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Chapter 7: Problem 5

Which has the larger second ionization energy, lithium or beryllium? Why?

Short Answer

Expert verified
Lithium has the larger second ionization energy as the electron being removed for the second ionization occupies a lower energy level (\(1s\)) and is closer to the nucleus, making it harder to remove than the electron in beryllium's outer and higher energy level (\(2s\)).

Step by step solution

01

Understanding Ionization Energy

Ionization energy is the amount of energy required to remove an electron from a neutral atom or positive ion, transforming it into a positive ion. The higher the ionization energy, the harder it is to remove an electron. The concept of second ionization energy refers to the energy required to remove an electron from an ion that has already lost one electron.
02

Knowing the Electronic Configuration of Lithium and Beryllium

Before we compare their second ionization energies, let's determine the electronic configurations of lithium (Li) and beryllium (Be). By knowing their atomic numbers, we can identify how many electrons they have and their distribution among atomic orbitals. - Lithium (Li), atomic number 3: Its electronic configuration is \(1s^2 2s^1\). - Beryllium (Be), atomic number 4: Its electronic configuration is \(1s^2 2s^2\).
03

Identifying the First Ionized States

To find the second ionization energies, we need to remove two electrons from each atom. Firstly, we will remove one electron from both atoms and identify their first ionized states. - Lithium (Li): After the first electron is removed, its configuration becomes \(1s^2\), forming a Li+ ion. - Beryllium (Be): After the first electron is removed, its configuration becomes \(1s^2 2s^1\), forming a Be+ ion.
04

Comparing the Second Ionization Energies

Now, let's analyze the second ionization energies, that is, the energy required to remove another electron from the first ionized states. - For Li+: Removing the second electron means taking an electron from the inner \(1s^2\) orbital. Since the \(1s\) orbital is closer to the nucleus and has a lower energy level, a significant amount of energy is required to remove an electron from it. - For Be+: Removing the second electron means taking an electron from the \(2s^1\) orbital. Since the \(2s\) orbital is farther from the nucleus and occupies a higher energy level than \(1s\), the energy required to remove an electron from this orbital is lower than removing the electron from the Li+ ion.
05

Evaluating and Explaining the Result

Comparing the energy required for both ions, we can conclude that lithium has the larger second ionization energy. This is because the electron being removed for the second ionization in the case of lithium occupies a lower energy level (\(1s\)) and it's closer to the nucleus, making it harder to remove than the electron in beryllium's outer and higher energy level (\(2s\)).

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Most popular questions from this chapter

In each of the following sets, which atom or ion has the smallest radius? a. H, He b. Cl, In, Se c. element 120, element 119, element 116 d. Nb, Zn, Si e. \(\mathrm{Na}^{-}, \mathrm{Na}, \mathrm{Na}^{+}\)

X rays have wavelengths on the order of \(1 \times 10^{-10} \mathrm{m} .\) Calcu- late the energy of \(1.0 \times 10^{-10} \mathrm{m} \mathrm{X}\) rays in units of kilojoules per mole of \(\mathrm{X}\) rays. AM radio waves have wavelengths on the order of \(1 \times 10^{4} \mathrm{m}\) . Calculate the energy of \(1.0 \times 10^{4} \mathrm{m}\) radio waves in units of kilojoules per mole of radio waves. Consider that the bond energy of a carbon-carbon single bond found in organic compounds is 347 \(\mathrm{kJ} / \mathrm{mol}\) . Would \(\mathrm{x}\) rays and/or radio waves be able to disrupt organic compounds by breaking carbon-carbon single bonds?

Are the following statements true for the hydrogen atom only, true for all atoms, or not true for any atoms? a. The principal quantum number completely determines the energy of a given electron. b. The angular momentum quantum number, \(\ell,\) determines the shapes of the atomic orbitals. c. The magnetic quantum number, \(m_{\ell},\) determines the direction that the atomic orbitals point in space.

The wave function for the 2\(p_{z}\) orbital in the hydrogen atom is $$\psi_{2 p_{z}}=\frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a_{0}}\right)^{3 / 2} \sigma \mathrm{e}^{-\sigma / 2} \cos \theta$$ where \(a_{0}\) is the value for the radius of the first Bohr orbit in meters \(\left(5.29 \times 10^{-11}\right), \sigma\) is \(Z\left(r / a_{0}\right), r\) is the value for the distance from the nucleus in meters, and \(\theta\) is an angle. Calculate the value of \(\psi_{2 p_{z}}^{2}\) at \(r=a_{0}\) for \(\theta=0^{\circ}\left(z \text { axis ) and for } \theta=90^{\circ}\right.\) (xy plane).

Which of the following statements is(are) true? a. F has a larger first ionization energy than does Li. b. Cations are larger than their parent atoms. c. The removal of the first electron from a lithium atom (electron configuration is 1\(s^{2} 2 s^{1} )\) is exothermic - that is, removing this electron gives off energy. d. The He atom is larger than the \(\mathrm{H}^{+}\) ion. e. The Al atom is smaller than the Li atom.
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