

The four quantum numbers are:
  • Principal quantum number (n): Indicates the energy level and size of the orbital. Larger values of n mean larger orbitals and higher energy.
  • Azimuthal quantum number (l): Defines the shape of the orbital (s, p, d, f), with values ranging from 0 to n-1.
  • Magnetic quantum number (m_l): Specifies the orientation of the orbital in space in relation to the other orbitals, with possible values from -l to +l.
  • Spin quantum number (m_s): Describes the direction of electron spin, with only two possible values, +\(\frac{1}{2}\) or -\(\frac{1}{2}\).
These quantum numbers collectively explain the arrangement of electrons in atoms and the chemical properties that result from this arrangement.

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

An unknown element is a nonmetal and has a valence electron configuration of \(n s^{2} n p^{4}\). a. How many valence electrons does this element have? b. What are some possible identities for this element? c. What is the formula of the compound this element would form with potassium? d. Would this element have a larger or smaller radius than barium? e. Would this element have a greater or smaller ionization energy than fluorine?

Assume that a hydrogen atom's electron has been excited to the \(n=5\) level. How many different wavelengths of light can be emitted as this excited atom loses energy?

Identify how many unpaired electrons are present in each of the following in the ground state: \(\mathrm{O}, \mathrm{O}^{+}, \mathrm{O}^{-}, \mathrm{Os}, \mathrm{Zr}, \mathrm{S}, \mathrm{F}, \mathrm{Ar}\).

Answer the following questions based on the given electron configurations and identify the elements. a. Arrange these atoms in order of increasing size: \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{6} ;[\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{1} ;[\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{3}\). b. Arrange these atoms in order of decreasing first ionization energy: \([\mathrm{Ne}] 3 s^{2} 3 p^{5} ;[\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{3} ;[\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{5}\).

Which of the following sets of quantum numbers are not allowed? For each incorrect set, state why it is incorrect. a. \(n=3, \ell=3, m_{\ell}=0, m_{s}=-\frac{1}{2}\) b. \(n=4, \ell=3, m_{\ell}=2, m_{s}=-\frac{1}{2}\) c. \(n=4, \ell=1, m_{\ell}=1, m_{s}=+\frac{1}{2}\) d. \(n=2, \ell=1, m_{\ell}=-1, m_{s}=-1\) e. \(n=5, \ell=-4, m_{\ell}=2, m_{s}=+\frac{1}{2}\) f. \(n=3, \ell=1, m_{\ell}=2, m_{s}=-\frac{1}{2}\)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

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.