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Chapter 7: Problem 17

Which is higher in energy: the \(2 s\) or \(2 p\) orbital in hydrogen? Is this also true for helium? Explain.

Short Answer

Expert verified
In hydrogen, the \(2 s\) and \(2 p\) orbitals have the same energy level due to the energy of orbitals depending only on the principal quantum number. However, in helium (a multi-electron atom), the energy of orbitals depends on both the principal and azimuthal quantum numbers. As a result, the \(2 p\) orbital is higher in energy than the \(2 s\) orbital in helium.

Step by step solution

01

Understand the relationship between quantum numbers and orbital energy levels

In atoms, the energy levels of orbitals are determined by the principal quantum number (n) and the azimuthal quantum number (l). The energy of an orbital can be given as: \[E = -\frac{Z^2}{n^2} \] where \(E\) is the energy, \(Z\) is the atomic number, and \(n\) is the principal quantum number. In hydrogen-like atoms (only one electron, like the H atom or He+ ion), the energy of an orbital depends solely on the principal quantum number, and orbitals with the same \(n\) have the same energy. In multi-electron atoms, the energy of an orbital also depends on the azimuthal quantum number (l).
02

Compare the energy levels of \(2 s\) and \(2 p\) orbitals in hydrogen atom

In the case of the hydrogen atom, the energy of an orbital depends only on the principal quantum number, and all orbitals with the same principal quantum number have the same energy. Therefore, the energy of the \(2 s\) orbital and the \(2 p\) orbital in hydrogen is the same.
03

Compare the energy levels of \(2 s\) and \(2 p\) orbitals in helium atom

In the case of the helium atom (Z=2), there are two electrons in the atom, so we need to consider the interaction between electrons. The energy of an orbital can be given as: \[E = -\frac{Z^2}{n^2} + \frac{l(l+1)}{2n^2}\] In this case, the energy of an orbital depends on both the principal quantum number and the azimuthal quantum number. For the \(2 s\) orbital (n=2, l=0), we have: \[E_{2s} = -\frac{2^2}{2^2} + \frac{0(0+1)}{2(2^2)} = -\frac{4}{4} = -1 \, eV\] For the \(2 p\) orbital (n=2, l=1), we have: \[E_{2p} = -\frac{2^2}{2^2} + \frac{1(1+1)}{2(2^2)} = -1 + \frac{2}{8} = -\frac34 \, eV\] Since the energy of the \(2 s\) orbital in helium is lower than the energy of the \(2 p\) orbital, the \(2 p\) orbital is higher in energy.
04

Conclusion

In hydrogen, the \(2 s\) and \(2 p\) orbitals have the same energy level. In helium, the \(2 p\) orbital is higher in energy than the \(2 s\) orbital.

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Key Concepts

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

Quantum Numbers
Quantum numbers play a fundamental role in determining the properties of electrons within an atom. Each electron in an atom can be identified by a unique set of quantum numbers, commonly referred to as n, l, ml, and ms.

The principal quantum number (n) indicates the electron's energy level and its relative distance from the nucleus; higher values of n mean higher energy levels and greater distance from the nucleus. The azimuthal quantum number (l), also known as the angular momentum quantum number, defines the shape of the electron's orbital and is dependent on n. For any given n, l can range from 0 to n-1.

The magnetic quantum number (ml) describes the orientation of the orbital in space and can take on integer values from -l to +l. Lastly, the spin quantum number (ms) describes the electron's spin direction, with possible values being +1/2 or -1/2. Understanding these quantum numbers helps explain why certain elements have the electron configurations they do, and how electrons move between different energy levels.
Hydrogen Atom Orbitals
In the hydrogen atom, the simplest of all atoms with just one electron orbiting the nucleus, orbitals represent the regions in space where the electron is likely to be found. These orbitals are designated as 1s, 2s, 2p, and so on, where the number signifies the principal quantum number (n) and the letter designates the azimuthal quantum number (l), with s, p, d, and f corresponding to l values of 0, 1, 2, and 3, respectively.

The energy level of these orbitals in a hydrogen atom is solely based on the principal quantum number due to the presence of a single electron, which leads to a simplification in the energy equation such that all orbitals with the same n have identical energy levels. This is why both 2s and 2p orbitals in hydrogen have the same energy despite differing in shape and orientation.
Helium Atom Electron Configuration
Moving to the helium atom, we face a more complex situation compared to hydrogen because helium contains two electrons. The electron configuration of helium is 1s2, indicating both electrons occupy the lowest energy level, the 1s orbital. The presence of two electrons introduces electron-electron repulsion, influencing the energy levels of orbitals.

In multi-electron atoms like helium, energy levels not only depend on the principal quantum number (n) but also on the azimuthal quantum number (l). It causes orbitals within the same principal quantum level to have slightly different energy levels due to sublevel splitting, known as the Zeeman effect in strong magnetic fields.

Therefore, in helium, the 2s orbital is lower in energy than the 2p because the 2p electrons experience greater shielding and penetration effects, rendering the 2p orbital higher in energy. This subtle yet important distinction is crucial for understanding how electron configurations evolve as we move through the periodic table.

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Most popular questions from this chapter

An electron is excited from the \(n=1\) ground state to the \(n=3\) state in a hydrogen atom. Which of the following statements are true? Correct the false statements to make them true. a. It takes more energy to ionize (completely remove) the electron from \(n=3\) than from the ground state. b. The electron is farther from the nucleus on average in the \(n=3\) state than in the \(n=1\) state. c. The wavelength of light emitted if the electron drops from \(n=3\) to \(n=2\) will be shorter than the wavelength of light emitted if the electron falls from \(n=3\) to \(n=1\). d. The wavelength of light emitted when the electron returns to the ground state from \(n=3\) will be the same as the wavelength of light absorbed to go from \(n=1\) to \(n=3\). e. For \(n=3\), the electron is in the first excited state.

How many valence electrons do each of the following elements have, and what are the specific valence electrons for each element? a. \(\mathrm{Ca}\) d. In b. O e. \(\mathrm{Ar}\) c. element 117 f. \(\mathrm{Bi}\)

Give a possible set of values of the four quantum numbers for the \(4 s\) and \(3 d\) electrons in titanium.

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 \(75 \mathrm{pm}\left(1 \mathrm{pm}=10^{-12} \mathrm{~m}\right)\)

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