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(a) How is the concept of effective nuclear charge used to simplify the numerous electron-electron repulsions in ? many-electron atom? (b) Which experiences a greater effective nuclear charge in a Be atom, the 1 s electrons or the 2 s electrons? Explain.

Short Answer

Expert verified
(a) The concept of effective nuclear charge (Z_eff) is used to simplify electron-electron repulsions in a many-electron atom by accounting for the attractive force between the electron and the nucleus, as well as the repulsive force from other electrons, resulting in a single value that describes and compares different many-electron atoms. (b) In a Beryllium (Be) atom, the 1s electrons experience a greater effective nuclear charge (Z_eff = 4) than the 2s electrons (Z_eff = 2.3), due to their closer proximity to the nucleus and less shielding by other electrons, resulting in a stronger attractive force between the 1s electrons and the nucleus.

Step by step solution

01

Understanding the concept of effective nuclear charge

Effective nuclear charge (Z_eff) is the net positive charge experienced by an electron in a many-electron atom, taking into account the attractive force between the electron and the nucleus, as well as the repulsive force from other electrons. The concept helps simplify the understanding of electron-electron repulsions, as it reduces these repulsive forces to a single value, making it easier to describe and compare different many-electron atoms.
02

Determining the atomic structure and electron configuration of Beryllium

Beryllium (Be) has an atomic number of 4, which means it has 4 protons in its nucleus and 4 electrons surrounding it. The electron configuration of Beryllium is 1s² 2s², which indicates that there are 2 electrons occupying the 1s subshell and 2 electrons occupying the 2s subshell.
03

Calculating the effective nuclear charge for 1s and 2s electrons in Beryllium

We can estimate the effective nuclear charge (Z_eff) experienced by an electron using Slater's rules. According to Slater's rules: Z_eff = Z - S where Z is the atomic number, and S is the shielding constant, representing the shielding effect caused by other electrons. For 1s electrons in Beryllium (Z = 4), the contribution to the shielding constant is: S = 0.30(0) = 0 (Note that there are no other electrons in the same subshell) So, Z_eff for 1s electrons = 4 - 0 = 4 For 2s electrons in Beryllium, the contribution to the shielding constant is: S = 0.85(2) = 1.7 (This accounts for the two 1s electrons' shielding effect) So, Z_eff for 2s electrons = 4 - 1.7 = 2.3
04

Comparing the effective nuclear charges of 1s and 2s electrons in Beryllium

The comparison of the calculated effective nuclear charges for the 1s and 2s electrons in Beryllium is: Z_eff(1s) = 4 Z_eff(2s) = 2.3 Since the effective nuclear charge experienced by the 1s electrons is greater than that experienced by the 2s electrons, we can conclude that the 1s electrons experience a greater effective nuclear charge in a Beryllium atom. This is because the 1s electrons are closer to the nucleus and are less shielded by other electrons, resulting in a stronger attractive force between the 1s electrons and the nucleus.

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Key Concepts

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

Electron Shielding
Electron shielding is a key concept in understanding how electrons in an atom experience forces from each other and the nucleus. When you have multiple electrons, each electron not only feels the attractive force from the positively charged nucleus but also repulsive forces from other electrons.
These repulsive forces act like a shield, reducing the net positive charge that an electron feels from the nucleus.
This effect is what we refer to as "electron shielding." Some important points to remember about electron shielding are:
  • Shielding reduces the effective nuclear charge felt by an electron.
  • Inner electrons shield outer electrons more effectively than electrons in the same shell.
  • As a result, electrons farther from the nucleus are less tightly held.
The concept of electron shielding helps in predicting which electrons in an atom will be more closely bound to the nucleus and explains variations in chemical properties across the periodic table.
Beryllium Atom
The Beryllium atom is an interesting example of how electron structure and shielding work together. Beryllium has four electrons and is represented by the atomic number 4.
Its electron configuration is 1s² 2s², meaning it has two electrons in the 1s subshell and two in the 2s subshell. Let's break down why this is significant:
  • The 1s electrons are closest to the nucleus and experience the full charge of the nucleus with minimal shielding.
  • The 2s electrons, however, are shielded by the 1s electrons. This reduces the effective nuclear charge they experience.
Understanding this configuration is crucial because it influences properties like ionization energy and reactivity. The structure impacts how easily an atom like Beryllium can lose its outer electrons in chemical reactions.
Slater's Rules
Slater's Rules are a set of guidelines that help calculate the effective nuclear charge, or Z_eff, that an electron experiences in an atom.
This includes considering the electron shielding effect, which we touched on earlier. According to Slater's Rules, the calculation of Z_eff involves two main components: 1. **Actual Nuclear Charge (Z):** This is simply the number of protons, or atomic number. 2. **Shielding Constant (S):** This accounts for the electron shielding effect, which reduces the net positive charge experienced by the electron. Using Slater's Rules for Beryllium:
  • For 1s electrons, the shielding constant is minimal because they don't shield each other, leading to a high Z_eff of 4.
  • For 2s electrons, the shielding constant is higher due to the presence of 1s electrons, giving a lower Z_eff of 2.3.
These rules allow chemists to understand and predict electron behavior across different atoms by simplifying complicated interactions within the atom into manageable calculations.

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Most popular questions from this chapter

Using only the periodic table, arrange each set of atoms in order of increasing radius: (a) \(\mathrm{Ca}, \mathrm{Mg}\), Be; (b) \(\mathrm{Ga}, \mathrm{Br}\), \(\mathrm{Ge} ;\) (c) \(\mathrm{Al}, \mathrm{T}\), \(\mathrm{Si}\).

Predict whether each of the following oxides is ionic or molecular. \(\mathrm{SO}_{2}, \mathrm{MgO}, \mathrm{Li}_{2} \mathrm{O}, \mathrm{P}_{2} \mathrm{O}_{5}, \mathrm{Y}_{2} \mathrm{O}_{3}, \mathrm{~N}_{2} \mathrm{O}\), and \(\mathrm{XeO}_{3}\) Explain the reasons for your choices.

Detailed calculations show that the value of \(Z_{\text {eff }}\) for \(\mathrm{Na}\) and \(K\) atoms is \(2.51+\) and \(3.49+\), respectively. (a) What value do you estimate for \(Z_{\text {eff }}\) experienced by the outermost electron in both Na and \(K\) by assuming core electrons contribute \(1.00\) and valence electrons contribute \(0.00\) to the screening constant? (b) What values do you estimate for \(Z_{\text {eff }}\) using Slater's rules? (c) Which approach gives a more accurate estimate of \(Z_{\text {eff }} ?\) (d) Does either method of approximation account for the gradual increase in \(Z_{\text {eff }}\) that occurs upon moving down a group?

Potassium superoxide, \(\mathrm{KO}_{2}\), is often used in oxygen masks (such as those used by firefighters) because \(\mathrm{KO}_{2}\) reacts with \(\mathrm{CO}_{2}\) to release molecular oxygen. Experiments indicate that \(2 \mathrm{~mol}\) of \(\mathrm{KO}_{2}(\mathrm{~s})\) react with each mole of \(\mathrm{CO}_{2}(g) .\) (a) The products of the reaction are \(\mathrm{K}_{2} \mathrm{CO}_{3}(s)\) and \(\mathrm{O}_{2}(g) .\) Write a balanced equation for the reaction between \(\mathrm{KO}_{2}(\mathrm{~s})\) and \(\mathrm{CO}_{2}(\mathrm{~g}) .\) (b) Indicate the oxidation number for each atom involved in the reaction in part (a). What elements are being oxidized and reduced? (c) What mass of \(\mathrm{KO}_{2}(s)\) is needed to consume \(18.0 \mathrm{~g}\) \(\mathrm{CO}_{2}(g) ?\) What mass of \(\mathrm{O}_{2}(g)\) is produced during this reaction?

Explain the following variations in atomic or ionic radii: (a) \(1^{-}>1>1^{+},(b) C a^{2+}>M g^{2+}>B e^{2+}\) (c) \(\mathrm{Fe}>\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+}\).

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