Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) Write the electron configuration for \(\mathrm{Li}\), and estimate the effective nuclear charge experienced by the valence electron. (b) The energy of an electron in a one-electron atom or ion equals \(\left(-2.18 \times 10^{-18} \mathrm{~J}\right)\left(\frac{Z^{2}}{n^{2}}\right)\) where \(Z\) is the nuclear charge and \(n\) is the principal quantum number of the electron. Estimate the first ionization energy of Li. (c) Compare the result of your calculation with the value reported in Table 7.4 and explain the difference. (d) What value of the effective nuclear charge gives the proper value for the ionization energy? Does this agree with your explanation in \((\mathrm{c}) ?\)

Short Answer

Expert verified
The electron configuration for Lithium (Li) is \(1s^{2}2s^{1}\), and the estimated effective nuclear charge experienced by the valence electron is 1. The first ionization energy of Li calculated using this is \(5.45\times10^{-19}\;\mathrm{J}\), which differs from the reported value, \(7.98\times10^{-19}\;\mathrm{J}\), due to imperfect shielding of inner-shell electrons. The effective nuclear charge that gives the correct ionization energy is approximately 1.40, confirming that the valence electron experiences a slightly higher effective nuclear charge.

Step by step solution

01

(a) Electron Configuration of Li

To write the electron configuration of Lithium (Li), we must know its atomic number, which is 3. This tells us that there are 3 electrons. The electron configuration follows the Aufbau principle, filling the lowest energy levels first. So, the electron configuration for Lithium (Li) is: \(1s^{2}2s^{1}\).
02

Effective Nuclear Charge

To estimate the effective nuclear charge (\(Z_{eff}\)) experienced by the valence electron, we must consider the shielding effect of inner-shell electrons, which diminishes the effective nuclear charge experienced by the outer electrons. For Lithium (Li), there is one valence electron in the 2s orbital, and the inner shell has 2 electrons in the 1s orbital. We will use the formula: \(Z_{eff} = Z - S\) Here, Z is the actual nuclear charge (3 for Li), and S is the shielding constant (which is approximately equal to the number of inner shell electrons, which is 2 for Li). \(Z_{eff} = 3 - 2 = 1\) So, the effective nuclear charge experienced by the valence electron of Lithium is 1.
03

(b) First Ionization Energy of Li

To find the first ionization energy of Lithium, we can use the given formula: \(E = \left(-2.18 \times 10^{-18} \mathrm{~J}\right)\left(\frac{\mathrm{Z}^{2}}{n^{2}}\right)\) where Z is the nuclear charge and n is the principal quantum number of the electron. In the case of the valence electron of Lithium, Z = 1 (as estimated) and n = 2 (since the electron is in the second shell). \(E = (-2.18 \times 10^{-18}\mathrm{~J})\left(\frac{1^2}{2^2}\right) = -5.45\times10^{-19}\;\mathrm{J}\) Since ionization energy is the negative of this value, the first ionization energy of Li = \(5.45\times10^{-19}\;\mathrm{J}\)
04

(c) Comparison of Calculated and Reported Ionization Energy

Comparing the calculated ionization energy with the value reported in Table 7.4: Our calculated value: \(5.45\times10^{-19}\;\mathrm{J}\) Reported value: \(7.98\times10^{-19}\;\mathrm{J}\) The calculated and reported values are different because our estimate of the effective nuclear charge was too low, and we assumed a perfect shielding effect of the inner shell electrons. In reality, the shielding effect is not perfect, and the valence electron feels a slightly higher effective nuclear charge.
05

(d) Effective Nuclear Charge for Correct Ionization Energy

To find the value of the effective nuclear charge that gives the proper value for the ionization energy, we can rearrange the energy formula as: \(Z_{eff} = \sqrt{\frac{E \times n^2}{-2.18 \times 10^{-18} \;\mathrm{J}}}\) Using the reported ionization energy value: \(Z_{eff} = \sqrt{\frac{7.98\times10^{-19}\;\mathrm{J} \times 2^2}{-2.18 \times 10^{-18} \;\mathrm{J}}}\) \(Z_{eff} \approx 1.40\) This value of the effective nuclear charge (1.40) gives the correct value for the ionization energy, and it is higher than our original estimate (1). This confirms that the valence electron experiences a slightly higher effective nuclear charge due to the imperfect shielding effect of the inner-shell electrons, as explained in part (c).

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 Vaia!

Key Concepts

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

Electron Configuration
Understanding the electron configuration of an atom, such as lithium, is fundamental in predicting its chemical behavior. An atom's electron configuration refers to the distribution of its electrons among the different orbitals. Orbitals are theoretical constructs that represent the probability of finding an electron in a particular region around the nucleus. These orbitals are filled according to certain rules, one of which is the Aufbau principle. Electrons are added one at a time to the lowest energy orbitals first before moving to higher energy levels.

For lithium (Li), with an atomic number of 3, the electrons occupy the first and second shells. The configuration is written as 1s^2 2s^1, indicating that two electrons fill the s orbital of the first energy level (1s), and one electron is in the s orbital of the second energy level (2s). This framework of electron configuration helps in predicting the reactivity and the type of bonds an atom can form.
Effective Nuclear Charge
Atoms consist of a positively charged nucleus surrounded by negatively charged electrons. The strength with which an electron feels the nucleus's pull is the effective nuclear charge (Zeff). It is a measure of the net positive charge experienced by an electron in a multi-electron atom. The formula Zeff = Z - S helps calculate this, where Z is the actual nuclear charge (the number of protons), and S is the shielding constant.

The effective nuclear charge is essential for understanding an atom's ionization energy since it affects how tightly an electron is held by the nucleus. Lithium's valence electron experiences an effective nuclear charge reduced by the inner-shell electrons, which illustrates the concept of electron shielding.
Shielding Effect
When dealing with multi-electron atoms like lithium, it's crucial to consider the shielding effect. Inner electrons can shield (or block) the outer electrons from the full charge of the nucleus. This is because inner electrons repel outer electrons due to their negative charges, reducing the net force exerted by the nucleus on outer electrons.

The shielding effect varies across different elements and depends on the electron distribution. It's a fundamental concept when estimating the effective nuclear charge, influencing how we predict atomic properties, such as ionization energies. For lithium's valence electron, the two 1s electrons provide a partial shield from the nucleus, resulting in a lower effective nuclear charge than the actual nuclear charge.
Principal Quantum Number
Electron configurations also involve principal quantum numbers, symbolized as n. These numbers denote the main energy levels or shells of an atom and are integral in determining the distance of an electron from the nucleus. The principal quantum number starts at 1 and increases with the energy level, determining the energy and size of orbitals.

For lithium, its valence electron's principal quantum number is 2, indicating it's in the second energy level. This is essential in calculations of ionization energy, as it defines the region of space where the valence electron is likely to be found and how tightly it's bound to the nucleus.
Aufbau Principle
The Aufbau principle comes from a German word meaning 'building up'. It represents a key rule for determining the electron configuration of an atom. According to this principle, electrons fill atomic orbitals starting at the lowest available energy levels before filling higher levels. This approach provides the most stable, lowest energy configuration for the atom.

The principle serves as a guideline for writing the electron configuration, such as that of lithium: the first two electrons go into the 1s orbital, and the third into the 2s orbital (as in 1s^2 2s^1). It's worth noting that the Aufbau principle is a model that works very well for light elements but requires modifications when applied to more complex, heavier elements.

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

The first ionization energy and electron affinity of Ar are both positive values. (a) What is the significance of the positive value in each case? (b) What are the units of electron affinity?

We can draw an analogy between the attraction of an electron to a nucleus and seeing a lightbulb -in essence, the more \(n u=\) clear charge the electron "sees," the greater the attraction. (a) Within this analogy, discuss how the screening by core electrons is analogous to putting a frosted-glass lampshade between the lightbulb and your eyes, as shown in the illustration. (b) Explain how we could mimic moving to the right in a row of the periodic table by changing the wattage of the lightbulb. (c) How would you change the wattage of the bulb and/or the frosted glass to mimic the effect of moving down a column of the periodic table? [Section 7.2]

(a) Why does the quantum mechanical description of manyelectron atoms make it difficult to define a precise atomic radius? (b) When nonbonded atoms come up against one another, what determines how closely the nuclear centers can approach?

Although the electron affinity of bromine is a negative quantity, it is positive for \(\mathrm{Kr}\). Use the electron configurations of the two elements to explain the difference.

Mercury in the environment can exist in oxidation states 0,+1 , and \(+2 .\) One major question in environmental chemistry research is how to best measure the oxidation state of mercury in natural systems; this is made more complicated by the fact that mercury can be reduced or oxidized on surfaces differently than it would be if it were free in solution. XPS, X-ray photoelectron spectroscopy, is a technique related to PES (see Exercise 7.107 ), but instead of using ultraviolet light to eject valence electrons, X-rays are used to eject core electrons. The energies of the core electrons are different for different oxidation states of the element. In one set of experiments, researchers examined mercury contamination of minerals in water. They measured the XPS signals that corresponded to electrons ejected from mercury's 4 forbitals at \(105 \mathrm{eV},\) from an X-ray source that provided \(1253.6 \mathrm{eV}\) of energy. The oxygen on the mineral surface gave emitted electron energies at \(531 \mathrm{eV}\), corresponding to the 1 s orbital of oxygen. Overall the researchers concluded that oxidation states were +2 for \(\mathrm{Hg}\) and -2 for \(\mathrm{O} .\) (a) Calculate the wavelength of the X-rays used in this experiment. (b) Compare the energies of the \(4 f\) electrons in mercury and the 1 s electrons in oxygen from these data to the first ionization energies of mercury and oxygen from the data in this chapter. (c) Write out the ground- state electron configurations for \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-}\); which electrons are the valence electrons in each case? (d) Use Slater's rules to estimate \(Z_{\text {eff }}\) for the \(4 f\) and valence electrons of \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-}\); assume for this purpose that all the inner electrons with \((n-3)\) or less screen a full + \(1 .\)

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.

Sign-up for free