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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

For hydrogen atoms, the wave function for the state \(n=3\) \(\ell=0, m_{\ell}=0\) is $$\psi_{300}=\frac{1}{81 \sqrt{3 \pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2}\left(27-18 \sigma+2 \sigma^{2}\right) e^{-\sigma \beta}$$ where \(\sigma=r / a_{0}\) and \(a_{0}\) is the Bohr radius \(\left(5.29 \times 10^{-11} \mathrm{m}\right) .\) Calculate the position of the nodes for this wave function.

Although no currently known elements contain electrons in g orbitals in the ground state, it is possible that these elements will be found or that electrons in excited states of known elements could be in \(g\) orbitals. For \(g\) orbitals, the value of \(\ell\) is \(4 .\) What is the lowest value of \(n\) for which \(g\) orbitals could exist? What are the possible values of \(m_{\ell} ?\) How many electrons could a set of \(g\) orbitals hold?

Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electromagnetic radiation is emitted in each transition? a. \(n=3 \rightarrow n=2\) b. \(n=4 \rightarrow n=2\) c. \(n=2 \rightarrow n=1\)

The elements Si, Ga, As, Ge, Al, Cd, S, and Se are all used in the manufacture of various semiconductor devices. Write the expected electron configuration for these atoms.

Which of the following orbital designations are incorrect: \(1 s,\) \(1 p, 7 d, 9 s, 3 f, 4 f, 2 d ?\)
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