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Chapter 44: Problem 10

State the quantum numbers \(\left(n, \ell, m_{\ell}, m_{s}\right)\) for each electron in the ground state of helium \((Z=2)\). Explain your answer.

Short Answer

Expert verified
First electron: \((1, 0, 0, +\frac{1}{2})\); Second electron: \((1, 0, 0, -\frac{1}{2})\).

Step by step solution

01

Determine the Electron Configuration of Helium

Helium has an atomic number \(Z = 2\), which means it has 2 electrons. In its ground state, helium's electron configuration is \(1s^2\), indicating both electrons occupy the 1s orbital.
02

Quantum Numbers for the First Electron

The first electron in helium is in the 1s orbital. For this electron, the principal quantum number \(n = 1\). The azimuthal quantum number \(\ell = 0\) since this is an 's' orbital. For the magnetic quantum number, we have \(m_{\ell} = 0\) because \(\ell = 0\) allows only one value of \(m_{\ell}\). The spin quantum number \(m_{s}\) can be either \(+\frac{1}{2}\) or \(-\frac{1}{2}\), but we choose \(+\frac{1}{2}\) for the first electron.
03

Quantum Numbers for the Second Electron

The second electron also occupies the 1s orbital. Therefore, \(n = 1\) and \(\ell = 0\) remain the same. The magnetic quantum number is still \(m_{\ell} = 0\). To satisfy the Pauli Exclusion Principle, the spin quantum number for this electron must be opposite to that of the first. Hence, we set \(m_{s} = -\frac{1}{2}\) for the second electron.
04

Summary of Quantum Numbers

For each electron in helium's ground state: - **First Electron:** \((n, \ell, m_{\ell}, m_{s}) = (1, 0, 0, +\frac{1}{2})\)- **Second Electron:** \((n, \ell, m_{\ell}, m_{s}) = (1, 0, 0, -\frac{1}{2})\)

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Key Concepts

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

Electron Configuration
When discussing helium's atomic structure, understanding electron configuration is crucial. The electron configuration describes how electrons are distributed in an atom's orbitals. For helium, with an atomic number of 2, the electron configuration is represented as \(1s^2\). This notation shows that both of helium's electrons are present in the first energy level within the s orbital.
  • "1" refers to the principal energy level, or shell, where the electrons are located.
  • "s" signifies the type of orbital. The s orbital can hold a maximum of 2 electrons.
  • "2" indicates that both electrons are occupying the 1s orbital.
By understanding helium's electron configuration, you can also easily identify its place in the periodic table, which leads to understanding its chemical properties. The simplicity of helium's electron configuration perfectly demonstrates its stability and why it does not easily participate in chemical reactions.
Pauli Exclusion Principle
The Pauli Exclusion Principle is a fundamental concept in quantum mechanics and is essential for explaining the behavior of electrons in atoms. According to this principle, no two electrons within an atom can have the same set of four quantum numbers. This is key to understanding how electrons fill orbitals in elements like helium.
In helium, both electrons are located in the same 1s orbital. They share the first three quantum numbers:
  • Principal quantum number \(n = 1\)
  • Azimuthal quantum number \(\ell = 0\)
  • Magnetic quantum number \(m_{\ell} = 0\)
However, to comply with the Pauli Exclusion Principle, their spin quantum numbers \(m_{s}\) must be different. This is why in helium:
  • The first electron has \(m_{s} = +\frac{1}{2}\)
  • The second electron has \(m_{s} = -\frac{1}{2}\)
This differentiation in spin allows both electrons to coexist in the same orbital, adhering to the exclusion principle's rule and maintaining the atom's overall stability.
Helium Atomic Structure
Helium is a noble gas and is unique in its atomic structure. Being one of the simplest elements, helium has two protons in its nucleus and naturally accommodates two electrons. In its ground state, these electrons fully occupy the 1s orbital, as depicted by the electron configuration \(1s^2\).
Helium鈥檚 atomic structure can be deeply understood through its quantum numbers, which define the properties and positions of its electrons:
  • For the first electron:
    • \(n = 1\), \(\ell = 0\), \(m_{\ell} = 0\), \(m_{s} = +\frac{1}{2}\)
  • For the second electron:
    • \(n = 1\), \(\ell = 0\), \(m_{\ell} = 0\), \(m_{s} = -\frac{1}{2}\)
This configuration makes helium incredibly stable. Helium's inertness in chemical reactions is due to its filled first energy level, satisfying the atom's need for a lower energy state. This stability is why helium is used in applications where chemical reactivity is undesirable, such as in balloons and cryogenics.

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Most popular questions from this chapter

List all four quantum numbers for the single hydrogen \((Z=1)\) electron in the ground state. [Hint: It is in the 1 s orbital of the \(K\) shell.]

Why is sodium \((Z=11)\) the next univalent atom after lithium? Sodium has a single electron in the \(n=3\) shell. To see why this is necessarily so, notice that the Pauli Exclusion Principle allows only two electrons in the \(n=1\) shell. The next eight electrons can fit in the \(n=2\) shell, as follows: $$\begin{array}{l} n=2, \quad \ell=0, \quad m_{\ell}=0, \quad m_{s}=\pm \frac{1}{2} \\ n=2, \quad \ell=1, \quad m_{\ell}=0, \quad m_{s}=\pm \frac{1}{2} \\ n=2, \quad \ell=1, \quad m_{\ell}=1, \quad m_{s}=\pm \frac{1}{2} \\ n=2, \quad \ell=1, \quad m_{\ell}=-1, \quad m_{s}=\pm \frac{1}{2} \end{array}$$ The eleventh electron must go into the \(n=3\) shell, from which it is easily removed to yield \(\mathrm{Na}^{+}\).

The single electron of a hydrogen atom can exist in a variety of excited states beyond the lowest-energy ground state. How many states would be available when the principal quantum number equals 4 and the orbital quantum number equals 2? [Hint: \(n=4\) and \(\ell=2 .]\)

Silicon is a semiconductor, as is carbon (C) for which \(Z=6\). Specify the ground state electron configuration. Why do they behave similarly?

Suppose electrons had no spin, so that the spin quantum number did not exist. If the Exclusion Principle still applied to the remaining quantum numbers, what would be the first three univalent atoms? The electrons would take on the following quantum numbers: $$\begin{array}{lllll} \text { Electron 1: } & n=1, & \ell=0, & m_{\ell}=0 & \text { (univalent) } \\\ \text { Electron 2: } & n=2, & \ell=0, & m_{\ell}=0 & \text { (univalent) } \\\ \text { Electron 3: } & n=2, & \ell=1, & m_{\ell}=0 & \\ \text { Electron 4: } & n=2, & \ell=1, & m_{\ell}=+1 & \\ \text { Electron 5: } & n=2, & \ell=1, & m_{\ell}=-1 & \\ \text { Electron 6: } & n=3, & \ell=0, & m_{\ell}=0 & \text { (univalent) } \end{array}$$ Each electron marked "univalent" is the first electron in a new shell. Since an electron is easily removed if it is the outermost electron in the atom, atoms with that number of electrons are univalent. They are the atoms with \(Z=1\) (hydrogen), \(Z=2\) (helium), and \(Z=6\) (carbon). Can you show that \(Z=15\) (phosphorus) would also be univalent?
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