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Chapter 7: Problem 4

Compare the first ionization energy of helium to its second ionization energy, remembering that both electrons come from the \(1 s\) orbital. Explain the difference without using actual numbers from the text.

Short Answer

Expert verified
The difference between the first and second ionization energies of helium can be explained by the change in the effective nuclear charge experienced by the electrons being removed. In the neutral helium atom, both electrons in the 1s orbital experience the same effective nuclear charge. After removing the first electron, the remaining electron in the He+ ion experiences a much higher effective nuclear charge and thus a stronger electrostatic attraction to the nucleus. This increased attraction results in a significantly higher ionization energy needed to remove the second electron compared to the first ionization energy.

Step by step solution

01

Understanding Ionization Energy

Ionization energy is the energy required to remove an electron from an atom or ion. In the case of helium, the first ionization energy is the energy needed to remove one electron from the neutral helium atom (He), while the second ionization energy is the energy needed to remove the second electron from the resulting helium ion (He+).
02

Important Factors for Ionization Energy

Two main factors influence the ionization energy of an atom: electron shielding and effective nuclear charge. Electron shielding refers to the phenomenon where inner-shell electrons reduce the net positive charge experienced by outer-shell electrons. Effective nuclear charge is the net positive charge that an electron experiences, taking into account both the positive charge of the protons in the nucleus and the negative charge of any shielding electrons.
03

Comparing Helium's First Ionization Energy

For the first ionization energy, we need to consider the energy required to remove the first electron from the neutral helium atom (He). The electron is found in the 1s orbital, which is the closest to the nucleus. There are no shielding electrons, as both electrons are in the 1s orbital, and the effective nuclear charge is the same for both. Hence, the first ionization energy will be determined by the strength of the electrostatic attraction between the electron and the two protons in the helium nucleus.
04

Comparing Helium's Second Ionization Energy

For the second ionization energy, we are concerned with the energy necessary to remove the remaining electron from the helium ion (He+). In this case, the electron is still in the 1s orbital and is the only electron remaining. With no shielding electrons present, the effective nuclear charge experienced by this electron will be significantly higher than the charge experienced by either electron in the neutral helium atom. Consequently, the electrostatic attraction between the electron and the two protons in the nucleus will be much stronger than in the case of the first ionization energy.
05

Explaining the Difference

The significant difference between the first and second ionization energies of helium can be attributed to the change in the effective nuclear charge experienced by the electrons being removed. For the first ionization, both electrons in the 1s orbital experience the same effective nuclear charge. However, once the first electron is removed, the second electron in the helium ion (He+) experiences a much higher effective nuclear charge and hence a significantly stronger electrostatic attraction to the nucleus. This increased attraction results in a much higher ionization energy required to remove the second electron as compared to the first ionization energy.

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Most popular questions from this chapter

Neutron diffraction is used in determining the structures of molecules. a. Calculate the de Broglie wavelength of a neutron moving at \(1.00 \%\) of the speed of light. b. Calculate the velocity of a neutron with a wavelength of \(75 \mathrm{pm}\left(1 \mathrm{pm}=10^{-12} \mathrm{~m}\right)\)

Give the maximum number of electrons in an atom that can have these quantum numbers: a. \(n=0, \ell=0, m_{\ell}=0\) b. \(n=2, \ell=1, m_{\ell}=-1, m_{s}=-\frac{1}{2}\) c. \(n=3, m_{s}=+\frac{1}{2}\) d. \(n=2, \ell=2\) e. \(n=1, \ell=0, m_{\ell}=0\)

Consider the following approximate visible light spectrum: Barium emits light in the visible region of the spectrum. If each photon of light emitted from barium has an energy of \(3.59 \times\) \(10^{-19} \mathrm{~J}\), what color of visible light is emitted?

For each of the following pairs of elements \((\mathrm{C}\) and \(\mathrm{N}) \quad(\mathrm{Ar}\) and \(\mathrm{Br})\) pick the atom with a. more favorable (exothermic) electron affinity. b. higher ionization energy. c. larger size.

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