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

Short Answer

Expert verified
Lithium has the larger second ionization energy compared to Beryllium. The second ionization of lithium involves removing an electron from the tightly held inner shell, while for beryllium, it requires removing an electron from the less tightly held \(2s\) shell. This difference in energy required is due to the strong attraction between the electrons in the \(1s\) shell and the nucleus in lithium, whereas in beryllium, the electron is farther away from the nucleus and experiences less attraction.

Step by step solution

01

Write the electronic configurations of lithium and beryllium atoms

: First, we need to write the electronic configurations of lithium and beryllium atoms by filling orbitals in order of increasing energy levels: Lithium (Li) - Atomic number: 3 Electronic configuration: \(1s^{2}2s^{1}\) Beryllium (Be) - Atomic number: 4 Electronic configuration: \(1s^{2}2s^{2}\)
02

Compare their first ionization energies

: Now let's compare their first ionization energies by considering the factors influencing ionization energy: size of the atom, nuclear charge, and electron shielding. First Ionization energy of Li: According to Li's electronic configuration, the electron removed during the first ionization is from the \(2s^{1}\) orbital. This electron is relatively easy to remove due to its distance from the nucleus and weak shielding effect from the \(1s^{2}\) electrons. First Ionization energy of Be: In the case of Be, the electron removed during the first ionization is also from the \(2s\) orbital, but due to a higher nuclear charge (+4 in Be vs. +3 in Li), the first ionization energy of Be is greater than that of Li.
03

Compare their second ionization energies

: To compare their second ionization energies, we will be removing another electron from each ion and consider the factors discussed earlier. Second Ionization energy of Li: After the first ionization, Li remains with electronic configuration \(1s^{2}\). Now, to remove the second electron, we need to overcome the strong attraction between the electrons in the \(1s\) shell and the nucleus, resulting in a large second ionization energy. Second Ionization energy of Be: After the first ionization, Be remains with electronic configuration \(1s^{2}2s^{1}\). To remove the second electron, which is in the \(2s\) shell, we need less energy compared to removing an electron from the inner shell as in Li's case.
04

Conclusion

: Comparing the second ionization energies of lithium and beryllium, Lithium's second ionization energy involves removing an electron from the tightly held inner shell, resulting in a larger second ionization energy. Beryllium's second ionization energy involves removing an electron from the less tightly held \(2s\) shell, requiring less energy than Li. Therefore, Lithium has the larger second ionization energy compared to Beryllium.

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Most popular questions from this chapter

The successive ionization energies for an unknown element are \(I_{1}=896 \mathrm{~kJ} / \mathrm{mol}\) \(I_{2}=1752 \mathrm{~kJ} / \mathrm{mol}\) \(I_{3}=14,807 \mathrm{~kJ} / \mathrm{mol}\) \(I_{4}=17,948 \mathrm{~kJ} / \mathrm{mol}\) To which family in the periodic table does the unknown element most likely belong?

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: \([\mathrm{Ne}] 3 s^{2} 3 p^{5} ;[\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{3} ;[\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{5}\)

Calculate the de Broglie wavelength for each of the following. a. an electron with a velocity \(10 . \%\) of the speed of light b. a tennis ball \((55 \mathrm{~g})\) served at \(35 \mathrm{~m} / \mathrm{s}(\sim 80 \mathrm{mi} / \mathrm{h})\)

From the information below, identify element \(\mathrm{X}\). a. The wavelength of the radio waves sent by an FM station broadcasting at \(97.1 \mathrm{MHz}\) is \(30.0\) million \(\left(3.00 \times 10^{7}\right)\) times greater than the wavelength corresponding to the energy difference between a particular excited state of the hydrogen atom and the ground state. b. Let \(V\) represent the principal quantum number for the valence shell of element \(X\). If an electron in the hydrogen atom falls from shell \(V\) to the inner shell corresponding to the excited state mentioned above in part a, the wavelength of light emitted is the same as the wavelength of an electron moving at a speed of \(570 . \mathrm{m} / \mathrm{s}\) c. The number of unpaired electrons for element \(\mathrm{X}\) in the ground state is the same as the maximum number of electrons in an atom that can have the quantum number designations \(n=2\), \(m_{\ell}=-1\), and \(m_{s}=-\frac{1}{2}\) d. Let \(A\) equal the charge of the stable ion that would form when the undiscovered element 120 forms ionic compounds. This value of \(A\) also represents the angular momentum quantum number for the subshell containing the unpaired electron(s) for element \(\mathrm{X}\).

Octyl methoxycinnamate and oxybenzone are common ingredients in sunscreen applications. These compounds work by absorbing ultraviolet (UV) B light (wavelength \(280-320 \mathrm{~nm}\) ), the UV light most associated with sunburn symptoms. What frequency range of light do these compounds absorb?

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